Solutions Manual for Functions of One Complex Variable I, Second Edition 1
© Copyright by Andreas Kleefeld, 2009 All Rights Reserved2
1 by
John B. Conway updated on January, 7th 2013
2 Last
PREFACE
Most of the exercises I solved were assigned homeworks in the graduate courses Math 713 and Math 714 ”Complex Analysis I and II” at the University of Wisconsin – Milwaukee taught by Professor Dashan Fan in Fall 2008 and Spring 2009. The solutions manual is intented for all students taking a graduate level Complex Analysis course. Students can check their answers to homework problems assigned from the excellent book “Functions of One Complex Variable I”, Second Edition by John B. Conway. Furthermore students can prepare for quizzes, tests, exams and final exams by solving additional exercises and check their results. Maybe students even study for preliminary exams for their doctoral studies. However, I have to warn you not to copy straight of this book and turn in your homework, because this would violate the purpose of homeworks. Of course, that is up to you. I strongly encourage you to send me solutions that are still missing to
[email protected] (LATEXpreferred but not mandatory) in order to complete this solutions manual. Think about the contribution you will give to other students. If you find typing errors or mathematical mistakes pop an email to
[email protected]. The recent version of this solutions manual can be found at http://www.math.tu-cottbus.de/INSTITUT/kleefeld/Files/Solution.html. The goal of this project is to give solutions to all of the 452 exercises.
CONTRIBUTION
I thank (without special order) Christopher T. Alvin Martin J. Michael David Perkins for contributions to this book.
2
Contents 1 The Complex Number System 1.1 The real numbers . . . . . . . . . . . . . . . . . . 1.2 The field of complex numbers . . . . . . . . . . . 1.3 The complex plane . . . . . . . . . . . . . . . . . 1.4 Polar representation and roots of complex numbers 1.5 Lines and half planes in the complex plane . . . . . 1.6 The extended plane and its spherical representation
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1 1 1 5 5 7 7
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9 9 12 13 15 17 20
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Elementary Properties and Examples of Analytic Functions 3.1 Power series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Analytic functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Analytic functions as mappings. Möbius transformations . . . . . . . . . . . . . . . . . .
21 21 24 31
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Complex Integration 4.1 Riemann-Stieltjes integrals . . . . . . . . . . . . . . . . . . . . . . . 4.2 Power series representation of analytic functions . . . . . . . . . . . . 4.3 Zeros of an analytic function . . . . . . . . . . . . . . . . . . . . . . 4.4 The index of a closed curve . . . . . . . . . . . . . . . . . . . . . . . 4.5 Cauchy’s Theorem and Integral Formula . . . . . . . . . . . . . . . . 4.6 The homotopic version of Cauchy’s Theorem and simple connectivity 4.7 Counting zeros; the Open Mapping Theorem . . . . . . . . . . . . . 4.8 Goursat’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
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42 42 48 58 60 61 63 66 67
Singularities 5.1 Classification of singularities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Residues . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 The Argument Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
68 68 75 82
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5
Metric Spaces and the Topology of C 2.1 Definitions and examples of metric spaces 2.2 Connectedness . . . . . . . . . . . . . . 2.3 Sequences and completeness . . . . . . . 2.4 Compactness . . . . . . . . . . . . . . . 2.5 Continuity . . . . . . . . . . . . . . . . . 2.6 Uniform convergence . . . . . . . . . . .
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6
7
The Maximum Modulus Theorem 6.1 The Maximum Principle . . . . . . . . . . . . . . . . . . 6.2 Schwarz’s Lemma . . . . . . . . . . . . . . . . . . . . . . 6.3 Convex functions and Hadamard’s Three Circles Theorem 6.4 The Phragmén-Lindelöf Theorem . . . . . . . . . . . . .
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84 84 86 88 90
Compactness and Convergence in the Space of Analytic Functions 7.1 The space of continuous functions C(G, Ω) . . . . . . . . . . . 7.2 Spaces of analytic functions . . . . . . . . . . . . . . . . . . . 7.3 Spaces of meromorphic functions . . . . . . . . . . . . . . . . 7.4 The Riemann Mapping Theorem . . . . . . . . . . . . . . . . . 7.5 The Weierstrass Factorization Theorem . . . . . . . . . . . . . 7.6 Factorization of the sine function . . . . . . . . . . . . . . . . . 7.7 The gamma function . . . . . . . . . . . . . . . . . . . . . . . 7.8 The Riemann zeta function . . . . . . . . . . . . . . . . . . . .
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93 93 97 100 103 106 110 112 120
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Runge’s Theorem 123 8.1 Runge’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123 8.2 Simple connectedness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 124 8.3 Mittag-Leffler’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125
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Analytic Continuation and Riemann Surfaces 9.1 Schwarz Reflection Principle . . . . . . . . . . . . . . . . 9.2 Analytic Continuation Along a Path . . . . . . . . . . . . 9.3 Monodromy Theorem . . . . . . . . . . . . . . . . . . . . 9.4 Topological Spaces and Neighborhood Systems . . . . . . 9.5 The Sheaf of Germs of Analytic Functions on an Open Set 9.6 Analytic Manifolds . . . . . . . . . . . . . . . . . . . . . 9.7 Covering spaces . . . . . . . . . . . . . . . . . . . . . . .
10 Harmonic Functions 10.1 Basic properties of harmonic functions . . . 10.2 Harmonic functions on a disk . . . . . . . . 10.3 Subharmonic and superharmonic functions . 10.4 The Dirichlet Problem . . . . . . . . . . . 10.5 Green’s Function . . . . . . . . . . . . . .
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130 130 131 132 132 133 134 135
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137 137 141 144 148 148
11 Entire Functions 151 11.1 Jensen’s Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151 11.2 The genus and order of an entire function . . . . . . . . . . . . . . . . . . . . . . . . . . 153 11.3 Hadamard Factorization Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 158 12 The Range of an Analytic Function 12.1 Bloch’s Theorem . . . . . . . . 12.2 The Little Picard Theorem . . . 12.3 Schottky’s Theorem . . . . . . . 12.4 The Great Picard Theorem . . .
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161 161 162 162 162
Chapter 1
The Complex Number System 1.1 The real numbers No exercises are assigned in this section.
1.2 The field of complex numbers Exercise 1. Find the real and imaginary parts of the following: √ 3 −1 + i 3 1 z−a 3 3 + 5i ; ; (a ∈ R); z ; ; z z+a 7i + 1 2
Solution. Let z = x + iy. Then a)
√ 6 ! −1 − i 3 n 1 + i n ; i ; √ 2 2 ! 1 z ! 1 Im z Re
for
2 ≤ n ≤ 8.
x Re(z) = 2 +y |z|2 y Im(z) = − 2 =− 2 2 x +y |z| =
x2
b) Re Im
z − a
z+a z − a z+a
= =
|z|2 − a2 x2 + y2 − a2 = 2 2 2 + y + 2ax + a |z| + 2aRe(z) + a2 2ya 2Im(z)a = 2 2 2 2 x + y + 2xa + a |z| + 2aRe(z) + a2 x2
c) Re z3 = x3 − 3xy2 = Re(z)3 − 3Re(z)Im(z)2 Im z3 = 3x2 y − y3 = 3Re(z)2 Im(z) − Im(z)3 1
d) ! 3 + 5i Re 7i + 1 ! 3 + 5i Im 7i + 1
19 25 8 = − 25 =
e)
f)
√ −1 + i 3 3 Re 2 √ −1 + i 3 3 Im 2
g)
√ −1 − i 3 6 Re 2 √ −1 − i 3 6 Im 2
Re (in )
=
Im (in )
=
0, 1, −1, 0, 1, −1,
= 1
= 0
= 1
= 0
n is odd n ∈ {4k : k ∈ Z} n ∈ {2 + 4k : k ∈ Z} n is even n ∈ {1 + 4k : k ∈ Z} n ∈ {3 + 4k : k ∈ Z}
2
h)
Re
Im
1+i √ 2
1+i √ 2
!n !
0, √ 2 − 2 , −1, √2 − 2 , 0, √ 2 2 , 1, 1, √ 2 2 , 0, √2 − 2 , −1, √ − 22 , 0,
=
!n !
=
n=2 n=3 n=4 n=5 n=6 n=7 n=8 n=2 n=3 n=4 n=5 n=6 n=7 n=8
Exercise 2. Find the absolute value and conjugate of each of the following: 3−i i −2 + i; −3; (2 + i)(4 + 3i); √ ; (1 + i)6 ; i17 . ; i + 3 2 + 3i Solution. It is easy to calculate: a) z = −2 + i,
|z| =
√ 5,
z¯ = −2 − i
b) z = −3,
|z| = 3,
c)
√ |z| = 5 5,
z = (2 + i)(4 + 3i) = 5 + 10i, d
3−i z= √ , 2 + 3i
|z| =
1 √ 110, 11
e) z=
1 3 i = + i, i + 3 10 10
z¯ = −3
|z| =
z¯ = 5 − 10i
3+i z¯ = √ 2 − 3i
1 √ 10, 10
z¯ =
3 1 − i 10 10
f) z = (1 + i)6 = −8i,
|z| = 8,
z¯ = 8i
g) i17 = i,
|z| = 1,
Exercise 3. Show that z is a real number if and only if z = z¯.
3
z¯ = −i
Solution. Let z = x + iy. ⇒: If z is a real number, then z = x (y = 0). This implies z¯ = x and therefore z = z¯. ⇔: If z = z¯, then we must have x + iy = x − iy for all x, y ∈ R. This implies y = −y which is true if y = 0 and therefore z = x. This means that z is a real number. Exercise 4. If z and w are complex numbers, prove the following equations: |z + w|2 |z − w|2
|z + w|2
= |z|2 + 2Re(zw) ¯ + |w|2 . 2 = |z| − 2Re(zw) ¯ + |w|2 . + |z − w|2 = 2 |z|2 + |w|2 .
Solution. We can easily verify that z¯ = z. Thus |z + w|2
¯ = z¯z + zw¯ + w¯z + ww¯ = (z + w)(z + w) = (z + w)(¯z + w) 2 2 2 = |z| + |w| + zw¯ + z¯w = |z| + |w|2 + zw¯ + z¯w¯ zw¯ + zw¯ = |z|2 + |w|2 + zw¯ + zw¯ = |z|2 + |w|2 + 2 2 = |z|2 + |w|2 + 2Re(zw) ¯ = |z|2 + 2Re(zw) ¯ + |w|2 .
|z − w|2
¯ = z¯z − zw¯ − w¯z + ww¯ = (z − w)(z − w) = (z − w)(¯z − w) 2 2 2 = |z| + |w| − zw¯ − z¯w = |z| + |w|2 − zw¯ − z¯w¯ zw¯ + zw¯ = |z|2 + |w|2 − zw¯ − zw¯ = |z|2 + |w|2 − 2 2 = |z|2 + |w|2 − 2Re(zw) ¯ = |z|2 − 2Re(zw) ¯ + |w|2 .
|z + w|2 + |z − w|2
= |z|2 + Re(zw) ¯ + |w|2 + |z|2 − Re(zw) ¯ + |w|2 = |z|2 + |w|2 + |z|2 + |w|2 = 2|z|2 + 2|w|2 = 2 |z|2 + |w|2 .
Exercise 5. Use induction to prove that for z = z1 + . . . + zn ; w = w1 w2 . . . wn : |w| = |w1 | . . . |wn |; z¯ = z¯1 + . . . + z¯n ; w¯ = w¯ 1 . . . w¯ n . Solution. Not available.
Exercise 6. Let R(z) be a rational function of z. Show that R(z) = R(¯z) if all the coefficients in R(z) are real. Solution. Let R(z) be a rational function of z, that is R(z) =
an zn + an−1 zn−1 + . . . a0 bm zm + bm−1 zm−1 + . . . b0
where n, m are nonnegative integers. Let all coefficients of R(z) be real, that is a0 , a1 , . . . , an , b0 , b1 , . . . , bm ∈ R. Then R(z) = =
an zn + an−1 zn−1 + . . . a0 an zn + an−1 zn−1 + . . . a0 = m m−1 bm z + bm−1 z + . . . b0 bm zm + bm−1 zm−1 + . . . b0 an zn + an−1 zn−1 + . . . a0
=
bm zm + bm−1 zm−1 + . . . b0 4
an z¯n + an−1 z¯n−1 + . . . a0 = R(¯z). bm z¯m + bm−1 z¯m−1 + . . . b0
1.3 The complex plane Exercise 1. Prove (3.4) and give necessary and sufficient conditions for equality. Solution. Let z and w be complex numbers. Then ||z| − |w||
= ||z − w + w| − |w|| ≤ ||z − w| + |w| − |w|| = ||z − w|| = |z − w|
Notice that |z| and |w| is the distance from z and w, respectively, to the origin while |z − w| is the distance between z and w. Considering the construction of the implied triangle, in order to guarantee equality, it is necessary and sufficient that ⇐⇒
||z| − |w|| = |z − w| (|z| − |w|)2 = |z − w|2
⇐⇒ ⇐⇒
|z||w| = Re(zw) ¯ |zw| ¯ = Re(zw) ¯
⇐⇒ ⇐⇒
(|z| − |w|)2 = |z|2 − 2Re(zw) ¯ + |w|2 |z|2 − 2|z||w| + |w|2 = |z|2 − 2Re(zw) ¯ + |w|2
Equivalently, this is zw¯ ≥ 0. Multiplying this by t = wz = |w|1 2 · |w|2 · wz . Then t ≥ 0 and z = tw.
w w,
we get zw¯ ·
w w
= |w|2 ·
z w
≥ 0 if w , 0. If
Exercise 2. Show that equality occurs in (3.3) if and only if zk /zl ≥ 0 for any integers k and l, 1 ≤ k, l ≤ n, for which zl , 0. Solution. Not available. Exercise 3. Let a ∈ R and c > 0 be fixed. Describe the set of points z satisfying |z − a| − |z + a| = 2c for every possible choice of a and c. Now let a be any complex number and, using a rotation of the plane, describe the locus of points satisfying the above equation. Solution. Not available.
1.4 Polar representation and roots of complex numbers Exercise 1. Find the sixth roots of unity. Solution. Start with z6 = 1 and z = rcis(θ), therefore r6 cis(6θ) = 1. Hence r = 1 and θ = 2kπ 6 with k ∈ {−3, −2, −1, 0, 1, 2}. The following table gives a list of principle values of arguments and the corresponding value of the root of the equation z6 = 1. θ0 = 0 z0 = 1 θ1 = π3 z1 = cis( π3 ) 2π z2 = cis( 2π θ2 = 3 3 ) θ3 = π z3 = cis(π) = −1 θ4 = −2π z4 = cis( −2π 3 3 ) −π z5 = cis( −π θ5 = 3 3 ) 5
Exercise 2. Calculate the following: a) the square roots of i b) the cube roots of i c) the square roots of
√ 3 + 3i
√ Solution. c) The square roots of q 3 + 3i. √ 2 √ √ Let z = 3 + 3i. Then r = |z| = 3 + 32 = 12 and α = tan−1 √33 = π3 . So, the 2 distinct roots of z √ are given by 2 r cos α+2kπ where k = 0, 1. Specifically, + i sin α+2kπ n n √4 √ z = 12 cos
π 3
+ 2kπ + i sin 2
π 3
! + 2kπ . 2
Therefore, the square roots of z, zk , are given by √4 √ √4 z0 = 12 cos π6 + i sin π6 = 12 23 + 21 i √4 √ √4 3 7π 1 12 − z1 = 12 cos 7π = + i sin − i 6 6 2 2 . So, in rectangular form, the second roots of z are given by
√4
108 2 ,
√ 4
12 2
√4 √ 4 12 and − 108 , − 2 2 .
Exercise 3. A primitive nth root of unity is a complex number a such that 1, a, a2 , ..., an−1 are distinct nth roots of unity. Show that if a and b are primitive nth and mth roots of unity, respectively, then ab is a kth root of unity for some integer k. What is the smallest value of k? What can be said if a and b are nonprimitive roots of unity? Solution. Not available. Exercise 4. Use the binomial equation (a + b)n =
n X k=0
where n k
!
=
! n n−k k a b, k
n! , k!(n − k)!
and compare the real and imaginary parts of each side of de Moivre’s formula to obtain the formulas: ! ! n n cos nθ = cosn θ − cosn−2 θ sin2 θ + cosn−4 θ sin4 θ − . . . 2 4 ! ! n n cosn−3 θ sin3 θ + . . . cosn−1 θ sin θ − sin nθ = 3 1 Solution. Not available. n−1 Exercise 5. Let z = cis 2π = 0. n for an integer n ≥ 2. Show that 1 + z + . . . + z
6
P Solution. The summation of the finite geometric sequence 1, z, z2 , . . . , zn−1 can be calculated as nj=1 z j−1 = n zn −1 2π n th n = . We want to show that z is an n root of unity. So, using de Moivre’s formula, z = cis z−1 n n z −1 1−1 2π 2 n−1 cis n · n = cis(2π) = 1. It follows that 1 + z + z + ... + z = z−1 = z−1 = 0 as required.
Exercise 6. Show that ϕ(t) = cis t is a group homomorphism of the additive group R onto the multiplicative group T = {z : |z| = 1}. Solution. Not available. Exercise 7. If z ∈ C and Re(zn ) ≤ 0 for every positive integer n, show that z is a non-negative real number.
Solution. Let n be an arbitrary but fixed positive integer and let z ∈ C and Re(zn ) ≥ 0. Since zn = rn (cos(nθ) + i sin(nθ)), we have Re(zn ) = rn cos(nθ) ≥ 0.
If z = 0, then we are done, since r = 0 and Re(zn ) = 0. Therefore, assume z , 0, then r > 0. Thus Re(zn ) = rn cos(nθ) ≥ 0
implies cos(nθ) ≥ 0 for all n. This implies θ = 0 as we will show next. Clearly, θ < [π/2, 3π/2]. If π ≤ θ < πk . If we choose n = k + 1, we have θ ∈ (0, π/2), then there exists a k ∈ {2, 3, . . .} such that k+1 π ≤ nθ <
(k + 1)π k
which is impossible since cos(nθ) ≥ 0. Similarly, we can derive a contradiction if we assume θ ∈ (3π/2, 2π). Then 2π − π/k ≤ θ < 2π − π/(k + 1) for some k ∈ {2, 3, . . .}.
1.5 Lines and half planes in the complex plane Exercise 1. Let C be the circle {z : |z − c| = r}, r > 0; let a = c + rcis α and put z − a Lβ = z : Im =0 b where b = cisβ. Find necessary and sufficient conditions in terms of β that Lβ be tangent to C at a. Solution. Not available.
1.6 The extended plane and its spherical representation Exercise 1. Give the details in the derivation of (6.7) and (6.8). Solution. Not available. Exercise 2. For each of the following points in C, give the corresponding point of S : 0, 1 + i, 3 + 2i. Solution. We have
If z1 = 0, then |z1 | = 0 and therefore
! 2y |z|2 − 1 2x . , , Φ(z) = |z|2 + 1 |z|2 + 1 |z|2 + 1 Φ(z1 ) = (0, 0, −1). 7
Thus, z1 = 0 corresponds√to the point (0, 0, −1) on the sphere S . If z2 = 1 + i, then |z2 | = 2 and therefore ! 2 2 1 Φ(z2 ) = , , . 3 3 3 Thus, z2 = 1 + i corresponds to the point 23 , 32 , 13 on the sphere S . √ If z3 = 3 + 2i, then |z3 | = 13 and therefore ! 3 2 6 Φ(z3 ) = , , . 7 7 7 Thus, z3 = 3 + 2i corresponds to the point 37 , 72 , 67 on the sphere S . Exercise 3. Which subsets of S correspond to the real and imaginary axes in C. Solution. If z is on the real axes, then z = x which implies |z|2 = x2 . Thus ! x2 − 1 2x . , 0, 2 Φ(z) = 2 x +1 x +1 Therefore the set
(
! ) 2x x2 − 1 |x ∈ R ⊂S , 0, x2 + 1 x2 + 1
corresponds to the real axes in C. That means the unit circle x2 + z2 = 1 lying in the xz−plane corresponds to the real axes in C. If z is on the imaginary axes, then z = iy which implies |z|2 = y2 . Thus ! 2y y2 − 1 Φ(z) = 0, 2 . , y + 1 y2 + 1 Therefore the set
(
0,
! ) 2y y2 − 1 |y ∈ R ⊂S , y2 + 1 y2 + 1
corresponds to the imaginary axes in C. That means the unit circle y2 + z2 = 1 lying in the yz−plane corresponds to the imaginary axes in C. Exercise 4. Let Λ be a circle lying in S . Then there is a unique plane P in R3 such that P ∩ S = Λ. Recall from analytic geometry that P = {(x1 , x2 , x3 ) : x1 β1 + x2 β2 + x3 β3 = l} where (β1 , β2 , β3 ) is a vector orthogonal to P and l is some real number. It can be assumed that β21 +β22 +β23 = 1. Use this information to show that if Λ contains the point N then its projection on C is a straight line. Otherwise, Λ projects onto a circle in C. Solution. Not available. Exercise 5. Let Z and Z 0 be points on S corresponding to z and z0 respectively. Let W be the point on S corresponding to z + z0 . Find the coordinates of W in terms of the coordinates of Z and Z 0 . Solution. Not available. 8
Chapter 2
Metric Spaces and the Topology of C 2.1 Definitions and examples of metric spaces Exercise 1. Show that each of the examples of metric spaces given in (1.2)-(1.6) is, indeed, a metric space. Example (1.6) is the only one likely to give any difficulty. Also, describe B(x; r) for each of these examples. Solution. Not available. Exercise 2. Which of the following subsets of C are open and which are closed: (a) {z : |z| < 1}; (b) the real axis; (c) {z : zn = 1 for some integer n ≥ 1}; (d) {z ∈ C is real and 0 ≤ z < 1}; (e) {z ∈ C : z is real and 0 ≤ z ≤ 1}? Solution. We have (a) A := {z ∈ C : |z| < 1}
S Let z ∈ A and set εz = 1−|z| z∈A B(z, εz ) is open also. A 2 , then B(z, εz ) ⊂ A is open and therefore A = cannot be closed, otherwise A and C − A were both closed and open sets yet C is connected.
(b) B := {z ∈ C : z = x + iy, y = 0} (the real axis) Let z ∈ C − B, then Imz , 0. Set εz = complement is open.
|Imz| 2 ,
then B(z, εz ) ⊂ C − B.Hence B is closed since its
For any real x and any ε > 0 the point x + i 2ε ∈ B(x, ε) but x + i 2ε ∈ C − R. Thus B is not open. (c) C := {z ∈ C : zn = 1 for some integer n ≥ 1} Claim: C is neither closed nor open.
C is not open because if zn = 1 then z = rcis(θ) with r = 1 and any ε-ball around z would contain an element z0 := (1 + 4ε )cis(θ). To show that C cannot be closed, note that for qp ∈ Q with p ∈ Z and q ∈ N the number z := cis qp 2π ∈ C since zq = cis(p2π) = cos(p2π) + i sin(p2π) = 1. Now fix x ∈ R − Q and let {xn }n be a rational sequence that converges to x. Let m be any natural number. Now zm = 1 implies that sin(mx2π) = 0 which in turn means that mx ∈ Z contradicting the choice x ∈ R − Q. We have constructed a sequence of elements of C that converges to a point that is not element of C. Hence C is not closed. 9
(d) D := {z ∈ C : z is real and 0 ≤ z < 1}
The set cannot be open by the argument given in 2.1) and it is not closed because zn := 1 − sequence in D that converges to a point outside of D.
1 n
is a
(e) E := {z ∈ C : z is real and 0 ≤ z ≤ 1}
This set E is not open by the observation in 2.1) and it is closed because its complement is open: If ) is contained in the complement of E, if z is imaginary then z , E and z real, then B(z, min{|x|,|x−1|} 2 B(z, |Imy| ) is completely contained in the complement of E. 2
Exercise 3. If (X, d) is any metric space show that every open ball is, in fact, an open set. Also, show that every closed ball is a closed set. Solution. Not available. Exercise 4. Give the details of the proof of (1.9c). Solution. Not available. Exercise 5. Prove Proposition 1.11. Solution. Not available. Exercise 6. Prove that a set G ⊂ X is open if and only if X − G is closed. Solution. Not available. Exercise 7. Show that (C∞ , d) where d is given (I. 6.7) and (I. 6.8) is a metric space. 0
| Solution. To show (C∞ , d) with d(z, z0 ) = [(1+|z|22|z−z for z, z0 ∈ C and d(z, ∞) = (1+|z|22 )1/2 , z ∈ C is a )(1+|z0 |2 )]1/2 metric space. i) d(z, z0 ) ≥ 0. 0 | ≥ 0 for all z, z0 ∈ C (the denominator is always positive). Since |z−z0 | ≥ 0, we have d(z, z0 ) = [(1+|z|22|z−z )(1+|z0 |2 )]1/2 Obviously, d(z, ∞) = (1+|z|22 )1/2 ≥ 0 for all z ∈ C. ii) d(z, z0 ) = 0 iff z = z0 . 0 | We have 2|z − z0 | = 0 iff z = z0 . Therefore, d(z, z0 ) = [(1+|z|22|z−z = 0 iff z = z0 . d(∞, ∞) = )(1+|z0 |2 )]1/2 2 limz→∞ (1+|z|2 )1/2 = 0. Thus, d(∞, 0) = 0 iff z = ∞. iii) d(z, z0 ) = d(z0 , z). 0 0 | −z| 2 0 0 We have d(z, z0 ) = [(1+|z|22|z−z = [(1+|z0 |2|z 2 )(1+|z|2 )]1/2 = d(z , z) for all z, z ∈ C. Also d(z, ∞) = (1+|z|2 )1/2 = )(1+|z0 |2 )]1/2 0 d(∞, z) by the symmetry of d(z, z ) in general. iv) d(z, z0 ) ≤ d(z, x) + d(x, z0 ). We have
2|z − z0 | [(1 + |z|2 )(1 + |z0 |2 )]1/2 2|z − x| 2|x − z0 | ≤ + [(1 + |z|2 )(1 + |x|2 )]1/2 [(1 + |x|2 )(1 + |z0 |2 )]1/2 = d(z, x) + d(x, z0 )
d(z, z0 ) =
for all x, z, z0 ∈ C. The first inequality will be shown next. First, we need the following equation (z − z0 )(1 + x¯ x) = (z − x)(1 + x¯z0 ) + (x − z0 )(1 + x¯z), 10
(2.1)
which follows from a simple computation (z − x)(1 + x¯z0 ) + (x − z0 )(1 + x¯z) = z − x + x¯z0 z + x x¯z0 + x − z0 + x x¯z − x¯z0 z = z + x x¯z − z0 − x x¯z0 = (z − z0 )(1 + x¯ x).
Using (2.1) and taking norms gives |(z − z0 )(1 + x¯ x)|
= |(z − z0 )(1 + |x|2 )| = |(z − z0 )|(1 + |x|2 )
= |(z − x)(1 + x¯z0 ) + (x − z0 )(1 + x¯z)| ≤ |z − x|(1 + |x|2 )1/2 (1 + |z0 |2 )1/2 + |x − z0 |(1 + |x|2 )1/2 (1 + |z|2 )1/2
(2.2)
where the last inequality follows by |z + x¯z0 | ≤ (1 + |x|2 )1/2 (1 + |z0 |2 )1/2
⇔ (1 + x¯z0 )(1 + x¯z0 ) ≤ (1 + |x|2 )(1 + |z0 |2 ) and |z + x¯z| ≤ (1 + |x|2 )1/2 (1 + |z|2 )1/2
⇔ (1 + x¯z)(1 + x¯z) ≤ (1 + |x|2 )(1 + |z|2 ) since (1 + x¯z)(1 + x¯z) ≤ (1 + |x|2 )(1 + |z|2 ) ⇔ x¯z + x¯z ≤ |x|2 + |y|2
⇔ 2Re(x¯z) ≤ |x|2 + |y|2
which is true by Exercise 4 part 2 on page 3. Thus, dividing (2.2) by (1 + |x|2 )1/2 yields |z − z0 |(1 + |x|2 )1/2 ≤ |z − x|(1 + |z0 |2 )1/2 + |x − z0 |(1 + |z|2 )1/2 . Multiplying this by (1 + gives the assertion above.
|x|2 )1/2 (1
2 + |z0 |2 )1/2 (1 + |z|2 )1/2
Exercise 8. Let (X, d) be a metric space and Y ⊂ X. Suppose G ⊂ X is open; show that G ∩ Y is open in (Y, d). Conversely, show that if G1 ⊂ Y is open in (Y, d), there is an open set G ⊂ X such that G1 = G ∩ Y. Solution. Set G1 = G ∩ Y, let G be open in (X, d) and Y ⊂ X. In order to show that G1 is open in (Y, d), pick an arbitrary point p ∈ G1 . Then p ∈ G and since G is open, there exists an > 0 such that BX (p; ) ⊂ G. But then BY (p; ) = BX (p; ) ∩ Y ⊂ G ∩ Y = G1 which proves that p is an interior point of G1 in the metric d. Thus G1 is open in Y (Proposition 1.13a). Let G1 be an open set in Y. Then, for every p ∈ G1 , there exists an −ball BY (p : ) ⊂ G1 . 11
Thus G1 =
[
BY (p; ).
p∈G1
Since we can write BY (p; ) = BX (p; ) ∩ Y, we get G1 =
[
BY (p; ) =
p∈G1
where G =
S
p∈G1
[
p∈G1
[ BX (p; ) ∩ Y = G ∩ Y BX (p; ) ∩ Y = p∈G1
BX (p; ) is open in (X, d). (Proposition 1.9c since each BX (p; ) is open).
Exercise 9. Do Exercise 8 with “closed” in place of “open.” Solution. Not available. Exercise 10. Prove Proposition 1.13. Solution. Not available. Exercise 11. Show that {cis k : k a non-negative integer } is dense in T = {z ∈ C : |z| = 1}. For which values of θ is {cis(kθ) : k a non-negative integer } dense in T? Solution. Not available.
2.2 Connectedness Exercise 1. The purpose of this exercise is to show that a connected subset of R is an interval. (a) Show that a set A ⊂ R is an interval iff for any two points a and b in A with a < b, the interval [a, b] ⊂ A. (b) Use part (a) to show that if a set A ⊂ R is connected then it is an interval. Solution. Not available. Exercise 2. Show that the sets S and T in the proof of Theorem 2.3 are open. Solution. Not available. Exercise 3. Which of the following subsets X of C are connected; nif X is not connected, what are its o components: (a) X = {z : |z| ≤ 1} ∪ {z : |z − 2| < 1}. (b) X = [0, 1) ∩ 1 + n1 : n ≥ 1 . (c) X = C − (A ∩ B) where A = [0, ∞) and B = {z = r cis θ : r = θ, 0 ≤ θ ≤ ∞}? Solution. a) Define X = {z : |z| ≤ 1} ∪ {z : |z − 2| < 1} := A ∪ B. It suffices to show that X is path-connected. Obviously A is path-connected and B is path-connected. Next, we will show that X is path-connected. Recall that a space is path-connected if for any two points x and y there exists a continuous function f from the interval [0, 1] to X with f (0) = x and f (1) = y (this function f is called the path from x to y). Let x ∈ A and y ∈ B and define the function f (t) : [0, 1] → X by (1 − 3t)x + 3tRe(x), 0 ≤ t ≤ 31 f (t) = (2 − 3t)Re(x) + (3t − 1)Re(y), 31 < t ≤ 32 2 (3 − 3t)Re(y) + (3t − 2)y, 3 < t ≤ 1. 12
This function is obviously continuous, since f (1/3) = limt→ 1 − f (t) = limt→ 1 + f (t) = Re(x) ∈ X and 3 3 f (2/3) = limt→ 2 − f (t) = limt→ 2 + f (t) = Re(y) ∈ X. In addition, we have f (0) = x and f (1) = y. Therefore 3 3 X is path-connected and hence X is connected. b) There is no way to connect {2} and { 23 }. Therefore X is not connected. The components are [0, 1), {2}, { 23 }, { 34 }, . . ., {1 + 1n }. c) X is not connected, since there is no way to connect (2, 1) and (1, −2). The k − th component is given by C − {A ∩ B} where A = [2πk − 2π, 2πk) and B = {z = r cis θ : r = θ, 2πk − 2π ≤ θ < 2πk},
k ∈ {1, 2, 3, . . .}.
Exercise 4. Prove the following generalization of Lemma 2.6. If {D j : j ∈ J} is a collection of connected S subsets of X and if for each j and k in J we have D j ∩ Dk , then D = {D j : j ∈ J} is connected. S Solution. Let D = j∈J D j and C = {D j : j ∈ J}. If D is connected, we could write D as the disjoint union A ∪ B where A and B are nonempty subsets of X. Thus, for each C ∈ C either C ⊂ A or C ⊂ B. We have C ⊂ A ∀C ∈ C or C ⊂ B ∀C ∈ C. If not, then there exist E, F ∈ C such that E ⊂ A and F ⊂ B. But, we assume A ∪ B is disjoint and thus E ∪ F is disjoint which contradicts the assumption E ∩ F , ∅. Therefore, all members of C are contained in either A or all B. Thus, either D = A and B = ∅ or D = B and A = ∅. Both contradicting the fact that A, B are assumed to be nonempty. Hence, [ D= Dj j∈J
is connected. Exercise 5. Show that if F ⊂ X is closed and connected then for every pair of points a, b in F and each > 0 there are points z0 , z1 , . . . , zn in F with z0 = a, zn = b and d(zk−1 , zk ) < for 1 ≤ k ≤ n. Is the hypothesis that F be closed needed? If F is a set which satisfies this property then F is not necessarily connected, even if F is closed. Give an example to illustrate this. Solution. Not available.
2.3 Sequences and completeness Exercise 1. Prove Proposition 3.4. Solution. a) A set is closed iff it contains all its limit points. Let S ⊂ X be a set. “⇐”: Assume S contains all its limit points. We have to show that S is closed or that S c is open. Let x ∈ S c . By assumption x is not a limit point and hence there exists an open −ball around x, B(x; ) such that B(x; ) ∩ S = ∅ (negation of Proposition 1.13f). So B(x; ) ⊂ S c and therefore S c is open. “⇒”: Let S be closed and x be a limit point. We claim x ∈ S . If not, S c (open) would be an open neighborhood of x, that does not intersect S . (Proposition 1.13f again) which contradicts the fact that x is a limit point of S . b) If A ⊂ X, then A− = A ∪ {x : x is a limit point of A} := A ∪ A0 . Let A ⊂ X be a set. ¯ We want to show x ∈ A ∪ A0 . If x ∈ A, then obviously x ∈ A ∪ A0 . Suppose x < A. Since “⊆”: Let x ∈ A. ¯ we have (Proposition 1.13f) that for every > 0, B(x; ) ∩ A , ∅. Because x < A, B(x; ) must x ∈ A, 13
intersect A in a point different from x. In particular, for every integer n there is a point xn in B(x; n1 ) ∩ A. Thus d(x, xn ) < 1n which implies xn → x (see Proof of Proposition 3.2). Then x ∈ A0 , so x ∈ A ∪ A0 . “⊇”: To show A ∪ A0 ⊆ A− . Let x ∈ A ∪ A0 . If x ∈ A, then x ∈ A− (A ⊂ A− ). Now, assume x ∈ A0 but not in A. Then there exists {xn } ⊂ A with limn→∞ xn = x. It follows, ∀ > 0 B(x; ) ∩ A , ∅ since {xn } ⊂ A. By Proposition 1.13f we get x ∈ A− . Exercise 2. Furnish the details of the proof of Proposition 3.8. Solution. Not available. Exercise 3. Show that diam A = diam A− . Solution. Not available. Exercise 4. Let zn , z be points in C and let d be the metric on C∞ . Show that |zn − z| → 0 if and only if d(zn , z) → 0. Also show that if |zn | → ∞ then {zn } is Cauchy in C∞ . (Must {zn } converge in C∞ ?) Solution. First assume that |zn − z| → 0, then 2 d(zn , z) = p |zn − z| → 0 (1 + |zn |2 )(1 + |z|2 )
p since the denominator (1 + |zn |2 )(1 + |z|2 ) ≥ 1 is bounded below away from 0. To see the converse, let d(zn , z) → 0 or equivalently d2 (zn , z) → 0. We need to show that if zn → z in the d-norm, then |zn | 9 ∞ because otherwise the denominator grows without bounds. In fact we will show that if |zn | → ∞, then d(zn , z) 9 0 for any z ∈ C. Then 4 |zn |2 − zn z¯ − z¯n z + |z|2 d2 (zn , z) = 1 + |zn |2 1 + |z|2 4 |zn |2 − 2Re(zn z¯) + |z|2 = 1 + |z|2 |zn |2 + 1 + |z|2 2 n z¯) 4 1 − 2Re(z + |z|z|n |2 |zn |2 = if |zn | , 0 2 1 + |z|2 + 1+|z| |zn |2 4 in particular if |zn | → ∞, d(zn , z) → 1+|z| 2 , 0. This shows that the denominator remains bounded as d(zn , z) → 0 and therefore the numerator 2|zn − z| → 0. Hence convergence in d-normpimplies convergence in |·|-norm for numbers zn , z ∈ C. Next assume that |zn | → ∞. Then clearly also 1 + |zn |2 → ∞ and therefore d(zn , ∞) = √ 2 2 → 0. The last thing to show is that if zn → ∞ in (C∞ , d), also |zn | → ∞. But p1+|zn | d(zn , ∞) → 0 implies 1 + |zn |2 → ∞ which is equivalent to |zn | → ∞.
Exercise 5. Show that every convergent sequence in (X, d) is a Cauchy sequence.
Solution. Let {xn } be a convergent sequence with limit x. That is, given > 0 ∃N such that d(xn , x) < n > N and d(x, xm ) < 2 if m > N. Thus d(xn , xm ) ≤ d(xn , x) + d(x, xm ) < and therefore {xn } is a Cauchy sequence. 14
+ = , 2 2
∀n, m ≥ N
2
if
Exercise 6. Give three examples of non complete metric spaces. qR 1 [ f (x) − g(x)]2 dx, f, g ∈ X. Consider Solution. Example 1: Let X = C[−1, 1] and the metric d( f, g) = −1 the Cauchy sequence 0, −1 ≤ x ≤ 0 fn (x) = nx, 0 < x ≤ n1 . 1 1, n < x ≤1
It is obvious that the limit function f is discontinuous. Hence, the metric space (X, d) is not complete. Example 2: Let X = (0, 1] with metric d(x, y) = |x − y|, x, y ∈ X. The sequence { n1 } is Cauchy, but converges to 0, which is not in the space. Thus, the metric space is not complete. Example 3: Let X = Q, the rationals, with metric d(x, y) = |x − y|, x, √ y ∈ X. The sequence defined by x1 = 1, xn+1 = x2n + x1n is a Cauchy sequence of rational numbers. The limit 2 is not a rational number. Therefore, the metric space is not complete. Exercise 7. Put a metric d on R such that |xn − x| → 0 if and only if d(xn , x) → 0, but that {xn } is a Cauchy sequence in (R, d) when |xn | → ∞. (Hint: Take inspiration from C∞ .) Solution. Not available. Exercise 8. Suppose {xn } is a Cauchy sequence and {xnk } is a subsequence that is convergent. Show that {xn } must be convergent. Solution. Since {xnk } is convergent, there is a x such that xnk → x as k → ∞. We have to show that xn → x as n → x. Let > 0. Then we have ∃N ∈ N such that d(xn , xm ) < 2 ∀n, m ≥ N since {xn } is Cauchy and ∃M ∈ N such that d(xnk , x) < 2 ∀nk ≥ M since xnk → x as k → ∞. Now, fix nk0 > M + N, then d(xn , x) ≤ d(xn , xnk0 ) + d(xnk0 , x) <
+ = , 2 2
∀n ≥ N.
Thus, xn → x as n → ∞ and therefore {xn } is convergent.
2.4 Compactness Exercise 1. Finish the proof of Proposition 4.4. Solution. Not available. Exercise 2. Let p = (p1 , . . . , pn ) and q = (q1 , . . . , qn ) be points in Rn with pk < qk for each k. Let R = [p1 , q1 ] × . . . × [pn , qn ] and show that 12 n X 2 diam R = d(p, q) = (qk − pk ) . k=1
Solution. By definition
diam(R) = sup d(x, y). x∈R,y∈R
Obviously, R is compact, so we have diam(R) = max d(x, y). x∈R,y∈R
15
Let x = (x1 , . . . , xn ) ∈ R and y = (y1 , . . . , yn ) ∈ R. Then clearly, pi ≤ xi ≤ qi and pi ≤ yi ≤ qi for all i = 1, . . . , n. We also have (yi − xi )2 ≤ (qi − pi )2 ,
∀i = 1, . . . , n.
(2.3)
Therefore, by (2.3) we obtain d(x, y) =
v t
n X (yi − xi )2 ≤ i=1
v t n X i=1
(qi − pi )2 = d(p, q).
Note that we get equality if for example xi = pi and yi = qi for all i = 1, . . . , n. Thus the maximum distance is obtained, so v t n X diam(R) = sup d(x, y) = (qi − pi )2 = d(p, q). x∈R,y∈R
i=1
n
Exercise 3. Let F = [a1 , b1 ] × . . . × [an , bn ] ⊂ R and let > 0; use Exercise 2 to show that there are S rectangles R1 , . . . , Rm such that F = m k=1 Rk and diam Rk < for each k. If xk ∈ Rk then it follows that Rk ⊂ B(xk ; ). Solution. Not available. Exercise 4. Show that the union of a finite number of compact sets is compact. S Solution. Let K = ni=1 Ki be a finite union of compact sets. Let {Gλ }λ∈Γ be an open cover of K, that is [ K⊂ Gλ . λ∈Γ
Of course, {Gλ }λ∈Γ is an open cover of each Ki , i = 1, . . . , n, that is [ Ki ⊂ Gλ , ∀i = 1, . . . , n. λ∈Γ
Since each Ki is compact, Ki ⊂
ki [
Gi, j
j=1
for each i = 1, . . . , n. Thus K=
n [ j=1
Ki ⊂
ki n [ [
Gi, j
i=1 j=1
which is a finite union and therefore K is compact. Exercise 5. Let X be the set of all bounded sequences of complex numbers. That is, {xn } ∈ X iff sup{|xn | : n ≥ 1} < ∞. If x = {xn } and y = {yn }, define d(x, y) = sup{|xn − yn | : n ≥ 1}. Show that for each x in X and ¯ ) is not totally bounded although it is complete. (Hint: you might have an easier time of it if > 0, B(x; you first show that you can assume x = (0, 0, . . .).) Solution. Not available. Exercise 6. Show that the closure of a totally bounded set is totally bounded. 16
Solution. Let (X, d) be a given metric space and let S ⊂ X be totally bounded. Let > 0. Since S is totally bounded, we have by Theorem 4.9 d) p. 22 that there exist a finite number of points x1 , . . . , xn ∈ S such that S ⊆
n [
B(xk ; /2).
k=1
Taking the closure, gives the desired result S¯ ⊆
n [
B(xk ; ).
k=1
2.5 Continuity Exercise 1. Prove Proposition 5.2. Solution. Not available. Exercise 2. Show that if f and g are uniformly continuous (Lipschitz) functions from X into C then so is f + g. Solution. The distance in C is ρ(x, y) = |x − y|. The distance in X is d(x, y). Let f : X → C be Lipschitz, that is, there is a constant M > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| ≤ Md(x, y),
∀x, y ∈ X
and g : X → C be Lipschitz, that is, there is a constant N > 0 such that ρ(g(x), g(y)) = |g(x) − g(y)| ≤ Nd(x, y),
∀x, y ∈ X.
We have ρ( f (x) + g(x), f (y) + g(y))
= | f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)| ≤ | f (x) − f (y)| + |g(x) − g(y)| ≤ Md(x, y) + Nd(x, y) = (M + N)d(x, y), ∀x, y ∈ X.
Since there is a constant K = M + N > 0 such that ρ( f (x) + g(x), f (y) + g(y)) ≤ Kd(x, y),
∀x, y ∈ X,
we have that f + g is Lipschitz. Now, let f, g be both uniformly continuous, that is, ∀1 > 0 ∃δ1 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < 1 whenever d(x, y) < δ1 and ∀2 > 0 ∃δ2 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < 2 whenever d(x, y) < δ2 . We have ρ( f (x) + g(x), f (y) + g(y))
= | f (x) + g(x) − f (y) − g(y)| = | f (x) − f (y) + g(x) − g(y)| ≤ | f (x) − f (y)| + |g(x) − g(y)| ≤ 1 + 2 ,
17
whenever d(x, y) < δ1 and d(x, y) < δ2 , that is, whenever d(x, y) < min(δ1 , δ2 ). So, choosing = 1 + 2 and δ = min(δ1 , δ2 ), we have shown that ∀ > 0 ∃δ > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < whenever d(x, y) < δ. Thus f + g is uniformly continuous. Exercise 3. We say that f : X → C is bounded if there is a constant M > 0 with | f (x)| ≤ M for all x in X. Show that if f and g are bounded uniformly continuous (Lipschitz) functions from X into C then so is f g. Solution. Let f be bounded, that is, there exists a constant M1 > 0 with | f (x)| < M1 for all x ∈ X and let g be bounded, that is, there exists a constant M2 > 0 with |g(x)| < M2 for all x ∈ X. Obviously f g is bounded, because | f (x)g(x)| ≤ | f (x)| |g(x)| ≤ M1 · M2 ∀x ∈ X. So, there exists a constant M = M1 M2 with | f (x)g(x)| ≤ M
∀x ∈ X.
Let f be Lipschitz, that is, there exists a constant N1 > 0 such that ρ( f (x), f (y)) ≤ N1 d(x, y)
∀x, y ∈ X
and let g be Lipschitz, that is, there exists a constant N2 > 0 such that ρ(g(x), g(y)) ≤ N1 d(x, y)
∀x, y ∈ X.
Now, ρ( f (x)g(x), f (y)g(y)) = | f (x)g(x) − f (y)g(y)| = = | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)| ≤ ≤
≤
| f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)| | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)|
M1 N2 d(x, y) + M2 N1 d(x, y) = (M1 N2 + M2 N1 )d(x, y)
∀x, y ∈ X.
Thus, there exists a constant K = M1 N2 + M2 N1 > 0 with ρ( f (x)g(x), f (y)g(y)) ≤ Kd(x, y)
∀x, y ∈ X,
so f g is Lipschitz and bounded. Now, let f and g be both bounded and uniformly continuous, that is ∃ ∃
∀ ∀
M1 such that | f (x)| ≤ M1 M2 such that |g(x)| ≤ M2
∀x ∈ X ∀x ∈ X
1 > 0 ∃δ1 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < 1 whenever d(x, y) < δ1 2 > 0 ∃δ2 > 0 such that ρ( f (x), f (y)) = | f (x) − f (y)| < 2 whenever d(x, y) < δ2 .
18
It remains to verify that f g is uniformly continuous, since we have already shown that f g is bounded. We have ρ( f (x)g(x), f (y)g(y))
= | f (x)g(x) − f (y)g(y)| =
= | f (x)g(x) − f (x)g(y) + f (x)g(y) − f (y)g(y)| ≤ | f (x)g(x) − f (x)g(y)| + | f (x)g(y) − f (y)g(y)| ≤ | f (x)| |g(x) − g(y)| + | f (x)| |g(y) − g(y)| ≤
M1 2 + M2 1 ,
whenever d(x, y) < min(δ1 , δ2 ). So choosing = M1 2 + M2 1 and δ = min(δ1 , δ2 ), we have ∀ > 0 ∃δ > 0 such that | f (x)g(x) − f (y)g(y)| < whenever d(x, y) < δ. Thus, f g is uniformly continuous and bounded. Exercise 4. Is the composition of two uniformly continuous (Lipschitz) functions again uniformly continuous (Lipschitz)? Solution. Not available. Exercise 5. Suppose f : X → Ω is uniformly continuous; show that if {xn } is a Cauchy sequence in X then { f (xn )} is a Cauchy sequence in Ω. Is this still true if we only assume that f is continuous? (Prove or give a counterexample.) Solution. Assume f : X → Ω is uniformly continuous, that is, for every > 0 there exists δ > 0 such that ρ( f (x), f (y)) < if d(x, y) < δ. If {xn } is a Cauchy sequence in X, we have, for every 1 > 0 there exists N ∈ N such that d(xn , xm ) < 1 for all n, m ≥ N. But then, by the uniform continuity, we have that ρ( f (xn ), f (xm )) <
∀n, m ≥ N
whenever d(xn , xm ) < δ which tells us that { f (xn )} is a Cauchy sequence in Ω. If f is continuous, the statement is not true. Here is a counterexample: Let f (x) = 1x which is continuous on (0, 1). The sequence xn = 1n is apparently convergent and therefore a Cauchy sequence in X. But { f (xn )} = { f ( n1 )} = {n} is obviously not Cauchy. Note that f (x) = 1x is not uniformly continuous on (0, 1). To see that pick = 1. Then there is no δ > 0 such that | f (x) − f (y)| < 1 whenever |x − y| < δ. Assume there exists such a δ. WLOG assume δ < 1 since the interval (0, 1) is considered. Let y = x + δ/2 and set x = δ/2, then δ/2 1 1 1 y − x | f (x) − f (y)| = − = = = > 1, x y xy δ/2· δ δ that is no matter what δ < 1 we choose, we always obtain | f (x) − f (y)| > 1. Therefore f (x) = uniformly continuous.
1 x
cannot be
Exercise 6. Recall the definition of a dense set (1.14). Suppose that Ω is a complete metric space and that f : (D, d) → (Ω; ρ) is uniformly continuous, where D is dense in (X, d). Use Exercise 5 to show that there is a uniformly continuous function g : X → Ω with g(x) = f (x) for every x in D. Solution. Not available. Exercise 7. Let G be an open subset of C and let P be a polygon in G from a to b. Use Theorems 5.15 and 5.17 to show that there is a polygon Q ⊂ G from a to b which is composed of line segments which are parallel to either the real or imaginary axes. 19
Solution. Not available. Exercise 8. Use Lebesgue’s Covering Lemma (4.8) to give another proof of Theorem 5.15. Solution. Suppose f : X → Ω is continuous and X is compact. To show f is uniformly continuous. Let > 0. Since f is continuous, we have for all x ∈ X there is a δ x > 0 such that ρ( f (x), f (y)) < /2 whenever d(x, y) < δ x . In addition, [ X= B(x; δ x ) x∈X
is an open cover of X. Since X is by assumption compact (it is also sequentially compact as stated in Theorem 4.9 p. 22), we can use Lebesgue’s Covering Lemma 4.8 p. 21 to obtain a δ > 0 such that x ∈ X implies that B(x, δ) ⊂ B(z; δz ) for some z ∈ X. More precisely, x, y ∈ B(z; δz ) and therefore ρ( f (x), f (z)) ≤ ρ( f (x), f (z)) + ρ( f (z), f (y)) <
+ = 2 2
and hence f is uniformly continuous on X. Exercise 9. Prove the following converse to Exercise 2.5. Suppose (X, d) is a compact metric space having the property that for every > 0 and for any points a, b in X, there are points z0 , z1 , . . . , zn in X with z0 = a, zn = b, and d(zk−l , zk ) < for 1 ≤ k ≤ n. Then (X, d) is connected. (Hint: Use Theorem 5.17.) Solution. Not available. Exercise 10. Let f and g be continuous functions from (X, d) to (Ω, p) and let D be a dense subset of X. Prove that if f (x) = g(x) for x in D then f = g. Use this to show that the function g obtained in Exercise 6 is unique. Solution. Not available.
2.6 Uniform convergence Exercise 1. Let { fn } be a sequence of uniformly continuous functions from (X, d) into (Ω, ρ) and suppose that f = u − − lim fn exists. Prove that f is uniformly continuous. If each fn is a Lipschitz function with constant Mn and sup Mn < ∞, show that f is a Lipschitz function. If sup Mn = ∞, show that f may fail to be Lipschitz. Solution. Not available.
20
Chapter 3
Elementary Properties and Examples of Analytic Functions 3.1 Power series Exercise 1. Prove Proposition 1.5. Solution. Not available. Exercise 2. Give the details of the proof of Proposition 1.6. Solution. Not available. Exercise 3. Prove that lim sup(an +bn ) ≤ lim sup an +lim sup bn and lim inf(an +bn ) ≥ lim inf an +lim inf bn for bounded sequences of real numbers {an } and {bn }. Solution. Let r > lim supn→∞ an (we know there are only finitely many by definition) and let s > lim supn→∞ (same here, there are only finitely many by definition). Then r + s > an + bn for all but finitely many n’s. This however, implies that r + s ≥ lim sup(an + bn ). n→∞
Since this holds for any r > lim supn→∞ an and s > lim supn→∞ bn , we have lim sup(an + bn ) ≤ lim sup an + lim sup bn . n→∞
n→∞
n→∞
Let r < lim inf n→∞ an (we know there are only finitely many by definition) and let s < lim inf n→∞ (same here, there are only finitely many by definition). Then r + s < an + bn for all but finitely many n’s. This however, implies that r + s ≤ lim inf (an + bn ). n→∞
Since this holds for any r < lim inf n→∞ an and s < lim inf n→∞ bn , we have lim inf (an + bn ) ≥ lim inf an + lim inf bn . n→∞
n→∞
n→∞
Exercise 4. Show that lim inf an ≤ lim sup an for any sequence in R. 21
Solution. Let m = lim inf n→∞ an and bn = inf{an , an+1 , . . .}. Let M = lim supn→∞ an . Take any s > M. Then, by definition of the lim supn→∞ an = M, we obtain that an < s for infinitely many n’s which implies that bn < s for all n and hence lim supn→∞ bn = m < s. This holds for all s > M. But the infimum of all these s’s is M. Therefore m ≤ M which is lim inf an ≤ lim sup an . n→∞
n→∞
Exercise 5. If {an } is a convergent sequence in R and a = lim an , show that a = lim inf an = lim sup an . Solution. Suppose that {an } is a convergent sequence in R with limit a = limn→∞ an . Then by definition, we have: ∀ > 0 ∃N > 0 such that ∀n ≥ N, we have |an − a| ≤ , that is a − ≤ an ≤ a + . This means that all but finitely many an ’s are ≤ a + and ≥ a − . This shows that a − ≤ lim inf an ≤ a + |n→∞ {z } =:m
and
a − ≤ lim sup an ≤ a + . |n→∞ {z } =:M
By the previous Exercise 4, we also have
a − ≤ m ≤ M ≤ a + . Hence, 0 ≤ M − m ≤ 2.
Since > 0 is arbitrary, we obtain m = M and further
a − ≤ lim inf an ≤ lim sup an ≤ a + , n→∞
n→∞
we obtain lim inf an = lim sup an = a. n→∞
n→∞
Exercise 6. Find the radius of convergence for each of the following power series: (a) P P∞ n n P∞ n! n2 n (b) ∞ n=0 a z , a ∈ C; (c) n=0 k z , k an integer , 0; (d) n=0 z . P∞ n n P∞ n k Solution. a) We have n=0 a z = k=0 bk z with bk = a , a ∈ C. We also have, lim sup |bk |1/k = lim sup |ak |1/k = lim sup |a| = |a|. k→∞
Therefore, R = 1/|a|, so
2
k→∞
k→∞
1 |a| , R= ∞,
a,0 . a=0
b) In this case, bn = an where a ∈ C. n2 2 an a bn 1 = lim = lim = lim R = lim n→∞ n2 +2n+1 n→∞ 2n+1 n→∞ a(n+1)2 n→∞ bn+1 a a 0, |a| > 1 1 = lim 2n+1 = 1, |a| = 1 . n→∞ |a| ∞, |a| < 1 22
P∞
n=0
an zn , a ∈ C;
c) Now, bn = kn , k is an integer , 0. We have R = lim sup |bn |1/n = lim sup |kn |1/n = lim sup |k| = |k|. n→∞
n→∞
n→∞
So
d) We can write
P∞
n=0
zn! =
Thus,
P∞
k=0
1 1k , = R= |k| − 1k ,
ak zk where 0, 2, ak = 1, 0,
k > 0, k integer . k < 0, k integer
k=0 k=1 k = n!, n ∈ N, n > 1 otherwise
lim sup |ak |1/k = lim sup |1|1/k! = 1. k→∞
k→∞
Therefore 1/R = 1 which implies R = 1. Exercise 7. Show that the radius of convergence of the power series ∞ X (−1)n
n
n=1
zn(n+1)
is 1, and discuss convergence for z = 1, −1, and i. (Hint: The nth coefficient of this series is not (−1)n /n.) Solution. Rewrite the power series in standard form, then n ∞ ∞ (−1) X (−1)n n(n+1) X k n z = ak z with ak = 0 n n=1 n=1
if ∃ n ∈ N s.t. k = n(n + 1) . else
To find the radius of convergence we use the root criterion and therefore need the estimates √ √ 1 ≤ n(n+1) n ≤ n n for n ∈ N. 1
The first inequality is immediate from the fact that n ≥ 1 and hence n n(n+1) ≥ 1. For the second inequality note that ⇔ ⇔
n ≤ nn+1 1 1 n n(n+1) ≤ n n √ √ n(n+1) n ≤ nn
.
Using this one obtains r
n(n+1)
|
(−1)n |= n
1 √ ≤1 n
n(n+1)
and r
n(n+1)
|
(−1)n |= n 23
1 1 √ ≥ √n . n n
n(n+1)
Vague memories of calculus classes tell me that
√n n → 1, thus
1 R
= lim sup
√ an = 1, i.e. R = 1.
n(n+1)
P (−1)n If z = 1 the series reduces to ∞ n=1 n which converges with the Leibniz Criterion. If z = −1 we note that the exponents n(n + 1) are always even integers and therefore the series is the same as in the previous case of z = 1. Now let z = i. The expression in(n+1) will always be real, so if the series converges at z = i, it converges to a real number. We also note that formally ∞ ∞ X ifn mod 4 ∈ {0, 1} (−1)n n(n+1) X 1n . i = cn with 1 n − n if n mod 4 ∈ {2, 3} n=1 n=1 Define the partial sums S k :=
Pk
n=0 ck .
We claim that the following chain of inequalities holds a)
b)
c)
d)
0 ≤ S 4k+3 < S 4k < S 4k+4 < S 4k+2 < S 4k+1 ≤ 1. To verify this, note that 16k2 + 8k − 1 < 0, hence a) (4k + 1)(4k + 2)(4k + 3) 1 1 S 4k+4 − S 4k = c4k+3 + c4k+4 = − + < 0, hence c) 4k + 3 4k + 4 S 4k+2 − S 4k+1 = c4k+2 < 0, hence d).
S 4k+3 − S 4k = c4k+1 + c4k+2 + c4k+3 = −
Relation b) is obvious and so are the upper bound c1 = 1 and the non-negativity constraint. We remark that {S 4k+l }k≥1 , l ∈ {0, 1, 2, 3} describe bounded and monotone subsequences that converge to some point. Now that |cn | & 0 the difference between S 4k+l and S 4k+m , l, m ∈ 0, 1, 2, 3 tends to zero, i.e. all subsequences converge to the same limit. Therefore the power series converges also in the case of z = i.
3.2 Analytic functions Exercise 1. Show that f (z) = |z|2 = x2 + y2 has a derivative only at the origin. Solution. The derivative of f at z is given by f 0 (z) = lim
h→0
f (z + h) − f (z) , h
h∈C
provided the limit exist. We have f (z + h) − f (z) h
= =
¯ − z¯z z¯z + h¯z + zh¯ + hh¯ − z¯z |z + h|2 − |z|2 (z + h)(¯z + h) = = h h h ¯ h z¯ + h¯ + z =: D. h
If the limit of D exists, it may be found by letting the point h = (x, y) approach the origin (0,0) in the complex plane C in any manner. 1.) Take the path along the real axes, that is y = 0. Then h¯ = h and thus D = z¯ + h + z
h = z¯ + h + z h
24
and therefore, if the limit of D exists, its value has to be z¯ + z. 2.) Take the path along the imaginary axes, that is x = 0. Then h¯ = −h and thus D = z¯ − h − z
h = z¯ − h − z h
and therefore, if the limit of D exists, its value has to be z¯ − z. Because of the uniqueness of the limit of D, we must have z¯ + z = z¯ − z ⇐⇒ z = −z ⇐⇒ z = 0, if the limit of D exists. It remains to show that the limit of D exists at z = 0. Since z = 0, we have that D = h¯ and thus the limit of D is 0. In summary, the function f (z) = |z|2 = x2 + y2 has a derivative only at the origin with value 0. Exercise 2. Prove that if bn , an are real and positive and 0 < b = lim bn < ∞, a = lim sup an then ab = lim sup(an bn ). Docs this remain true if the requirement of positivity is dropped? Solution. Let a = lim sup an < ∞. Then there exists a monotonic subsequence {ank } of {an } that converges n→∞
to a. Since lim bnk = b, lim ank bnk = ab. Hence, {ank bnk } is a subsequence of an bn that converges to ab. So
ab ≤
k→∞ lim sup an bn . n→∞
n→∞
Hence, lim sup an bn ≥ b lim sup an . n→∞
n→∞
Now, let a = lim sup an = ∞. Then there exists a subsequence {ank } of {an } such that lim ank = a > 0. k→∞
n→∞
And, since lim bn > 0, lim ank bnk = ∞. Hence lim sup an bn = ∞. In this second case, lim sup an bn ≥ n→∞
k→∞
n→∞
n→∞
b lim sup an . n→∞
In both cases, we have established that ab ≤ lim sup an bn . Now, since for all n ∈ N, an > 0 and bn > 0, n→∞
1 1 = . Applying the inequality we have established replacing bn with n→∞ bn b with an bn :
consider lim
lim sup an = lim sup n→∞
n→∞
1 bn
and an replaced
1 1 (an bn ) ≥ lim sup an bn . bn b n→∞
Rearranging, lim sup an bn ≤ b lim sup an . n→∞
n→∞
It follows that ab = lim sup an bn as required. n→∞
Now, consider the case where we drop the positivity requirement. Let bn = 0, − 12 , 0, − 13 , 0, − 14 , . . . and note 0 = b = lim bn < ∞. Also let an = 0, −2, 0, −3, 0, −4 . . . and note lim sup an = a = 0. In this case, n→∞
n→∞
ab = 0 , 1 = lim sup an bn . n→∞
Exercise 3. Show that lim n1/n = 1.
25
Solution. Let n ∈ N. Also let a = n1/n . Then a = n1/n ⇐⇒ log a = log n1/n log n ⇐⇒ log a = n log n ⇐⇒ lim log a = lim n→∞ n→∞ n log(x) log n . Then lim f (x) = 0 by L’Hopital’s Rule. Thus, lim = lim f (n) = 0 also. So n→∞ n→∞ n n→∞ x 1/n lim a = lim n = 1.
Now, let f (x) = n→∞
n→∞
Exercise 4. Show that (cos z)0 = − sin z and (sin z)0 = cos z. Solution. We have by definition (cos z)0 = and similarly
!0 i 1 iz 1 iz e + e−iz = eiz − e−iz = − e − e−iz = − sin z 2 2 2i
!0 1 1 iz i iz −iz (sin z) = e −e = e + e−iz = eiz + e−iz = cos z. 2i 2i 2 0
Exercise 5. Derive formulas (2.14). Solution. Not available. Exercise 6. Describe the following sets: {z : ez = i}, {z : ez = −1}, {z : ez = −i}, {z : cos z = 0}, {z : sin z = 0}. Solution. Using the definition we obtain ! ) ( 1 + 2k πi , {z : ez = i} = 2 ( ! ) 1 {z : ez = −i} = − + 2k πi , 2
{z : ez = −1} = {(1 + 2k) πi} , {z : cos z = 0} =
and {z : sin z = 0} = {kπ} where k ∈ Z. Exercise 7. Prove formulas for cos(z + w) and sin(z + w). Solution. We have cos(z) = sin(z) = 26
eiz + e−iz 2 eiz − e−iz . 2i
(
! ) 1 +k π , 2
1. Claim: cos(z) cos(w) − sin(z) sin(w) = cos(z + w). Proof: cos(z) cos(w) − sin(z) sin(w) = = = = =
eiz + e−iz eiw + e−iw eiz − e−iz eiw − e−iw − 2 2 2i 2i 1 iz iw 1 e e + e−iz e−iw + eiz eiw + e−iz e−iw 4 4 1 iz iw 1 −iz −iw e e + e e 2 2 1 iz iw e e + e−iz e−iw 2 cos(z + w).
2. Claim: sin(z) cos(w) + cos(z) sin(w) = sin(z + w). Proof: sin(z) cos(w) + cos(z) sin(w) = = = = = Exercise 8. Define tan z =
sin z cos z ;
eiz − e−iz eiw + e−iw eiz + e−iz eiw − e−iw + 2i 2 2 2i 1 1 iz iw −iz −iw iz iw e e −e e + e e − e−iz e−iw 4i 4i 1 iz iw 1 −iz −iw e e − e e 2i 2 1 iz iw e e − e−iz e−iw 2i sin(z + w).
where is this function defined and analytic?
Solution. Since both sin z and cos z are analytic in the entire complex plane, it follows from the discussion sin z in the text following Definition 2.3 that tan z = cos z is analytic wherever cos z , 0. Now, cos z = 0 implies π that z is real and equal to an odd multiple of 2 . Thus let ( ) (2k + 1)π G≡ k ∈ Z . 2 Then tan z is defined and analytic on C−G. If z ∈ G, then cos z = 0 so tan z is undefined on the non-extended complex plane.
Exercise 9. Suppose that zn , z ∈ G = C − {z : z ≤ 0} and zn = rn eiθn , z = reiθ where −π < θ, θn < π. Prove that if zn → z then θn → θ and rn → r. Solution. Not available. Exercise 10. Prove the following generalization of Proposition 2.20. Let G and Ω be open in C and suppose f and h are functions defined on G, g : Ω → C and suppose that f (G) ⊂ Ω. Suppose that g and h are analytic, g0 (ω) , 0 for any ω, that f is continuous, h is one-one, and that they satisfy h(z) = g( f (z)) for z in G. Show that f is analytic. Give a formula for f 0 (z). Solution. Not available. Exercise 11. Suppose that f : G → C is a branch of the logarithm and that n is an integer. Prove that zn = exp(n f (z)) for all z in G. 27
Solution. Let f (z) be a branch of the logarithm so that e f (z) = z. Let n ∈ Z and consider en f (z) . In the following cases we shall apply several of the properties of the complex exponential function developed in the discussion on page 38 in the text. CASE 1: Assume n > 0. Then we note that en f (z) = e f (z)+ f (z)+···+ f (z) (n times) = e f (z) e f (z) · · · e f (z) (n times) n = e f (z)
= xn .
CASE 2: Assume n < 0. Then let m = −n so that m > 0 and en f (z) =
1 1 = zn , = em f (z) zm
the middle step following from Case 1. CASE 3: Assume n = 0. Then en f (z) = e0 = 1 = z0 = zn . 1
Exercise 12. Show that the real part of the function z 2 is always positive. Solution. We know that we can write z = reiθ , 0, −π < θ < π and Log(z) = ln(r) + iθ. Thus θ θ 1 1 1 1 1 1 1 + i sin . z 2 = eLogz 2 = e 2 Logz = e 2 (lnr+iθ) = e 2 lnr e 2 θi = e 2 lnr cos 2 2 Hence, Re(z) = e 2 lnr cos 1
since e 2 lnr > 0 and cos 1
θ 2
θ 2
> 0, 1
> 0 since −π < θ < π. Thus, the real part of the function z 2 is always positive.
Exercise 13. Let G = C − {z ∈ R : z ≤ 0} and let n be a positive integer. Find all analytic functions f : G → C such that z = ( f (z))n for all z ∈ G. Solution. Let Log(z) be the principal branch, then log(z) = Log(z) + 2kπi for some k ∈ Z. Thus, we can write f (z) = z1/n = elog(z)/n = e(Log(z)+2kπi)/n = eLog(z)/n · e2kπi/n . We know that the latter factor are the n-th roots of unity and depend only on k and n. They correspond to the n distinct powers of the expression ζ = e2πi/n . Therefore, the branches of z1/n on the set U are given by f (z) = ζ k · eLog(z)/n , where k = 0, . . . , n − 1 and therefore they are all constant multiples of each other. Exercise 14. Suppose f : G → C is analytic and that G is connected. Show that if f (z) is real for all z in G then f is constant. Solution. First of all, we can write f : G → C as f (z) = u(z) + iv(z) 28
where u, v are real-valued functions. Since f : G → C is analytic, that is f is continuously differentiable (Definition 2.3), we have that u and v have continuous partial derivatives. By Theorem 2.29, this implies that u, v satisfy the Cauchy-Riemann equations. That is, ∂u ∂v = ∂x ∂y
∂u ∂v =− . ∂y ∂x
and
Since f (z) is real ∀z ∈ G, this implies
(3.1)
v(z) ≡ 0
and therefore f (z) = u(z). So, since v(z) = 0, we have ∂v ∂v = =0 ∂x ∂y and by (3.1) we obtain ∂u ∂u = =0 ∂x ∂y and thus u0 (z) = 0 (see reasoning of equation 2.22 and 2.23 on page 41). Hence f 0 (z) = 0. Since G is connected and f : G → C is differentiable with f 0 (z) = 0 ∀z ∈ G, we have that f is constant. n o Exercise 15. For r > 0 1et A = ω : ω = exp 1z where 0 < |z| < r ; determine the set A. Solution. Define the set S = {z : 0 < |z| < r} where r > 0. The image of this set under 1/z is clearly ) ( 1 T = z : < |z| . r
To find the image of A is the same as finding the image of T under ez . Claim: The image of A is C − {0} (thus not depending on r). To proof the claim, we need to show that for w , 0, the equation ez = w has a solution z ∈ T . Using polar coordinates we can write w = |w|eiθ . We have to find a complex number z = x + iy such that x2 + y2 > 1r and e x eiy = w = |w|eiθ . We want e x = |w| and y = θ + 2kπ, for some k ∈ Z. Using x = log |w| and k 0 such that (log |w|)2 + (θ + 2kπ)2 > 1r , then we found z = x + iy. Exercise 16. Find an open connected set G ⊂ C and two continuous functions f and g defined on G such that f (z)2 = g(z)2 = 1 − z2 for all z in G. Can you make G maximal? Are f and g analytic? Solution. Not available. Exercise 17. Give the principal branch of
√
1 − z.
Solution. Not available. Exercise 18. Let f : G → C and g : G → C be branches of za and zb respectively. Show that f g is a branch of za+b and f /g is a branch of za−b . Suppose that f (G) ⊂ G and g(G) ⊂ G and prove that both f ◦ g and g ◦ f are branches of zab . Solution. Not available. Exercise 19. Let G be a region and define G∗ = {z : z¯ ∈ G}. If f : G → C is analytic prove that f ∗ : G∗ → C, defined by f ∗ (z) = f ( x¯), is also analytic. 29
Solution. Let z = x + iy and let f (z) = u(x, y) + iv(x, y). By assumption f is analytic and therefore u and v have continuous partial derivatives. In addition the Cauchy-Riemann Equations u x = vy and uy = −v x are satisfied. Since f ∗ (z) = f ( x¯), we have f ∗ (z) = u∗ (x, y) + iv∗ (x, y) where u∗ (x, y) = u(x, −y) and v∗ (x, y) = −v(x, −y). Hence, we have u∗x (x, y) = u x (x, −y) = vy (x, −y), u∗y (x, y) = −uy (x, −y) = v x (x, −y), v∗x (x, y) = −v x (x, −y) and v∗y (x, y) = vy (x, −y) and therefore u∗x = v∗y and u∗y = −v∗x so f ∗ is analytic.
Exercise 20. Let z1 , z2 , . . . , zn be complex numbers such that Re zk > 0 and Re(zl . . . zk ) > 0 for 1 ≤ k ≤ n. Show that log(z1 . . . zn ) = log z1 + . . . + log zn , where log z is the principal branch of the logarithm. If the restrictions on the zk are removed, does the formula remain valid? Solution. Let z1 , . . . , zn ∈ C such that Re(z j ) > 0 and Re(z1 · · · z j ) > 0 for 1 ≤ j ≤ n. The proof will be by induction. Consider first the case where n = 2. Let z1 , z2 ∈ C as above. Then − π2 < arg z1 < π2 , − π2 < arg z2 < π2 and − π2 < arg(z1 z2 ) < π2 . But note arg(z1 z2 ) = arg z1 + arg z2 implies that −π < arg z1 + arg z2 < π. Now, recall log(z1 z2 ) = ln |z1 z2 | + i arg(z1 z2 ) = ln |z1 | + ln |z2 | + i arg(z1 ) + i arg(z2 ) = log(z1 ) + log(z2 )
Assume this formula is true for n = k − 1. Let z1 , . . . , zk ∈ C as above. Then log(z1 . . . zk ) = log((z1 · · · zk−1 )zk )
= log(z1 · · · zk−1 ) + log zk by the case when n = 2 = log z1 + log z2 + . . . + log zk−1 + log zk by the case when n = k − 1
Hence, this is true for all n such that zi , 1 ≤ i ≤ n, satisfy the restrictions. If the restriction on the zk are removed, does the formula remain valid? Consider the complex numbers z1 = −1 + 2i, z2 = −2 + i. Then z1 z2 = 0 − 5i. Clearly, z1 , z2 , and z1 z2 do not meet the restrictions as stated above. Now, log(z1 ) = ln |z1 | + i arg z1
log(z2 ) = ln |z2 | + i arg z2
√ √ 3π 3π Thus log z1 + log z2 = ln 5 + ln 5 + i( 3π 2 ) = ln 5 + i( 2 ), but note that 2 < (−π, π) and log(z1 z2 ) = π ln 5 + i(− 2 ). Thus log z1 + log z2 , log(z1 z2 ) where log z is the principal branch of the logarithm. Hence, the formula is invalid. Exercise 21. Prove that there is no branch of the logarithm defined on G = C − {0}. (Hint: Suppose such a branch exists and compare this with the principal branch.) Solution. Define the subset Gˆ of G by Gˆ = C − {z ∈ R : z ≤ 0}. We use the notation Log for the principal ˆ that is part of the log on G, Log(z) = log |z| + i arg(z)
where arg(z) ∈ (−π, π). We will prove the statement above by contradiction. Assume f (z) is a branch of the logarithm defined on G. Then restricting f to Gˆ gives us a branch of the log ˆ Therefore its only difference to the principal branch is 2πik for some k ∈ Z. This yields on G. f (z) = log |z| + i arg(z) + 2πik
ˆ Since f is analytic in G, it is continuous at −1. But we can check that this is not the case, thus where z ∈ G. we have derived a contradiction 30
3.3 Analytic functions as mappings. Möbius transformations Exercise 1. 1. Find the image of {z : Re z < 0, |Im z| < π} under the exponential function. Solution. We have {z : Re z < 0, |Im z| < π} = {z = x + iy : x < 0, −π < y < π}. The image of {z = x + iy : x < 0, −π < y < π} under the exponential function is given by {e x+iy : x < 0, −π < y < π} = {e x eiy : x < 0, −π < y < π}
= {e x (cos(y) + i sin(y)) : x < 0, −π < y < π} = {r(cos(y) + i sin(y)) : 0 < r < 1, −π < y < π}.
If 0 < r < 1 is fixed, then r(cos(y) + i sin(y)) describes a circle with radius r without the point (−r, 0) centered at (0, 0). Since r varies between 0 and 1, we get that the image is a solid circle with radius 1, where the boundary does not belong to it, the negative x-axis does not belong to it and the origin does not belong to is. Exercise 2. Do exercise 1 for the set {z : |Im z| < π/2}. Solution. We have
π π < y < }. 2 2 The image of {z = x + iy : x ∈ R, − π2 < y < π2 } under the exponential function is given by {z : |Im z| < π/2} = {z = x + iy : x ∈ R, −
π π π π < y < } = {e x (cos(y) + i sin(y)) : x ∈ R, − < y < } 2 2 2 2 π π = {r(cos(y) + i sin(y)) : r > 0, − < y < }. 2 2 If r > 0 is fixed, then r(cos(y) + i sin(y)) describes a half circle with radius r centered at (0, 0) lying in the right half plane not touching the imaginary axis. Since r varies between 0 and infinity, we get that the image is the right half plane without the imaginary axis. {e x+iy : x ∈ R, −
Exercise 3. Discuss the mapping properties of cos z and sin z. Solution. Not available. Exercise 4. Discuss the mapping properties of zn and z1/n for n ≥ 2. (Hint: use polar coordinates.) Solution. Not available. Exercise 5. Find the fixed points of a dilation, a translation and the inversion on C∞ . Solution. In general, we have az + b . cz + d To get a dilation S (z) = az, we have that b = 0, c = 0, d = 1 and a > 0. To find a fixed point, we have to find all z such that S (z) = z. In this case az = z. Obviously z = 0 is a fixed point. Also z = ∞ is a fixed point, since a· ∞ = ∞(a > 0) or S (∞) = ac = 0a = ∞. To get a translation S (z) = z + b, we have that a = 1, c = 0, d = 1 and b ∈ R. To find a fixed point, we have to find all z such that S (z) = z. In this case z + b = z, which is true if z = ∞. To see that S (∞) = ac = 01 = ∞. To get a inversion S (z) = 1z , we have that a = 0, b = 1, c = 1 and d = 0. To find a fixed point, we have to find all z such that S (z) = z. In this case 1z = z which is equivalent to z2 = 1 and thus z = 1 and z = −1 are fixed points. S (z) =
31
Exercise 6. Evaluate the following cross ratios: (a) (7 + i, 1, 0, ∞) (b) (2, 1 − i, 1, 1 + i) (c) (0, 1, i, −1) (d) (i − 1, ∞, 1 + i, 0). Solution. We have S (z) = S (z) = S (z) =
z − z3 z2 − z3 / , if z2 , z3 , z4 ∈ C z − z4 z2 − z4 z − z3 , if z2 = ∞ z − z4 z − z3 , if z4 = ∞. z2 − z3
(3.2) (3.3) (3.4)
a) By (3.4) we get (7 + i, 1, 0, ∞) = b) By (3.2) we get (2, 1 − i, 1, 1 + i) =
7+i−0 = 7 + i. 1−0
1−i−1 1 −i 2 1+i 1+i 2−1 / = / = =2 =2 = 1 + i. 2 − 1 − i 1 − i − 1 − i 1 − i −2i 1 − i (1 − i)(i + 1) 2
c) By (3.2) we get (0, 1, i, −1) =
0−i 1−i 1−i 2 1+i 2 / = −i/ = −i = −i (1 + i) = −i − i2 = 1 − i. 0+1 1+1 2 1−i1+i 2
d) By (3.3) we get (i − 1, ∞, 1 + i, 0) = Exercise 7. If T z =
az+b cz+d ,
−2 i + 1 −2 i−1−1−i = = (i + 1) = 1 + i. i−1−0 i − 1 i + 1 −2
find z2 , z3 , z4 (in terms of a, b, c, d) such that T z = (z, z2 , z3 , z4 ).
Solution. The inverse of T is given by T −1 (z) = as shown on p. 47. We have T −1 (1) = z4 = − dc to obtain T z = (z, z2 , z3 , z4 ). Exercise 8. If T z =
az+b cz+d
d−b a−c ,
dz − b −cz + a
T −1 (0) = − ba and T −1 (∞) = − dc . Set z2 =
d−b a−c ,
z3 = − ba and
show that T (R∞ ) = R∞ iff we can choose a, b, c, d to be real numbers.
az + b . Assume T (R∞ ) = R∞ . Then let z0 ∈ R∞ such that T z0 = 0. Observe that this cz + d b b implies az0 = −b, so − ∈ R∞ and r1 ≡ ∈ R∞ as well. Likewise, if z∞ ∈ R∞ such that T z∞ = ∞, then a a
Solution. Let T z =
32
r2 ≡
d ∈ R∞ . Now let z1 ∈ R∞ such that T z1 = 1. Then c az1 + b =1 cz1 + d az1 + b = cz1 + d c d b = − z1 1 − a a a z1 z1 r2 r1 − − + =0 c a a c z1 + r1 z1 + r2 = c a z1 + r1 c = ∈ R∞ . z1 + r2 a
Let r3 =
d d c c . Then = × = r2 r3 ∈ R∞ . Thus a a c a Tz =
z+ b az + b z + r1 = c ad = cz + d r z 3 + r2 r3 az + a
and we have thus found real coefficients for T . Now to prove the converse, assume T (R∞ ) , R∞ . Then recognizing that R∞ is a circle in C∞ , by Theorem 3.14 we conclude that T (R∞ ) is some other circle in C∞ . In particular, this means that T (R∞ ) ∩ (C∞ − R∞ ) , ∅. In other words, there must be some value zc ∈ R∞ for which T zc < R∞ . Now, suppose by way of contradiction that there is some representation of T in which a, b, c, and d are all real. Then azc + b T zc = czc + d is clearly an element of R∞ , contradicting the observation that T zc is not real. We see, therefore, that there is a real choice for a, b, c, and d simultaneously if and only if T (R∞ ) = R∞ . Exercise 9. If T z = {z : |z| = 1}.
az+b cz+d ,
find necessary and sufficient conditions that T (Γ) = Γ where Γ is the unit circle
Solution. We want if z¯z = 1, then T (z)T (z) = 1. T (z)T (z) = 1
⇐⇒ ⇐⇒ ⇐⇒
¯ az + b az + b (az + b)(¯az¯ + b) = 1 ⇐⇒ =1 ¯ cz + d cz + d (cz + d)(¯cz¯ + d) ¯ + dd¯ az¯az¯ + b¯az¯ + azb¯ + bb¯ = cz¯cz¯ + dc¯ z¯ + dcz ¯ + z¯(b¯a + dc¯ ) − dd¯ = 0. z¯z(a¯a − c¯c) + z(ab¯ − cd)
This is the same as the equation z¯z − 1 = 0 if ab¯ − cd¯ = 0
b¯a − dc¯
= 0
a¯a − c¯c
= 1
and bb¯ − dd¯ = −1 33
which is equivalent to |a|2 − |c|2 = 1 and 1 = |d|2 − |b|2 . Hence, we have the two conditions ab¯ − cd¯ = 0
|a|2 + |b|2 = |c|2 + |d|2 .
and
¯ then ab¯ − cd¯ = 0 yields ab¯ − λb¯ d¯ = 0 iff a = λd¯ iff d = a¯¯ . These are the sufficient conditions. Let c = λb, λ 2 2 2 2 Insert this into |c| + |d| = |a| + |b| yields |λ|2 |b|2 + and therefore we can take |λ| = 1. Then
1 λ¯
T (z) =
|a|2 = |a|2 + |b|2 , |λ|2
= λ, and so the form of the Möbius transformation is az + b , ¯ + a¯ ) λ(bz
or T (z) = λ¯
where |λ| = 1
az + b , ¯ + a¯ bz
where |λ| = 1.
Take λ = e−iθ , then
az + b , ¯ + a¯ bz This mapping transforms |z| = 1 into |T (z)| = 1. T (z) = eiθ
for some θ.
Exercise 10. Let D = {z : |z| < 1} and find all Möbius transformations T such that T (D) = D. Solution. Take an α ∈ D = {z : |z| < 1} such that T (α) = 0. Its symmetric point with respect to the unit circle is 1 α∗ = α¯ since R2 z∗ − a = z¯ − a¯ (see page 51) with a = 0 and R = 1 (the unit circle). Therefore T (α∗ ) = ∞. Thus T looks like z−α αz ¯ −1
T (z) = K
where K is a constant. (It is easy to check that T (α) = 0 and T (α∗ ) = T ( α1¯ ) = ∞). Finally, we are going to choose the constant K in such a way that |T (z0 )| = 1 where z0 = eiθ . We have T (z0 ) = K
eiθ − α αe ¯ iθ − 1
and therefore eiθ − α e−iθ − α¯ eiθ − α e−iθ − α¯ |eiθ − α| = |K|. 1 = |T (z0 )| = |K| iθ = |K| = |K| α¯ − e−iθ α − eiθ |αe ¯ − 1| |eiθ |· |α¯ − e−iθ |
So |K| = 1 implies K = eiθ for some real θ. We arrive at T (z) = eiθ
z−α , αz ¯ −1
for some real θ. 34
Exercise 11. Show that the definition of symmetry (3.17) does not depend on the choice of points z2 , z3 , z4 . That is, show that if ω2 , ω3 , ω4 are also in Γ then equation (3.18) is satisfied iff (z∗ , ω2 , ω3 , ω4 ) = (z, ω2 , ω3 , ω4 ). (Hint: Use Exercise 8.) Solution. Let Γ be a circle in C∞ containing points z2 , z3 , z4 , w2 , w3 , and w4 , with all zi distinct and all wi distinct. Let z ∈ C∞ and consider z∗ , as defined in the text and established by the points zi . We know that (z∗ , z2 , z3 , z4 ) = (z, z2 , z3 , z4 ) Recall that the above left-hand cross-ratio implies a unique Möbius transformation T for which T z2 = 1, T z3 = 0, and T z4 = ∞. By Proposition 3.10, T maps Γ to R∞ . In fact, the definition of symmetry may be rewritten as T z∗ = T z. or z∗ = T −1 T z
(3.5)
Likewise, there is some transformation S for which S w2 = 1, S w3 = 0, and S w4 = ∞. Again, S maps Γ to R∞ . We wish to show that z∗ = S −1 S z. (3.6) Now, we proceed by showing that the right hand sides of (3.5) and (3.6) must be equal: T −1 T z = T −1 T S −1 S z Observe here that T S −1 must have real coefficients by Exercise 8 because R∞ 7→ R∞ under T S −1 . Thus we observe that T S −1 = T S −1 so T −1 T z = T −1 T S −1 S z = S −1 S z. Hence we have the desired result. Exercise 12. Prove Theorem 3.4. Solution. Not available. Exercise 13. Give a discussion of the mapping f (z) = 21 (z + 1/z). 2 Solution. The function f (z) = 12 z + 1z = z 2z+1 can be defined for all z ∈ C − {0} and therefore also in the punctured disk 0 < |z| < 1. To see that in this domain the function is injective, let z1 , z2 be two numbers in the domain of f with the same image, then 0 = f (z1 ) − f (z2 ) =
z21 + 1 z22 + 1 z21 z2 + z2 − z1 z22 − z1 (z1 z2 − 1)(z1 − z2 ) − = = . z1 z2 z1 z2 z1 z2
With the assumption 0 < |zi | < 1, i = 1, 2 the factor z1 z2 −1, is always nonzero and we conclude that z1 = z2 ; hence f (z) is injective. The range of the function is C − {z ∈ C | < z ∈ [−1, 1] and Im z = 0}. To see this, write z = reiθ in polar coordinates and let f (z) = w = a + ib, a, b ∈ R. ! " ! ! # 1 1 1 1 1 f (z) = f (reiθ ) = reiθ + e−iθ = r+ cos θ + i r − sin θ . 2 r 2 r r
35
For the real and imaginary part of w the following equations must hold a = 21 r + 1r cos θ b = 21 r − 1r sin θ.
(3.7)
If f (z) = w has imaginary part b = 0 then sin θ = 0 and |cos θ| = 1. Therefore points of the form a + ib, a ∈ [−1, 1], b = 0 cannot be in the range of f . For all other points the equations (3.7) can be solved for r and θ uniquely (after restricting the argument to [−π, π)). Given any value of r ∈ (0, 1), the graph of f (reiθ ) as a function of θ looks like an ellipse. In fact from 2 2 a b formulas (3.7) we see that 1 (r+ + = 1. 1 1 ) (1− 1 ) 2
r
2
r
If we fix the argument θ and let r vary in (0, 1) it follows from from equation (3.7) that the graph of f (reiθ ) is a hyperbola and it degenerates to rays if z is purely real or imaginary. In the case θ ∈ {(2k + 1)π | k ∈ Z} the graph of f in dependence on r is on the imaginary axis and for θ ∈ {2kπ | k ∈ Z} the graph of f (reiθ ) is 2 2 either (−∞, −1) or (1, ∞). If cos θ , 0 and sin θ , 0 then cosa θ − sinb θ = 1. Exercise 14. Suppose that one circle is contained inside another and that they are tangent at the point a. Let G be the region between the two circles and map G conformally onto the open unit disk. (Hint: first try (z − a)−1 .)
Solution. Using the hint, define the Möbius transformation T (z) = (z − a)−1 which sends the region G between two lines. Afterward applying a rotation followed by a translation, it is possible to send this region to any region bounded by two parallel lines we want. Hence, choose S (z) = cz + d where |c| = 1 such that π . S (T (G)) = x + iy : 0 < y < 2 Applying the exponential function to this region yields the right half plane exp(S (T (G))) = {x + iy : x > 0}. Finally, the Möbius transformation z−1 z+1 maps the right half plane onto the unit disk (see page 53). Hence the function f defined by R(exp(S (T (z)))) maps G onto D and is the desired conformal mapping ( f is a composition of conformal mappings). Doing some simplifications, we obtain c e z−a +d − 1 f (z) = c +d e z−a + 1 where the constants c and d will depend on the circle location. R(z) =
Exercise 15. Can you map the open unit disk conformally onto {z : 0 < |z| < 1}? Solution. Not available. Exercise 16. Map G = C − {z : −1 ≤ z ≤ 1} onto the open unit disk by an analytic function f . Can f be one-one? Solution. Not available. Exercise 17. Let G be a region and suppose that f : G → C is analytic such that f (G) is a subset of a circle. Show that f is constant. 36
Solution. Not available. Exercise 18. Let −∞ < a < b < ∞ and put Mz = z−ia z−ib . Define the lines L1 = {z : Im z = b}, L2 = {z : Im z = a} and L3 = {z : Re z = 0}. Determine which of the regions A, B, C, D, E, F in Figure 1, are mapped by M onto the regions U, V, W, X, Y, Z in Figure 2. Solution. We easily see that we have M(ia) = 0. Therefore the region B, C, E and F which touch the line ia are mapped somehow to the region U, V, X and Y which touch 0. Similarly we have M(ib) = ∞ and therefore the region B and E which touch the line ib are mapped somehow to the region U and X which touch ∞. Thus we conclude that C or F goes to either V or Y. Let us find out. Let x, y be small positive real numbers such that the poin z = x + iy + ia ∈ E. Thus, the imaginary part of Mz is a positive number multiplied by x(b − a) and therefore also positive. Therefore we conclude that M maps E to U and B to X. Because B and C meet at the line ia, we conclude that X and M(C) do, too. Hence, M maps C to Y and F to V. By a similar argument, we obtain that M maps A to Z and D to W. Exercise 19. Let a, b, and M be as in Exercise 18 and let log be the principal branch of the logarithm. (a) Show that log(Mz) is defined for all z except z = ic, a ≤ c ≤ b; and if h(z) = Im [log Mz] then 0 < h(z) < π for Re z > 0. (b) Show that log(z − ic) is defined for Re z > 0 and any real number c; also prove that |Im log(z − ic)| < π2 if Re z > 0. (c) Let h be as in (a) and prove that h(z) = Im [log(z − ia) − log(z − ib)]. (d) Show that Z b dt = i[log(z − ib) − log(z − ia)] a z − it (Hint: Use the Fundamental Theorem of Calculus.) (e) Combine (c) and (d) to get that h(x + iy) =
Z
a
b
y − a x y−b − arctan dt = arctan 2 2 x x x + (y − t)
!
(f) Interpret part (e) geometrically and show that for Re z > 0, h(z) is the angle depicted in the figure. Solution. Not available. Exercise 20. Let S z = az+b cz+d and T z = that α = λa, β = λb, γ = λc, δ = λd.
αz+β γz+δ ,
show that S = T iff there is a non zero complex number λ such
Solution. ⇐: Let λ , 0 be a complex number such that α = λa β = λb γ = λc δ = λd. Then T (z) =
αz + β λaz + λb λ(az + b) (az + b) = = = = S (z). γz + δ λcz + λd λ(cz + d) (cz + d)
Thus, S = T . ⇒: Let S = T , that is S (z) = T (z). Then S (0) = T (0), S (1) = T (1) and S (∞) = T (∞) which is equivalent 37
to b d a+b c+d a c
β = λ1 ⇐⇒ b = λ1 d, β = λ1 δ δ α+β γ+δ α = λ3 ⇐⇒ a = λ3 c, α = λ3 γ. γ
= = =
(3.8) (3.9) (3.10)
Insert b = λ1 d, β = λ1 δ, a = λ3 c and α = λ3 γ into (3.9), then
⇐⇒
⇐⇒ ⇐⇒ ⇐⇒
λ3 c + λ1 d λ3 γ + λ1 δ = c+d γ+δ λ3 cγ + λ1 dγ + λ3 cδ + λ1 dδ = λ3 γc + λ1 δc + λ3 γd + λ1 δd λ1 dγ + λ3 cδ − λ1 δc − λ3 γd = 0 λ1 (dγ − δc) − λ3 (dγ − δc) = 0 (λ1 − λ3 )(dγ − δc) = 0.
Thus, either λ1 − λ3 = 0 or dγ − δc = 0. If λ1 − λ3 = 0, then from (3.8) and (3.9), we get λ3 = ac = db = λ1 which implies ad − bc = 0 a contradiction (Not possible, otherwise we do not have a Möbius transformation). Thus, dγ − δc = 0 ⇐⇒ dc = γδ ⇐⇒ γc = dδ := λ, λ , 0 or γ δ
= cλ = dλ.
(3.11) (3.12)
Insert (3.11) and (3.12) into (3.8) to obtain b β b β = ⇒ = ⇒ β = λb. d δ d dλ Insert (3.11) and (3.12) into (3.10) to obtain a α a α = ⇒ = ⇒ α = λa. c γ c cλ Therefore, we have α = λa, β = λb, γ = λc, δ = λd. Exercise 21. Let T be a Möbius transformation with fixed points z1 and z2 . If S is a Möbius transformation show that S −1 T S has fixed points S −1 z1 , and S −1 z2 . Solution. To show S −1 T S (S −1 z1 ) = S −1 z1 . We have S −1 T S (S −1 z1 ) = S −1 T S S −1 z1 = S −1 T z1 = S −1 z1 , where the first step follows by the associativity of compositions, the second step since S S −1 = id and the last step by T z1 = z1 since z1 is a fixed point of T . To show S −1 T S (S −1 z2 ) = S −1 z2 . 38
We have S −1 T S (S −1 z2 ) = S −1 T S S −1 z2 = S −1 T z2 = S −1 z2 , where the first step follows by the associativity of compositions, the second step since S S −1 = id and the last step by T z2 = z2 since z2 is a fixed point of T . Note that S −1 T S is a well defined Möbius transformation, since S and T are Möbius transformation. The composition of Möbius transformation is a Möbius transformation. Exercise 22. (a) Show that a Möbius transformation has 0 and ∞ as its only fixed points iff it is a dilation, but not the identity. (b) Show that a Möbius transformation has ∞ as its only fixed point iff it is a translation, but not the identity. Solution.
a) Let M be a Möbius transformation with exactly two fixed points, at 0 and ∞. We know that Mz =
az + b cz + d
b with a, b, c, d ∈ C. We know that M(0) = 0 so = 0, whence we conclude that b = 0. But we also d a az know that M(∞) = ∞ so = ∞, meaning c = 0. Thus Mz = = αz for some α ∈ C. Since both c d b and c were 0, we know that both a and d are nonzero so α is likewise nonzero and finite. Suppose by way of contradiction that α = 1. Then for any z˜ ∈ C∞ , M˜z = z˜ and M has infinitely many fixed points, a contradiction. Thus α , 1 and M is a (complex) dilation not equal to the identity. On the other hand, assume M is a dilation not equal to the identity. In this case, we are free to express M as Mz = αz with α , 1. It is clear from this representation that M has fixed points at 0 and ∞ and, of course, at no other points. b) Let M be a Möbius transformation with exactly one fixed point at ∞. Again, we know that Mz =
az + b cz + d
and we can see that c = 0 as before. This time, however, we also examine the lack of a second fixed point. Recall that if z˜ is a fixed point of M, then z˜ satisfies c˜z2 + (d − a)˜z − b = 0, which in the case b . We note that b , 0 because 0 is not a fixed point of c = 0 reduces to (d − a)˜z − b = 0 or z˜ = d−a of M. Suppose that d , a. In such a case, there is a finite value of z˜ that is a fixed point of M, a contradiction. Thus, it must be the case that d = a and b Mz = z + . d Thus M is a translation. Now assume M is a translation. Then we can express M as Mz = z + β. Here, in the convention of the text, we have a = 1, b = β, c = 0, and d = 1. The fixed points z˜ of M satisfy c˜z2 + (d − a)˜z − b = 0, but no finite z˜ satisfies this equation for the given coefficients. Clearly, however, M has ∞ as a fixed point, so ∞ is the only fixed point of M. Exercise 23. Show that a Möbius transformation T satisfies T (0) = ∞ and T (∞) = 0 iff T z = az−1 for some a in C. Solution. Not available. 39
Exercise 24. Let T be a Möbius transformation, T , the identity. Show that a Möbius transformation S commutes with T if S and T have the same fixed points. (Hint: Use Exercises 21 and 22.) Solution. Let T and S have the same fixed points. To show T S = S T , T , id. ? : Suppose T and S have two fixed points, say z1 and z2 . Let M be a Möbius transformation with M(z1 ) = 0 and M(z2 ) = ∞. Then MS M −1 (0) = MS M −1 Mz1 = MS z1 = Mz1 = 0 and MS M −1 (∞) = MS M −1 Mz2 = MS z2 = Mz2 = ∞.
Thus MS M −1 is a dilation by exercise 22 a) since MS M −1 has 0 and ∞ as its only fixed points. Similar, we obtain MT M −1 (0) = 0 and MT M −1 (∞) = ∞ and therefore is also a dilation. It is easy to check that dilations commute (define C(z) = az, a > 0 and D(z) = bz, b > 0, then CD(z) = abz = baz = DC(z)), thus (MT M −1 )(MS M −1 ) = (MS M −1 )(MT M −1 ) ⇒ MT S M −1 = MS T M −1
⇒ T S = S T.
? : Suppose T and S have one fixed points, say z. Let M be a Möbius transformation with M(z) = ∞. Then MS M −1 (∞) = MS M −1 Mz = MS z = Mz = ∞.
Thus MS M −1 is a translation by exercise 22 b) since MS M −1 has ∞ as its only fixed point. Similar, we obtain MT M −1 (∞) = ∞ and therefore is also a translation. It is easy to check that translations commute (define C(z) = z + 1 and D(z) = z + b, b > 0, then CD(z) = z + a + b = z + b + a = DC(z)), thus (MT M −1 )(MS M −1 ) = (MS M −1 )(MT M −1 ) ⇒ MT S M −1 = MS T M −1 ⇒ T S = S T. Exercise 25. Find all the abelian subgroups of the group of Möbius transformations. Solution. Not available. Exercise 26. 26. (a) Let GL2 (C) = all invertible 2 × 2 matrices with entries in C and let M be the group ! a b of Möbius transformations. Define ϕ : GL2 (C) → M by ϕ = az+b cz+d . Show that ϕ is a group c d homomorphism of GL2 (C) onto M. Find the kernel of ϕ. (b) Let S L2 (C) be the subgroup of GL2 (C) consisting of all matrices of determinant 1. Show that the image of S L2 (C) under ϕ is all of M. What part of the kernel of ϕ is in S L2 (C)? Solution. a) We have to check that if A=
a b c d
!
and
B=
aˆ bˆ cˆ dˆ
!
then ϕ(AB) = ϕ(A) ◦ ϕ(B). A simple calculation shows that this is true. To find the kernel of the group 2 homomorphism we have to find all z such that az+b cz+d = z. This is equivalent to az + b = cz + dz and by comparing coefficients we obtain b = c = 0 and a = d. Therefore, the kernel is given by N = ker(ϕ) = {λI : λ ∈ C× }. Note that the kernel is a normal subgroup of GL2 (C). b) Restricting ϕ to S L2 (C) still yields a surjective map since for any matrix A ∈ GL2 (C) both A and 1 A have the same image and the modification matrix M has by construction the modification M = √det A determinant 1. The kernel of the restriction is simply N ∩ S L2 (C) = {±I}. 40
Exercise 27. If G is a group and N is a subgroup then N is said to be a normal subgroup of G if S −1 T S ∈ N whenever T ∈ N and S ∈ G. G is a simple group if the only normal subgroups of G are {I} (I = the identity of G) and G itself. Prove that the group M of Möbius transformations is a simple group. Solution. Not available. Exercise 28. Discuss the mapping properties of (1 − z)i . Solution. Not available. Exercise 29. For complex numbers α and β with |α|2 + |β|2 = 1 uα,β =
αz − β¯ βz + α¯
U = {uα,β : |α|2 + |β|2 = 1}
and let
(a) Show that U is a group under composition. (b) If S U2 is the set of all unitary matrices with determinant 1, show that S U2 is a group under matrix multiplication and that for each A in S U2 there are unique complex numbers α and β with |α|2 + |β|2 = 1 and ! α β A= −β¯ α¯ ! α β 7→ uα,β is an isomorphism of the group S U2 onto U. What is its kernel? (c) Show that −β¯ α¯ (d) If l ∈ {0, 12 , 1, 32 , . . .} let Hl = all the polynomials of degree ≤ 2l. For uα,β = u in U define T u(l) : Hl → Hl , by (T u(l) f )(z) = (βz + α) ¯ 2l f (u(z)). Show that T u(l) is an invertible linear transformation on Hl , and u 7→ T u(l) is an injective homomorphism of U into the group of invertible linear transformations of Hl , onto Hl . Solution. Not available. Exercise 30. For |z| < l define f (z) by !#1/2 " 1+z f (z) = exp −i log i 1−z
(a) Show that f maps D = {z : |z| < 1} conformally onto an annulus G. (b) Find all Möbius transformations S (z) that map D onto D and such that f (S (z)) = f (z) when |z| < 1. Solution. Not available.
41
Chapter 4
Complex Integration 4.1 Riemann-Stieltjes integrals Exercise 1. Let γ : [a, b] → R be non decreasing. Show that γ is of bounded variation and V(γ) = γ(b) − γ(a). Solution. Let γ : [a, b] → R be a monotone, non-decreasing function. For a given partition of [a, b], {a = t0 < t1 < . . . < tm = b}, for all n ∈ N, γ(tn ) ≥ γ(tn−1 ). Therefore, |γ(tn ) − γ(tn−1 )| = γ(tn ) − γ(tn−1 ) ≥ 0. Let P = {a = t0 < t1 < . . . < tm = b}, any partition of [a, b]. Then, m X n=1
|γ(tn ) − γ(tn−1 )| = [γ(tm ) − γ(tm−1 )] + [γ(tm−1 ) − γ(tm−2 )] + . . . + [γ(t2 ) − γ(t1 )] + [γ(t1 ) − γ(t0 )] = [γ(b) − γ(tm−1 )] + [γ(tm−1 ) − γ(tm−2 )] + . . . + [γ(t2 ) − γ(t1 )] + [γ(t1 ) − γ(a)] = γ(b) − γ(a)
Since γ(a), γ(b) ∈ R, γ(b) − γ(a) < ∞. It follows that γ is of bounded variation for any partition P. And also v(γ; P) = sup{v(γ; P) : P a partition of [a, b]} = V(γ) = γ(b) − γ(a). Exercise 2. Prove Proposition 1.2. Solution. Let γ : [a, b] → C be of bounded variation. (a) If P and Q are partitions of [a, b] and P ⊂ Q then v(γ; P) ≤ v(γ; Q). Proof will be by induction. Let P = {a = t0 < t1 < . . . < tm = b}. Suppose Q = P ∪ {x}, where tk < x < tk+1 . Then, X v(γ; P) = |γ(ti ) − γ(ti−1 )| + |γ(tk+1 ) − γ(tk )| i,k+1
≤
X
i,k+1
|γ(ti ) − γ(ti−1 )| + |γ(x) − γ(tk )| + |γ(tk+1 ) − γ(x)|
= v(γ; Q) Now consider the general case on the number of elements of Q \ P. The smallest partition set for Q \ P contains the trivial partition composed of {a, b}. Using the case defined above, we have v(γ; Q) ≥ v(γ; P) ≥ v(γ; Q \ P) ≥ |γ(b) − γ(a)|, whenever P ⊂ Q. (b) If σ : [a, b] → C is also of bounded variation and α, β ∈ C, then αγ + βσ is of bounded variation and 42
V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ). Let P = {a = t0 < t1 < . . . < tm = b} be a partition. Since γ and σ are of bounded variation for [a, b] ⊂ R, m X |γ(tk ) − γ(tk−1 )| ≤ Mγ and there exists constants Mγ > 0 and Mσ > 0, respectively, such that v(γ; P) = v(γ; P) =
m X k=1
k=1
|σ(tk ) − σ(tk−1 )| ≤ Mσ . Then,
v(αγ + βσ; P) = =
=
≤ =
=
m X
k=1 m X
k=1 m X
k=1 m X
k=1 m X
k=1 m X k=1
= |α|
|(αγ + βσ)(tk ) − (αγ + βσ)(tk−1 )| |αγ(tk ) + βσ(tk ) − αγ(tk−1 ) − βσ(tk−1 )| |(αγ(tk ) − αγ(tk−1 )) + (βσ(tk ) − βσ(tk−1 ))| |αγ(tk ) − αγ(tk−1 )| + |βσ(tk ) − βσ(tk−1 )| |αγ(tk ) − αγ(tk−1 )| +
m X
|α| |(γ(tk ) − γ(tk−1 ))| +
m X k=1
|βσ(tk ) − βσ(tk−1 )|
k=1 m X
|β| |(σ(tk ) − σ(tk−1 ))|
k=1 m X
|γ(tk ) − γ(tk−1 )| + |β|
k=1
|σ(tk ) − σ(tk−1 )|
= |α|v(γ; P) + |β|v(σ; P) ≤ |α|sup{v(γ; P) : P a partition of [a, b]} + |β|sup{v(σ; P) : P a partition of [a, b]}
= |α|V(γ) + |β|V(σ) <∞
The result is two-fold, V(αγ + βσ) ≤ |α|V(γ) + |β|V(σ) and it follows αγ + βσ is of bounded variation by |α|V(γ) + |β|V(σ). Exercise 3. Prove Proposition 1.7. Solution. Not available. Exercise 4. Prove Proposition 1.8 (Use induction). Solution. Not available. Exercise 5. Let γ(t) = exp((−1 + i)/t−1 ) for 0 < t ≤ 1 and γ(0) = 0. Show that γ is a rectifiable path and find V(γ). Give a rough sketch of the trace of γ. Solution. Not available. Exercise 6. Show that if γ : [a, b] → C is a Lipschitz function then γ is of bounded variation. 43
Solution. Let γ : [a, b] → C be Lipschitz, that is ∃ a constant C > 0 such that |γ(x) − γ(y)| ≤ C|x − y|,
∀x, y ∈ [a, b].
Then, for any partition P = {a = t0 < t1 < . . . < tm = b} of [a, b], we have v(γ; P) =
m X k=1
|γ(tk ) − γ(tk−1 )| ≤
m X
C|tk − tk−1 | = C
k=1
m X k=1
|tk − tk−1 | = C(b − a) =: M > 0.
Hence, the function γ : [a, b] → C, for [a, b] ⊂ R, is of bounded variation, since there is a constant M > 0 such that for any partition P = {a = t0 < t1 < . . . < tm = b} of [a, b] v(γ; P) =
m X k=1
|γ(tk ) − γ(tk−1 )| ≤ M.
Exercise 7. Show that γ : [0, 1] → C, defined by γ(t) = t + it sin 1t for t , 0 and γ(0) = 0, is a path but is not rectifiable. Sketch this path. Solution. Not available. Exercise R8. Let γ and σ be the two polygons [1, i] and [1, 1 + i, i]. Express γ and σ as paths and calculate R f where f (z) = |z|2 . f and σ γ
Solution. For γ we have γ(t) = (1 − t) + it for 0 ≤ t ≤ 1. So γ0 (t) = −1 + i. Therefore, Z
γ
|z|2 dz =
Z
1 0
((1 − t)2 + t2 )(−1 + i)dt
= (−1 + i)
Z
0
"
1
(2t2 − 2t + 1)dt
2 3 2 t −t +t 3 ! 2 −1+1 = (−1 + i) 3 2 2 =− +i . 3 3 = (−1 + i)
#1 0
Since σ is composed of multiple lines, the path is given by {γ1 (t) = 1 + it : 0 ≤ t ≤ 1; γ2 (t) = (1 − t) + i :
44
0 ≤ t ≤ 1}. So γ10 (t) = i and γ20 (t) = −1. Therefore, Z Z Z f = f+ f γ γ γ2 Z1 Z = |z|2 dz + |z|2 dz Z
γ1 1
γ2
Z 1 (t2 + 1)(i)dt + ((1 − t)2 + 1)(−1)dt 0 0 Z 1 Z 1 =i (t2 + 1)dt − (t2 − 2t + 2)dt =
0
0
"
#1
"
t3 − t2 + 2t − 3 0 ! 4 1 = i− −1+2 3 3 4 4 = i− 3 3 4 = (−1 + i) 3 t3 +t =i 3
#1 0
Exercise 9.R Define γ : [0, 2π] → C by γ(t) = exp(int) where n is some integer (positive, negative, or zero). Show that γ 1z dz = 2πin. Solution. Clearly, γ(t) = eint is continuous and smooth on [0, 2π]. Thus Z
z
−1
dz =
γ
Z
2π −int
e
int
ine
dt =
0
Z
2π
R
zn dz for every integer n.
0
Exercise 10. Define γ(t) = eit for 0 ≤ t ≤ 2π and find
γ
in dt = in(2π − 0) = 2πin.
Solution. Clearly, γ(t) = eint is continuous and smooth on [0, 2π] (It is the unit circle). Case 1: n = −1 Z Z 2π Z 2π z−1 dz = e−it ieit dt = i dt = 2πi. γ
Case 2: n , −1 Z
γ
zn dz
=
Z
0
2π
eint ieit dt = i 0
0
Z
0
2π
"
ei(n+1)t i(n + 1) h i ei(n+1)2π − 1
ei(n+1)t dt = t
#2π 0
1 h i(n+1)t i2π 1 = = e 0 n+1 n+1 1 1 [cos((n + 1)2π) − i sin((n + 1)2π)1] = [1 − i· 0 − 1] = n+1 n+1 = 0. R Exercise 11. Let γ be the closed polygon [1 − i, 1 + i, −1 + i, −1 − i, 1 − i]. Find γ 1z dz.
45
Solution. Define γ1 (t) γ2 (t) γ3 (t) γ4 (t)
t from − 1 to 1 t from 1 to − 1 t from 1 to − 1 t from − 1 to 1
= 1 + it, = t + i, = −1 + it, = t − i,
→ → → →
γ10 (t) γ20 (t) γ30 (t) γ40 (t)
= i = 1 = i = 1.
Thus Z
γ
Z
1 dz z
Z −1 Z −1 Z 1 1 1 1 1 i dt + dt + i dt + dt t+i −1 + it −1 1 + it 1 1 −1 t − i ! ! Z 1 Z 1 −1 1 1 1 = dt + i dt + − t−i −1 + it −1 t + i −1 1 + it Z 1 Z 1 −t + i + t + i −1 + it − 1 − it = dt + i dt −1 (t + i)(t − i) −1 (1 + it)(−1 + it) Z 1 Z 1 1 1 dt − 2i dt = 2i 2 2 −1 −1 − t −1 t + 1 Z 1 Z 1 1 1 = 2i dt + 2i dt 2+1 2+1 t t −1 −1 Z 1 π π π 1 − − = 4i = 2πi. dt = 4i [arctan(z)]1−1 = 4i = 4i 2 4 4 2 t + 1 −1 R eiz Exercise 12. Let I(r) = γ z dz where γ : [0, π] → C is defined by γ(t) = reit . Show that limr→∞ I(r) = 0. =
1
Solution. To show limr→∞ I(r) = 0 which is equivalent to ∀ > 0, ∃n ∈ N such that if r > N, then |I(r)| < . Let > 0, then Z π Z iz Z π ireit Z π e ireit e ≤ it ireit rie dt dz = |I(r) − 0| = |I(r)| = i = e dt e dt. 0 0 reit γ z 0 Now, recall that |ez | = eRe(z) (p. 38 eq. 2.13), so it eire = eRe(ireit ) = eRe(ir cos(t)+ir·i sin(t)) = eRe(ir cos(t)−r sin(t)) = e−r sin(t) .
Hence,
Z
π 0
Z it eire dt =
π
e−r sin(t) dt =
Z
δ1
e−r sin(t) dt +
π−δ2
e−r sin(t) dt +
δ1
0
0
Z
Z
π
e−r sin(t) dt π−δ2
for δ1 > 0 and δ2 > 0. Since e−r sin(t) is continuous, so δ1 > 0 can be chosen such that Z δ1 ∀r > 0. e−r sin(t) dt < , 3 0 Similar, δ2 > 0 can be chosen such that Z π
e−r sin(t) dt <
π−δ2
46
, 3
∀r > 0.
Finally, because of the Lebesgue’s dominated convergence Theorem Z π−δ2 Z π−δ2 −r sin(t) lim e−r sin(t) dt e dt = lim r→∞
δ1
δ1
we have lim
r→∞
Z
π−δ2
r→∞
e−r sin(t) dt <
δ1
. 3
Hence, lim I(r) = 0.
r→∞
R 1 Exercise 13. Find γ z− 2 dz where: (a) γ is the upper half of the unit circle from +1 to −1: (b) γ is the lower half of the unit circle from +1 to −1. Solution. Not available. Exercise 14. Prove that if ϕ : [a, b] → [c, d] is continuous and ϕ(a) = c, ϕ(b) = d then ϕ is one-one iff ϕ is strictly increasing. Solution. Not available. Exercise 15. Show that the relation in Definition 1.16 is an equivalence relation. Solution. Not available. Exercise 16. Show that if γ and σ are equivalent rectifiable paths then V(γ) = V(σ). Solution. Not available. Exercise 17. Show that if γ : [a, b] → C is a path then there is an equivalent path σ : [0, 1] → C. Solution. Not available. Exercise 18. Prove Proposition 1.17. Solution. Not available. Exercise 19. Let γ = 1 + eit for 0 ≤ t ≤ 2π and find Solution. Not available. Exercise 20. Let γ = 2eit for −π ≤ t ≤ π and find Solution. Not available.
R
γ
R
γ
(z2 − 1)−1 dz.
(z2 − 1)−1 dz.
Exercise 21. Show that if F1 and F2 are primitives for f : G → C and G is open end connected then there is a constant c such that F1 (z) = c + F2 (z) for each z in G. Solution. Let F1 and F2 be primitives for f : G → C. Recall then that F10 = f = F20 . Let h = F1 − F2 for all z ∈ G. Then for all z ∈ G, h0 (z) = F10 (z) − F20 (z) = f (z) − f (z) = 0 Since G is open and connected with h0 (z) = 0 for all z ∈ G, then h is a constant, say c: h = c. So h = F1 − F2 = c and F1 (z) = F2 (z) + c for all z ∈ G. 47
RExercise 22. Let γ be a closed rectifiable curve in an open set G and a < G. Show that for n ≥ 2, (z − a)−n dz = 0. γ
(z − a)1−n on G. Since Solution. Let f (z) = (z − a)−n for all z ∈ G. Then f has the antiderivative F(z) = 1−n 0 F (z) = f (z) where n ≥ 2 (note F is undefined for n = Z 1). Now, let γ be a closed rectifiable curve in G with endpoints z1 , z2 ∈ C. Since z1 = z2 , F(z1 ) = F(z2 ). So,
γ
(z − a)−n dz = F(z2 ) − F(z1 ) = 0.
Exercise 23. Prove the following integration by parts formula. Let fR and g be analytic in G and let γ be a R rectifiable curve from a to b in G. Then γ f g0 = f (b)g(b) f (a)g(a) − γ f 0 g. Solution. Let f, g be analytic in G and let γ be a rectifiable curve from a to b in G. Then Z
f g0 +
γ
Z
f 0g
= (def)
γ
Z
b
Z
b
f (γ(t))g0 (γ(t)) dγ(t) +
a
Z
b
f 0 (γ(t))g(γ(t)) dγ(t)
a
f (γ(t))g0 (γ(t)) + f 0 (γ(t))g(γ(t)) dγ(t) Za Z f g0 + f 0 g = ( f g)0
= Prop. 1.7 (a)
= (def)
γ
γ
Since ( f g)0 is continuous ( f and g are analytic) by Theorem 1.18 we obtain Z ( f g)0 = ( f g)(b) − ( f g)(a) = f (b)g(b) − f (a)g(a). γ
Hence,
Z
0
γ
f g = f (b)g(b) − f (a)g(a) −
Z
f 0 g.
γ
4.2 Power series representation of analytic functions Exercise 1. Show that the function defined by (2.2) is continuous. Solution. Not available. Exercise 2. Prove the following analogue of Leibniz.s rule (this exercise will be frequently used in the later sections.) Let G be an open set and let γ be a rectifiable curve in C. Suppose that ϕ : {y} × G → C is a continuous function and define g : G → C by Z g(z) = ϕ(w, z) dw γ
then g is continuous. If
∂ϕ ∂z
exists for each (w, z) in {γ} × G and is continuous then g is analytic and 0
g (z) =
Z
γ
∂ϕ (w, z) dw. ∂z
Solution. Not available.
48
Exercise 3. Suppose that γ is a rectifiable curve in C and ϕ is defined and continuous on {γ}. Use Exercise 2 to show that Z ϕ(w) dw g(z) = w −z γ is analytic on C − {γ} and
(n)
g (z) = n!
Z
γ
ϕ(w) dw. (w − z)n+1
Solution. Not available. Exercise 4. (a) Prove Abel’s Theorem: Let P an converges to A. Prove that
P
an (z − a)n have radius of convergence 1 and suppose that
lim−
r→1
X
an rn = A.
(Hint: Find a summation formula which is the analogue of integration by parts.) (b) Use Abel’s Theorem to prove that log 2 = 1 − 12 + 13 − . . .. Solution. Not available. Exercise 5. Give the power series expansion of log z about z = i and find its radius of convergence. Solution. Assume a ∈ C is not zero, then 1 1 1 = = z a+z−a a
1 a+z−a a
=
∞
∞
X (−1)n 1 1 1 X (−1)n n (z − a)n , = (z − a) = n+1 a 1 + z−a a n=0 an a a n=0
where the radius of convergence is |a|. Let a = i, then integrate term by term yields n ∞ ∞ ∞ i2 π X π X (1 + iz)n π X in+1 n+1 n+1 log(z) = i + i − (z − i) = i − , (z − i) + 2 2 n+z 2 n (n + 1)in+1 n=0 n=0 n=1 since Log(i) = π2 i. The convergence radius is |i| = 1. √ Exercise 6. Give the power series expansion of z about z = 1 and find its radius of convergence. Solution. From the definition of numbers with rational exponents za = exp(a log z) we see that the square Q root cannot be analytic on all of C. In fact, taking the first derivatives we see that f (n) (z) = 2−n n−1 k=0 (1 − 1 2k)z−n+ 2 and therefore n−1 1 Y 1 (n) , f (1) = an = n! n!2n k=0 and with these an , f (z) =
∞ ∞ n−1 X X √ 1 Y an (z − 1)z = (1 − 2k)(z − 1)n . z= n n!2 n=0 k=0 k=0
Note that the logarithm is undefined at zero. Thus, of all branch cuts of the logarithm, the best radius of convergence we can achieve is 1.
49
Exercise 7. Use the results of this section to evaluate the following integrals: (a) Z iz e dz, γ(t) = eit , 0 ≤ t ≤ 2π; 2 γ z (b)
Z
dz , z−a
Z
sin(z) dz, z3
γ
(c)
γ
(d)
Z
γ
log z dz, zn
γ(t) = a + reit ,
0 ≤ t ≤ 2π;
γ(t) = eit ,
0 ≤ t ≤ 2π;
1 γ(t) = 1 + eit , 2
0 ≤ t ≤ 2π and n ≥ 0.
Solution. a) Let a = 0, r = 1, n = 1, f (z) = eiz and f 0 (z) = ieiz . Clearly f (z) is analytic on C and ¯ 1) ⊂ C. Now use Corollary 2.13. B(0; Z iz Z Z iz 1! 1 e eiz e f 0 (0) = dz ⇐⇒ i = dz ⇐⇒ dz = −2π. 2 2πi γ (z − 0)2 2πi γ z2 γ z ¯ r) ⊂ C. Now use Corollary b) Let a = a, r = r, n = 0, and f (z) = 1. Clearly f (z) is analytic on C and B(a; 2.13. Z Z Z 1 1 1 0! 1 (0) dz ⇐⇒ dz = 2πi. f (a) = dz ⇐⇒ 1 = 2πi γ (z − a)1 2πi γ z − a z − a γ c) Let a = 0, r = 1, n = 2 and f (z) = sin(z). Then f 0 (z) = cos(z) and f 00 (z) = − sin(z). Clearly f (z) is ¯ 1) ⊂ C. Now use Corollary 2.13. analytic on C (p. 38) and B(0; Z Z Z sin(z) sin(z) sin(z) 2! 1 1 f 00 (0) = dz ⇐⇒ − sin(0) = dz ⇐⇒ 0 = dz 2πi γ (z − 0)3 πi γ z3 πi γ z3 Z sin(z) ⇐⇒ dz = 0. z3 γ d) We have that f (z) =
log(z) zn
is analytic in the disk B(1; 12 ) where n ≥ 0. Furthermore, γ is a closed rectifiable curve (γ(t) is a circle with radius r = 1/2 and center (1, R0) in C). Obviously γ(t) = 1 + 12 eit ⊆ B(1; 12 ). Hence, by Proposition 2.15, f has a primitive and therefore γ f = 0. Therefore, Z
γ
log(z) dz = 0. zn
Exercise 8. Use a Möbuis transformation to show that Proposition 2.15 holds if the disk B(a; R) is replaced by a half plane. Solution. Not available. 50
Exercise 9. Evaluate the following integrals: (a) Z z e − e−z dz where n is a positive integer and γ(t) = eit , 0 ≤ t ≤ 2π; zn γ (b)
Z
γ
(c) Z
γ
1 dz n where n is a positive integer and γ(t) = + eit , 0 ≤ t ≤ 2π; 2 z − 12
dz where γ(t) = 2eit , 0 ≤ t ≤ 2π. (Hint: expand (z2 + 1)−1 by means of partial fractions); z2 + 1
(d)
Z
γ
(e)
Z
γ
sin z dz where γ(t) = eit , 0 ≤ t ≤ 2π; z
z1/m 1 dz where γ(t) = 1 + eit , 0 ≤ t ≤ 2π; m (z − 1) 2
Solution. a) Let a = 0, r = 1, n = m − 1 and f (z) = ez − e−z f 0 (z) = ez + e−z f 00 (z) = ez − e−z .. . f (m−1) (z) = ez − (−1)m+1 e−z . ¯ 1) ⊂ C. Now use Corollary 2.13. Clearly f (z) is analytic on C and B(0; Z z e − e−z (m − 1)! (m−1) f (0) = dz m 2πi γ (z − 0) Z z (m − 1)! e − e−z 0 , m odd ⇐⇒ dz = m 2 , m even 2πi z γ Z z , m odd e − e−z 0 dz = ⇐⇒ (m > 0). 4πi m z , m even γ (m−1)! b) Let a = 1/2, r = 1, n = m − 1 and
f (z) = 1 f 0 (z) = 0 f 00 (z) = 0 .. . (m−1) f (z) = 0.
51
¯ 1 ; 1) ⊂ C. Now use Corollary 2.13. Clearly f (z) is analytic on C and B( 2 ! Z (m − 1)! 1 (m−1) 1 f = m dz 2 2πi γ z− 1 2 Z (m − 1)! 1 1 , m = 1 ⇐⇒ m dz = 0 , m > 1 1 2πi γ z− 2 Z 1 2πi , m = 1 . ⇐⇒ m dz = 0 1 ,m > 1 γ z− 2 c) Using partial fraction decomposition, we obtain
1 B Az + Ai + Bz − Bi A ! + = = z2 + 1 z − i z + i z2 + 1 which implies A+B = 0 Ai − Bi
(4.1)
= 1.
From (4.1), we get A = −B inserted into (4.2) yields −2Bi = 1 and therefore B = − 2i1 and A = Z Z Z dz dz dz 1 1 − . = 2+1 2i z − i 2i z z γ γ +i γ
(4.2) 1 2i .
Thus
For the first integral on the right side, use Proposition 2.6. Let a = 0, r = 2 and f (w) = 1, γ(t) = 2eit , ¯ 2) ⊂ C. Then 0 ≤ t ≤ 2π. Clearly f (z) is analytic on C. We also have B(0; Z Z Z 1 f (w) 1 1 1 f (z) = dw ⇐⇒ 1 = dw ⇐⇒ dz = 2πi, 2πi γ w − i 2πi γ w − i γ z−i where | − i − 0| < 2 ⇐⇒ 1 < 2. For the second integral on the right side, use Proposition 2.6. Let a = 0, r = 2 and f (w) = 1, γ(t) = 2eit , ¯ 2) ⊂ C. Then 0 ≤ t ≤ 2π. Clearly f (z) is analytic on C. We also have B(0; Z Z Z f (w) 1 1 1 1 dw ⇐⇒ 1 = dw ⇐⇒ dz = 2πi, f (z) = 2πi γ w + i 2πi γ w + i γ z+i where |i| < 2 ⇐⇒ 1 < 2. Hence Z Z Z dz dz 1 1 1 dz 1 − = 2πi − 2πi = π − π = 0. = 2 2i γ z − i 2i γ z + i 2i 2i γ z +1 So
Z
γ
dz = 0. z2 + 1
¯ 1) ⊂ C. Now use d) Let a = 0, r = 1, n = 0 and f (z) = sin(z). Clearly f (z) is analytic on C and B(0; Corollary 2.13. Z Z Z sin(z) sin(z) sin(z) 1 0! dz ⇐⇒ dz = 0. dz ⇐⇒ 0 = f (0) (0) = 1 2πi γ (z − 0) 2πi γ z z γ 52
e) Let a = 1, r = 1/2, n = m − 1 and f (z) = z1/m 1 1/m−1 f 0 (z) = z m ! 1 1 00 − 1 z1/m−2 f (z) = m m .. . ! ! 1 1 1 (m−1) f (z) = − 1 ·...· − (m − 2) z1/m−(m−1) . m m m ¯ 1 ) ⊂ C. Now use Corollary 2.13. Clearly f (z) is analytic on C and B(1; 2
⇐⇒ ⇐⇒ ⇐⇒ Exercise 10. Evaluate 2 < r < ∞. Solution. Let f (z) =
Z
z1/m dz m γ (z − 1) ! ! Z z1/m (m − 1)! 1 1 1 − 1 ·...· − (m − 2) · 1 = dz m m m m 2πi γ (z − 1) ! ! ! Z 1 − (m − 1)m 2πi 1 1 − m 1 − 2m z1/m ·...· dz = m (m − 1)! m m m m γ (z − 1) Z m−2 Y z1/m 2πi (1 − im). dz = m (m − 1)!mm−1 i=0 γ (z − 1) f (m−1) (1) =
R
z2 +1 γ z(z2 +4)
(m − 1)! 2πi
dz where γ(t) = reit , 0 ≤ t ≤ 2π, for all possible values of r, 0 < r < 2 and
z2 + 1 . Then f (z) can be decomposed using partial fractions as z(z2 + 4)
3/8 3/8 z2 + 1 1/4 + + . = z z − 2i z + 2i z(z2 + 4) Z Z Z Z Z 1 z2 + 1 1 1 1 3 3 1 dz = It follows, dz + dz + dz. By Proposition 2.6, dz = 2 + 4) 4 z 8 z − 2i 8 z + 2i z(z γ γ γ γ γ z 1 2πi f (0) = 2πi where f ≡ 1 as f ≡ 1 is analytic on B(0, 2) and γ = 0 + reit with 0 < r < 2. Now, note z−2i is Z 1 dz = 0. Similarly, analytic on B(0, 2) and γ for 0 < r < 2 is in B(0, 2). Thus, by Proposition 2.15, γ z − 2i Z Z 1 1 z2 + 1 π dz = 2πi = i. Now, for 2 < r < ∞, note γ(t) = reit dz = 0. It follows for 0 < r < 2, 2 4 2 γ z + 2i γ z(z + 4) so γ0 (t) = ireit . Calculating the integrals in turn, Z 2π Z 2π it Z ire 1 dt = i dz = dt = 2πi. reit 0 0 γ z f (z) =
For the second integral, we use Proposition 2.6 with a = 0, 2 < r < ∞, f (w) = 1, and γ(t) = 0 + reit for
53
¯ 2 < r < ∞) ⊂ C. Then, 0 ≤ t ≤ 2π. Clearly f (z) is analytic on C and B(0; Z Z f (w) f (w) 1 1 dw ⇐⇒ 1 = dw f (−2i) = 2πi γ w + 2i 2πi γ w + 2i Z 1 ⇐⇒ dz = 2πi z + 2i γ if 2 < r < ∞. Similarly, 1 f (2i) = 2πi
Z
γ
f (w) dw ⇐⇒ w − 2i
Z
γ
1 dz = 2πi. z − 2i
It follows that, Z
γ
Z Z Z 1 z2 + 1 1 1 1 3 3 dz = dz + dz + dz 2 4 γ z 8 γ z − 2i 8 γ z + 2i z(z + 4) 3 3 1 = (2πi) + (2πi) + (2πi) 4 8 8 = 2πi
Exercise 11. Find the domain of analyticity of f (z) =
! 1 1 + iz log ; 2i 1 − iz
also, show that tan f (z) = z (i.e., f is a branch of arctan z). Show that f (z) =
∞ X k=0
(−1)k
z2k+1 , 2k + 1
for
|z| < 1
(Hint: see Exercise III. 3.19.) Solution. Not available. Exercise 12. Show that sec z = 1 +
∞ X E2k 2k z (2k)! k=1
for some constants E2 , E4 , · · · . These numbers are called Euler’s constants. What is the radius of convergence of this series? Use the fact that 1 = cos z sec z to show that ! ! ! 2n 2n 2n n−1 E2 + (−1)n = 0. E2n−4 + · · · + (−1) E2n−2 + E2n − 2 2n − 4 2n − 2 Evaluate E2 , E4 , E6 , E8 . (E10 = 50521 and E12 = 2702765). Solution. It is easily seen that sec z = 1 +
∞ X E2k 2k z , (2k)! k=1
54
(for some E2 , E4 , . . .),
1 = 11 = 1 and sec z = cos1 z is an even function. The radius of convergence is the closest since sec(0) = cos(0) distance from 0 to the nearest non-analytic point which is precisely π2 . Since 1 = cos z sec z, we can multiply the series for sec z and cos z together to obtain n ∞ X X 2i E2k i−k 1= z . (−1) (2k)!(2i − 2k)! i=0 k=0
Finally, we can compare the coefficients of z2i and multiply by (2i)! to obtain the formula ! n X 2n =0 (−1)n−k E2k 2k i=0
which is equivalent to E2n −
2n 2n − 2
!
E2n−2 +
2n 2n − 4
!
E2n−4 + · · · + (−1)n−1
2n 2
!
E2 + (−1)n = 0.
In addition, we have E2 = 1, E4 = 5, E6 = 61 and E8 = 1285. Exercise 13. Find the series expansion of f (z) = ez z−1 and let
ez −1 z
about zero and determine its radius of convergence. Consider ∞ X ak
f (z) =
k=0
k!
zk
be its power series expansion about zero. What is the radius of convergence? Show that ! ! n+1 n+1 an . a1 + · · · + 0 = a0 + n 1 Using the fact that f (z) + 12 z is an even function show that ak = 0 for k odd and k > 1. The numbers B2n = (−1)n−1 a2n are called the Bernoulli numbers for n ≥ 1. Calculate B2 , B4 , · · · , Bl0 . Solution. The Bernoulli numbers Bn are defined by Bn = lim
z→∞
dn ez z−1 , dzn
n = 0, 1, 2, . . .
We determine the largest R such that ∞
X Bk z = zk z e − 1 k=0 k!
for 0 < |z| < R.
The function f (z) = ez z−1 has a removable singularity at z = 0, because it can be extended to 0 by an analytic function g on a small neighborhood B(0, r) of 0, setting ez z−1 if 0 < |z| < r g(z) := . 1 if x = 0
Furthermore the function f has essential singularities exactly at z = 2πk, k ∈ Z − {0}. Therefore f (z) has a Laurent series expansion on the punctured disk ann(0; 0, 2π) and the negative indices disappear because the singularity at the origin is a removable one. The uniqueness of the Laurent expression together with the 55
formula for the coefficients in the expansion (Theorem 2.8, p. 72) give the desired result. First we compute B0 = limz→0 ez z−1 = 1 and note that the result in question does not hold for B0 . Also z P Bk k −1 compute B1 . Taking the limits, B1 = limz→0 (1−z)e = − 21 . Consider the equation ez z−1 = ∞ k=0 k! z for (ez −1)2 z ∈ ann(0; 0, 2π). This is equivalent to ∞ ∞ X zl X Bk k z . z = l! k! l=1 k=0 Divide by z , 0, then
∞ ∞ ∞ ∞ X zl−1 X Bk X zl X Bk 1 = = l! k=0 k! l=0 (l + 1)! k=0 k! l=1 =
n ∞ X X zn−m Bm m z m! (n + 1 − m)! n=0 m=0
=
∞ X
zn
∞ X
! n zn X n + 1 Bm (n + 1)! m=0 m
n=0
=
n=0
= B0 +
n X
Bm m!(n + 1 − m)! m=0
∞ X
! n zn X n + 1 Bm . (n + 1)! m=0 m
0=
∞ X
n=1
Since B0 = 0 conclude that
n=1
! n zn X n + 1 Bm (n + 1)! m=0 m
for all nonzero z ∈ ann(0; 0, 2π). This can only happen if the inner summation equals zero for all n ∈ N. This prove the first part of the statement. For the second claim, one only needs to show that f (z) + 21 z is an even function. It is true that 2z + (ez − 1)z z(ez + 1) 1 = . f (z) + z = 2 2(ez − 1) 2(ez − 1)
as well as t’
1 −z(ez + 1) −z(1 + ez ) z(1 + ez ) 1 f (−z) − z = −z = = = f (z) + z. 2 e −1 2(1 − ez ) 2(ez − 1) 2
With the power series expansion of f and recalling that B1 = − 21 this is the same as saying 1+
∞ ∞ X X Bk k Bk (−z)k = 1 + (z) k! k! k=2 k=2
or, yet in other terms, Bk = −Bk = 0
for k odd , k > 1.
56
In part b) the values of B0 = 1 and B1 = − 12 have been computed already. Also B3 = B5 = B7 = 0 by the second statement of part b). It remains to compute B2 , B4 , B6 and B8 . The formula involving binomial coefficients in the part above can be solved for Bn and n−1 ! 1 X n + 1 Bn = − Bk . k n+1 k=0
This gives
! 1 3 1 1 B2 = − (B0 + 3B1 ) = − 1 − = , 3 3 2 6 1 1 B4 = − (B0 + 5B1 + 10B2 ) = − , 5 30 1 1 B6 = − (B0 + 7B1 + 21B2 + 35B4 ) = , 7 42 and 1 1 B8 = − (B0 + 9B1 + 36B2 + 126B4 + 84B6 ) = − . 9 30 Exercise 14. Find the power series expansion of tan z about z = 0, expressing the coefficients in terms of Bernoulli numbers. (Hint: use Exercise 13 and the formula cot 2z = 12 cot z − 12 tan z.) Solution. To find the power series expansion of tan z about z = 0, we replace z by 2iz in the function f (z) =
z(ez + 1) 2(ez − 1)
which we obtained in the previous problem. We obtain f (2iz) =
cos(z) 2iz(e2iz + 1) iz(eiz + e−iz ) = =z = z cot(z). iz −iz 2iz e −e sin(z) 2(e − 1)
Next, we replace z by 2iz of the power series expansion for f (z) to obtain z cot(z) = 1 − Multiplying the formula cot 2z =
1 2
∞ X 4n B2n n=1
(2n)!
z2n .
cot z − 12 tan z with 2z gives z tan z = z cot z − 2z cot 2z
and therefore
Hence,
∞ ∞ ∞ 2n n X X X 4n (4n − 1)B2n 2n 4 B 4 B 2n 2n 2n 2n z − 1 − z = z . z tan z = 1 − (2n)! (2n)! (2n)! n=1 n=1 n=1 tan(z) =
∞ X 4n (4n − 1)B2n n=1
(2n)!
57
z2n−1 .
4.3 Zeros of an analytic function Exercise 1. Let f be an entire function and suppose there is a constant M, an R > 0, and an integer n ≥ 1 such that | f (z)| ≤ M|z|n for |z| > R. Show that f is a polynomial of degree ≤ n. Solution. Since f is an entire function, that is f ∈ A(C), we have by Proposition 3.3 p. 77, that f has a power series expansion ∞ X f (k) (0) ak zk , whereak = f (z) = k! k=0 with infinite radius of convergence. We have to show that f (k) (0) = 0 for k ≥ n + 1, then we are done. Let r > R and let γ(t) = reit , 0 ≤ t ≤ 2π. Note that γ is a closed and rectifiable curve. Then Z Z k! f (z) k! | f (z)| (k) | f (0)| = dz ≤ |dz|. k+1 k+1 Corollary 2.13 p.73 2πi 2π z γ γ |z| Since r > R, |γ(t)| = |reit | = r > R, 0 ≤ t ≤ 2π. So | f (z)| ≤ M|z|n , z ∈ {γ}, we get Z Z Z k! k!M k! | f (z)| M|z|n |dz| ≤ |dz| = |z|n−k−1 |dz| 2π γ |z|k+1 2π γ |z|k+1 2π γ Z k!M n−k−1 k!M n−k−1 r r 2πr |dz| = = |z|=r 2π 2π γ 1 = k!Mrn−k = k!M k−n → 0 r
as r → ∞ since k ≥ n + 1 ⇐⇒ k − n ≥ 1. Hence, | f (k) (0)| = 0 ∀k ≥ n + 1 and thus f (z) =
n X f (k) (0) k z; k! k=0
a polynomial of degree less or equal n. Exercise 2. Give an example to show that G must be assumed to be connected in Theorem 3.7. Solution. Not available. Exercise 3. Find all entire functions f such that f (x) = e x for x in R. Solution. Let f (z) = ez . Clearly f is analytic on C since it is entire. Then for all x ∈ R, f (x) = e x . Now let g be an entire function such that g(x) = e x for all x ∈ R. Now, set H = {z ∈ G : f (z) = g(z)}. So, R ⊆ H. Clearly 0 is a limit point of H and 0 ∈ G. Hence, by Corollary 3.8, f (z) ≡ g(z). So f (z) = ez is the set of all entire functions f such that f (x) = e x for all x ∈ R. Exercise 4. Prove that ez+a = ez ea by applying Corollary 3.8. Solution. Not available. Exercise 5. Prove that cos(a + b) = cos a cos b − sin a sin b by applying Corollary 3.8. Solution. Not available. Exercise 6. Let G be a region and suppose that f : G → C is analytic and a ∈ G such that | f (a)| ≤ | f (z)| for all z in G. Show that either f (a) = 0 or f is constant. 58
Solution. Let f : G → C be analytic, where G is a region. Let a ∈ G such that | f (a)| ≤ | f (z)|
∀z ∈ G.
(4.3)
Obviously, f (a) satisfies (4.3) since 0 ≤ | f (z)| ∀z ∈ G is true. Now, suppose f (a) , 0 ∀a ∈ G. Define 1 . g(z) = f (z) Clearly, g ∈ A(G) (since f (a) , 0 ∀a ∈ G) and |g(z)| =
(def)
1 1 ≤ = |g(a)|, | f (z)| (4.3) | f (a)| (def)
∀z ∈ G.
According to the Maximum Modulus Theorem 3.11 p. 79, g is constant and thus f is constant. Thus, we conclude: Either f (a) = 0 or f is constant. Exercise 7. Give an elementary proof of the Maximum Modulus Theorem for polynomials. Solution. Not available. Exercise 8. Let G be a region and let f and g be analytic functions on G such that f (z)g(z) = 0 for all z in G. Show that either f ≡ 0 or g ≡ 0. Solution. It is easy to verify the following claim: If f (z) =constant in B(a; R) ⊂ G, then f (z) =same constant for all z ∈ G. Proof: Suppose f (z) = c in B(a; R). Define g(z) = f (z) − c, then g(z) ∈ A(G). g(n) (a) = 0 for all n = 0, 1, 2, . . .. So g(z) = f (z) − c ≡ 0 on G. Let G be a region and let f, g ∈ A(G) such that f (z)g(z) = 0,
∀z ∈ G.
Assume f (z) , 0, ∀z ∈ G. Let a ∈ G such that f (a) , 0, then (since f is continuous, because f ∈ A(G)) there is an R > 0 such that f (z) , 0 whenever z ∈ B(a; R). Hence g(z) = 0 ∀z ∈ B(a; R) to fulfill f (z)g(z) = 0 ∀z ∈ B(a; R). By the claim above we obtain that ∀z ∈ G.
g(z) = 0,
So g ≡ 0. Similar, assume g(z) , 0 ∀z ∈ G, then by the same reasoning, we obtain f ≡ 0. Thus, we conclude: Either f ≡ 0 or g ≡ 0. Exercise 9. Let U : C → R be a harmonic function such that U(z) ≥ 0 for all z in C; prove that U is constant. Solution. There exists an analytic function f : C → C such that U(z) = Re( f (z)), since U : C → R is assumed to be harmonic. Hence, Re( f (z)) ≥ 0 ∀z ∈ C by assumption. For all z ∈ C, define g(z) = e− f (z) ∈ A(C). Then |g(z)| = |e− f (z) | = eRe(− f (z)) = e−Re( f (z)) ≤ 1, (def)
∀z ∈ C,
where the last step follows since Re( f (z)) ≥ 0 by assumption. Thus, g(z) is bounded and an entire function, and therefore g is constant according to Liouville’s Theorem 3.4 p. 77 and hence f (z) is constant, too. It follows that U is constant. 59
Exercise 10. Show that if f and g are analytic functions on a region G such that f¯g is analytic then either f is constant or g ≡ 0. Solution. Not available.
4.4 The index of a closed curve Exercise 1. Prove Proposition 4.3. Solution. Not available. Exercise 2. Give an example of a closed rectifiable curve γ in C such that for any integer k there is a point a < {γ} with n(γ; a) = k. Solution. Not available. Exercise 3. Let p(z) be a polynomial of degree n and R let0 (z)R > 0 be sufficiently large so that p never vanishes dz = 2πin. in {z : |z| ≥ R}. If γ(t) = Reit , 0 ≤ t ≤ 2π, show that γ pp(z) Solution. Since p(z) is a polynomial of degree n, we can write p(z) = c
n Y k=1
(z − ak )
for some constant c (see Corollary 3.6 p. 77) where a1 , . . . , an are the zeros of p(z). By the product rule we obtain n n n Y n Y X X (z − al ) = c (z − al ). p0 (z) = c (z − ak )0 l=1
k=1
k=1 l=1
l,k
l,k
Thus Z
γ
p0 (z) dz = p(z)
Z c γ
Pn Qn k=1
c
Qn
l=1 (z l,k
k=1 (z
− al )
− ak )
dz
! 1 1 1 = + + ... + dz z − a2 z − an γ z − a1 Z Z Z −1 −1 = (z − a1 ) dz + (z − a2 ) dz + . . . + (z − an )−1 dz Z
γ
γ
γ
= n(γ, a1 )2πi + n(γ, a2 )2πi + . . . + n(γ, an )2πi =
n X
2πi = 2πin,
k=1
since by Definition 4.2 p. 81 γ(t) = Reit , 0 ≤ t ≤ 2π or γ(t) = Re2πit , 0 ≤ t ≤ 1 is a closed and rectifiable curve and a1 , . . . , an < {γ} since R > 0 sufficiently large so that p never vanishes in {z : |z| ≥ R}. In addition, we have n(γ; ai ) = 1 ∀1 ≤ i ≤ n. Hence, Z 0 p (z) dz = 2πin. γ p(z) Exercise 4. Fix w R= reiθ , 0 and let γ be a rectifiable path in C − {0} from 1 to w. Show that there is an integer k such that γ z−1 dz = log r + iθ + 2πik. Solution. Not available.
60
4.5 Cauchy’s Theorem and Integral Formula Exercise 1. Suppose f : G → C is analytic and define ϕ : G × G → C by ϕ(z, w) = [ f (z) − f (w)](z − w)−1 if z , w and ϕ(z, z) = f 0 (z). Prove that ϕ is continuous and for each fixed w, z → ϕ(z, w) is analytic. Solution. Not available. Exercise 2. Give the details of the proof of Theorem 5.6. Solution. Not available. 1 ¯ Exercise 3. Let B± = B(±1; 2 ), G = B(0; 3)−(B+ ∪ B− ). Let γ1 , γ2 , γ3 be curves whose traces are |z−1| = 1, |z+1| = 1, and |z| = 2, respectively. Give γ1 , γ2 , and γ3 orientations such that n(γ1 ; w)+n(γ2 ; w)+n(γ3 ; w) = 0 for all w in C − G.
Solution. Not available. Exercise 4. Show that the Integral Formula follows from Cauchy’s Theorem. Solution. Assume the conditions of Cauchy’s Theorem, that is: Let G be an open subset of the plane and f : GR → C an analytic function. If γ is a closed rectifiable curve in G such that n(γ; w) = 0 ∀w ∈ G − C, then γ f = 0. Now, let a ∈ G, where a < {γ}. Define f (z)− f (a) z−a , z , a . g(z) = f 0 (a), z=a Clearly, g ∈ A(G). (Check the power series of f (z) like in the proof of Theorem 3.7 c) ⇒ b) p. 78) By Cauchy’s Theorem, we obtain Z Z f (z) − f (a) g=0 ⇒ dz = 0 a<γ z−a γ γ Z Z f (z) f (a) ⇒ dz − dz = 0 γ z−a γ z−a Z Z f (z) 1 ⇒ dz = f (a) dz γ z−a γ z−a Z f (z) dz = f (a)n(γ; a)2πi ⇒ Def. 4.2 p.81 γ z−a Z 1 f (z) ⇒ dz = n(γ; a) f (a). 2πi γ z − a
This is exactly Cauchy’s Integral Formula. Exercise 5. Let γ be a closed rectifiable curve in C and a < {γ}. Show that for n ≥ 2
R
γ
(z − a)−n dz = 0.
Solution. Since γ is a closed rectifiable curve in C, {γ} is bounded. So there exists R > 0 such that {γ} ⊂ B(a; R). Hence C\B(a; R) belongs to the unbounded component of B(a; R). Therefore n(γ; w) = 0 ∀w ∈ C\B(a; R) by Theorem 4.4 p.82. We can apply Corollary 5.9 p.86 if we let G = B(a; R) the open set and f : G → C given by f (z) = 1 is analytic. Further γ is a closed and rectifiable curve in G such that ∀w ∈ C\G,
n(γ; w) = 0, 61
then for a in B(a; R)\{γ} Z 1 (n − 1)! dz f (a)n(γ; a) = 2πi (z − a)n γ Z (n − 1)! 1 ⇒ 0· n(γ; a) = dz 2πi (z − a)n γ (n−1)
since f (n−1) (a) = 0 if n ≥ 2. Hence Z
γ
(z − a)−n dz = 0,
for n ≥ 2.
Exercise 6. Let f be analytic on D = B(0; 1) and suppose | f (z)| ≤ 1 for |z| < 1. Show | f 0 (0)| ≤ 1. Solution. We can directly apply Cauchy’s estimate 2.14 p. 73 with a = 0, R = 1, M = 1 and N = 1, that is by assumption f is analytic on B(0; 1) and | f (z)| ≤ 1 for all z in B(0; 1) (since |z| < 1). Then | f 0 (0)| ≤ So | f 0 (0)| ≤ 1. Exercise 7. Let γ(t) = 1 + eit for 0 ≤ t ≤ 2π. Find
1!· 1 = 1. 11
R γ
z n z−1
dz for all positive integers n.
Solution. Let G = B(1; R) (open) where 1 < R < ∞ and let f : G → C be given by f (z) = zn which is analytic on G. By Theorem 4.4 p. 82, n(γ; w) = 0 ∀w ∈ C\G since C\G belongs to the unbounded component of G and γ is a closed and rectifiable curve in C. Now, we can apply Corollary 5.9 with G and f defined above such that n(γ; w) = 0 ∀w ∈ C\G and γ is a closed rectifiable curve in G, then for a = 1 in G\{γ} Z Z zn (n − 1)! z n (n − 1)! dz = dz f (n−1) (1)n(γ; a) = n 2πi 2πi γ (z − 1) γ z−1 Z (n − 1)! z n ⇒ n! = dz 2πi γ z−1 Z z n ⇒ dz = 2πin, γ z−1 since f (n−1) (1) = n! and n(γ; a) = 1. Exercise 8. Let G be a region and suppose fn : G → C is analytic for each n ≥ 1. Suppose that { fn } converges uniformly to a function f : G → C Show that f is analytic. Solution. Let G be a region and suppose fn : G → C is analytic for each n ≥ 1, that us fn is continuous for each n ≥ 1. Let T be an arbitrary triangular path in G. Then T is closed and rectifiable. We also have, that n(T ; w) = 0 ∀w ∈ C\G, since w is an element of the unbounded component of G (Theorem 4.4 p. 82). By Cauchy’s Theorem (First Version), p. 85 with m = 1, we have Z fn = 0, ∀n ∈ N, (4.4) T
62
since we suppose fn : G → C is analytic for each n ≥ 1 and T is closed and rectifiable. We also have Z Z fn = lim 0 = 0 f = lim n→∞ T (4.4) n→∞ T The first equality follows by Lemma 2.7 p. 71 since T is closed and rectifiable and since each { fn } ∈ A(G), we have fn is continuous for each fn and by the uniform convergence to a function f , we get f is continuous (Theorem 6.1 p. 29). Finally, R we can apply Morera’s Theorem 5.10 p. 86 to obtain that f is analytic in G, since f is continuous and γ f = 0 for every triangular path T in G.
Exercise 9. Show that if f : C → C is a continuous function such that f is analytic off [−1, 1] then f is an entire function. R Solution. By Morera’s Theorem, it suffice to show that for every triangular path T in C, we have T f = 0, since we already assume f is continuous. There are several cases to consider: R Case 1: A triangle does not intersect I. In this case, we obtain by Cauchy’s Theorem, that T f = 0, since we can find an open neighborhood G containing T (is closed and rectifiable) such that f ∈ A(G) (by assumption) and n(T ; w) = 0 ∀w ∈ C\G (by Theorem 4.4 p. 82). Case 2: A triangle touches I exactly at one point P. This single point ofR intersection P is a removable singularity, since f is continuous. Again apply Cauchy’s Theorem to obtain T f = 0. (Procedure: Translate R the triangle by ±i depending if it lies above or below the x-axis. By Cauchy’s Theorem T f = 0 over the translated triangle and let → 0 since f is continuous). Case 3: One edge of the triangle touches I. Like in case 2 we can translate the triangle by ± We can also argue this way: Let G be an open neighborhood containing the triangle T . Let {T n } be a sequence of triangles that are not intersecting I, but whose limit is the given T . Then by the continuity of f , we get Z Z f = lim 0 = 0, f = lim T
n→∞
Tn
n→∞
where the first equality follows by Lemma 2.7 and the second one by case 1. Case 4: A triangle T is cut into 2 parts by I. Then, we can always decompose the triangle T in three parts. Two trianglesR are of the kind explained in case 3 and one is of the kind explained in case 2 (a sketch might help). Thus, T f = 0 again. Case 5: A triangle T contains parts of I (also here a sketch might help). In this instance, we can decompose the triangle into 5 parts. Two triangles Rare of the kind explained in case 3 and the other three triangles are of the kind explained in case 2. Hence, T f = 0. R Summary: Since T f = 0 for every triangular path in C and f is continuous, we get by Morera’s Theorem: f ∈ A(C), that is f is an entire function. Exercise 10. Use Cauchy’s Integral Formula to prove the Cayley–Hamilton Theorem: If A is an n × n matrix over C and f (z) = det(z − A) is the characteristic polynomial of A then f (A) = 0. (This exercise was taken from a paper by C. A. McCarthy, Amer. Math. Monthly, 82 (1975), 390–391). Solution. Not available.
4.6 The homotopic version of Cauchy’s Theorem and simple connectivity Exercise 1. Let G be a region and let σ1 , σ2 : [0, 1] → G be the constant curves σ1 (t) ≡ a, σ2 (t) ≡ b. Show that if γ is closed rectifiable curve in G and γ ∼ σ1 , then γ ∼ σ2 . (Hint: connect a and b by a curve.) 63
Solution. Assume γ is a closed and rectifiable curve in G and γ ∼ σ1 . Since σ1 (t) ≡ a and σ2 (t) ≡ b, we surely have that σ1 and σ2 are closed and rectifiable curves in G. If we can show that σ1 ∼ σ2 , then we get γ ∼ σ2 , since ∼ is an equivalence relation, that is: γ ∼ σ1
and
σ1 ∼ σ2 ⇒ γ ∼ σ2 .
So, now we show σ1 ∼ σ2 , that is, we have to find a continuous function Γ : [0, 1] × [0, 1] → G such that Γ(s, 0) = σ1 (s) = a
and
Γ(s, 1) = σ2 (s) = b,
for
0≤s≤1
and Γ(0, t) = Γ(1, t) for 0 ≤ t ≤ 1. Let Γ(s, t) = tb + (1 − t)a. Clearly Γ is a continuous function. (it is constant with respect to s and a line with respect to t). In addition, it satisfies Γ(s, 0) = a ∀0 ≤ s ≤ 1 and Γ(s, 1) = b ∀0 ≤ s ≤ 1 and Γ(0, t) = tb + (1 − t)a = Γ(1, t) ∀0 ≤ t ≤ 1. So σ1 ∼ σ2 . Exercise 2. Show that if we remove the requirement “Γ(0, t) = Γ(1, t) for all t” from Definition6.1 then the curve γ0 (t) = e2πit , 0 ≤ t ≤ 1, is homotopic to the constant curve γ1 (t) ≡ 1 in the region G = C − {0}. Solution. Not available. Exercise 3. 3. Let C = all rectifiable curves in G joining a to b and show that Definition 6.11 gives an equivalence relation on C. Solution. Not available. Exercise 4. Let G = C − {0} and show that every closed curve in G is homotopic to a closed curve whose trace is contained in {z : |z| = 1}. Solution. Not available. Exercise 5. Evaluate the integral
R
dz γ z2 +1
where γ(θ) = 2| cos 2θ|eiθ for 0 ≤ θ ≤ 2π.
Solution. A sketch shows that the two zeros of z2 + 1 (they are ±i) are inside of the closed and rectifiable curve γ (The region looks like a clover with four leaves ). Using partial fraction decompositions gives Z Z Z 1 1 1 dz 1 dz − dz = 2+1 2i z − i 2i z + i z γ γ γ Z Z 1 1 1 1 = π dz − π dz 2πi γ z − i 2i γ z + i = π (n(γ; i) − n(γ; −i)) Def 4.2 p. 81
=
π· 0 = 0,
since n(γ; i) = n(γ; −i) since i and −i are contained in the region generated by γ. Hence Z dz = 0. 2+1 z γ Exercise 6. Let γ(θ) = θeiθ for 0 ≤ θ ≤ 2π and γ(θ) = 4π − θ for 2π ≤ θ ≤ 4π. Evaluate
R
dz . γ z2 +π2
Solution. A sketch reveals that the zero −πi is inside the region and the zero πi is outside the region. Using partial fraction decomposition yields Z Z Z 1 dz 1 1 1 = dz − dz 2 + π2 2πi z − iπ 2πi z + iπ z γ γ γ = n(γ; iπ) − n(γ; −iπ) = 0 − 1 = −1, Def 4.2 p. 81
64
since iπ is not contained in the region generated by γ, so n(γ; iπ) = 0 and −iπ is contained in the region, so n(γ; −iπ) = 1. Hence, Z dz = −1. 2 2 γ z +π 3 i 3 1 −1 Exercise R 7. Let f (z) = [(z− 2 −i)· (z−1− 2 i)· (z−1− 2 )· (z− 2 −i)] and let γ be the polygon [0, 2, 2+2i, 2i, 0]. Find γ f .
Solution. Not available.
Exercise 8. Let G = C − {a, b}, a , b, and let γ be the curve in the figure below. (a) Show that n(γ; a) = n(γ; b) = 0. (b) Convince yourself that γ is not homotopic to zero. (Notice that the word is “convince” and not “prove”. Can you prove it?) Notice that this example shows that it is possible to have a closed curve γ in a region such that n(γ; z) = 0 for all z not in G without γ being homotopic to zero. That is, the converse to Corollary 6.10 is false. Solution. Let γ be the path depicted on p. 96. We can write it as a sum of 6 paths. Two of them will be closed and have a and b in their unbounded component. Therefore, two integrals will be zero and we will have another 2 pair of non-closed paths. The first pair begins at the leftmost crossing pair and goes around a in opposite direction. The second pair begins at the middle crossing pair and goes around b in opposite direction. They Ralso meet at the rightmost crossing point. If we integrate over the path around a 1 dz where γ1 (t) = a + reit , 0 ≤ t ≤ π and γ2 (t) = a + reit , π ≤ t ≤ 2π for is equivalent to evaluate γ −γ z−a 1 2 some r > 0. It is easily seen that we have Z Z 1 1 dz = dz = πi z − a z − a γ2 γ1 and therefore n(γ; a) = 0 and similarly we obtain n(γ; b) = 0. Exercise 9. Let G be a region and let γ0 and γ1 be two closed smooth curves in G. Suppose γ0 ∼ γ1 , and Γ satisfies (6.2). Also suppose that γt (s) = Γ(s, t) is smooth for each t. If w ∈ C − G define h(t) = n(γt ; w) and show that h : [0, 1] → Z is continuous. Solution. Not available. Exercise 10. Find all possible values of through ±i.
R
dz , γ 1+z2
where γ is any closed rectifiable curve in C not passing
Solution. Let γ be a closed rectifiable curve in C not passing through ±i. Using partial fraction decomposition and the definition of the winding number we obtain ! Z Z Z 1 1 1 1 dz = dz − dz = (2πin(γ; i) − 2πin(γ; −i)) = π (n(γ; i) − n(γ; −i)) . 2 2i z − 1 z + i 2i 1 + z γ γ γ R ez −e−z Exercise 11. Evaluate γ z4 dz where γ is one of the curves depicted below. (Justify your answer.) Solution. Using Corollary 5.9 p. 86 we have ∀a ∈ G − {γ} Z 1 f (z) dz = 2πi f (k) (a)n(γ; a) , k+1 k! γ (z − a) 65
where γ is a closed rectifiable curve in G (open) such that n(γ; w) = 0 for all w ∈ C − G and f : G → C is analytic. In our case a = 0 and γ satisfies the above assumptions for exercise a), b) and c). Let f (z) = ez − e−z which is clearly analytic on any region G. We have f (3) (0) = e0 + e−0 = 2 and using the formula above for k = 3 we obtain Z z 1 2πi e − e−z dz = 2πi· 2n(γ; 0) = n(γ; 0). 4 6 3 z γ 2πi 3 .
Hence, n(γ; 0) = 1 for part a) and thus the result is is 4πi 3 . We obtain the same result for part c).
We have n(γ; 0) = 1 for part b) and thus the result
4.7 Counting zeros; the Open Mapping Theorem Exercise 1. Show that if f : G → C is analytic and γ is a rectifiable curve in G then f ◦ γ is also a rectifiable curve. Solution. Not available. Exercise 2. Let G be open and suppose that γ is a closed rectifiable curve in G such that γ ≈ 0. Set r = d({γ}, ∂G) and H = {z ∈ C : n(γ; z) = 0}. (a) Show that {z : d(z, ∂G) < 12 r} ⊂ H. (b) Use part (a) to show that if f : G → C is analytic then f (z) = α has at most a finite number of solutions z such that n(γ; z) , 0. Solution. Not available. Exercise 3. Let f be analytic in B(a; R) and suppose that f (a) = 0. Show that a is a zero of multiplicity m iff f (m−1) (a) = . . . = f (a) = 0 and f (a) (a) , 0. Solution. ⇐: Let f (n) (a) = 0 for all n = 1, 2, . . . , m − 1 (we also have f (a) = 0) and f (m) (a) , 0. Since f ∈ A(B(a; R)), we can write f (z) =
∞ X k=0
ak (z − a)k
where
ak =
f (k) (a) and |z − a| < R k!
∞ ∞ X X f (k) (a) f (k) (a) = (z − a)k = (z − a)k k! k! k=0 k=m
= (z − a)m
∞ X f (k) (a) (z − a)k−m = k! g(a),0, since k=0 | {z }
f (m) ,0
(z − a)m g(z)
g(z)
where g ∈ A(B(a; R)). Hence, a is a zero of multiplicity m. ⇒: Suppose a is a zero of multiplicity m ≥ 1 of f . By definition, there exists g ∈ A(B(0; R)) such that f (z) = (z − a)m g(z),
g(a) , 0 (p. 97).
Then after a lengthy calculation, we obtain f
(n)
(z) =
n X k=0
n k
!
m! (z − a)m−(n−k) g(k) (z), (m − (n − k))! 66
n = 1, 2, . . . , m.
If z = a, then clearly f (n) (a) = 0 for m − (n − k) > 0 and f (n) (a) , 0 for m − (n − k) = 0. If k = 0 and m = n, then f (m) (a) = f (n) (a) = m!g(a) , 0 since g(a) , 0 by assumption. Hence, f (n) (a) = 0 for n = 1, 2, . . . , m − 1 and f (n) (a) , 0 for n = m. Exercise 4. Suppose that f : G → C is analytic and one-one; show that f 0 (z) , 0 for any z in G. Solution. Assume f 0 (a) = 0 for some a ∈ G. Then f (a) = α where α is a constant. Define F(z) = f (z) − α. Then F(a) = f (a) − α = α − α = 0 and F 0 (a) = f 0 (a) = 0 by assumption. Hence, F has a as a root of multiplicity m ≥ 2. Then, we can use Theorem 7.4 p. 98 to argue that there is an > 0 and δ > 0 such that for 0 < |ξ − a| < δ, the equation f (z) = ξ has exactly m simple roots in B(a; ), since f is analytic in B(a; R) and α = f (a). Also f (z) − α has a zero of order m ≥ 2 at z = a. Since f (z) = ξ has exactly m (our case m ≥ 2) simple roots in B(a; ), we can find at least two distinct points z1 , z2 ∈ B(a; ) ⊂ G such that f (z1 ) = ξ = f (z2 ) contradicting that f is 1 − 1. So f 0 (z) , 0 for any z in G. Exercise 5. Let X and Ω be metric spaces and suppose that f : X → Ω, is one-one and onto. Show that f is an open map iff f is a closed map. (A function f is a closed map if it takes closed sets onto closed sets.) Solution. Not available. Exercise 6. Let P : C → R be defined by P(z) = Re z; show that P is an open map but is not a closed map. (Hint: Consider the set F = {z : Im z = (Re z)−1 and Re z , 0}.) Solution. Not available. Exercise 7. Use Theorem 7.2 to give another proof of the Fundamental Theorem of Algebra. Solution. Not available.
4.8 Goursat’s Theorem No exercises are assigned in this section.
67
Chapter 5
Singularities 5.1 Classification of singularities Exercise 1. Each of the following functions f has an isolated singularity at z = 0. Determine its nature; if it is a removable singularity define f (0) so that f is analytic at z = 0; if it is a pole find the singular part; if it is an essential singularity determine f ({z : 0 < |z| < δ}) for arbitrarily small values of δ. a) sin z ; f (z) = z b) cos z ; f (z) = z c) cos z − 1 f (z) = ; z d) f (z) = exp(z−1 ); e) f (z) =
log(z + 1) ; z2
f) f (z) =
cos(z−1 ) ; z−1
f (z) =
z2 + 1 ; z(z − 1)
g)
h)
f (z) = (1 − ez )−1 ; i) 1 f (z) = z sin ; z
68
j) 1 f (z) = zn sin . z Solution. a) According to Theorem 1.2 p.103, z = 0 is a removable singularity, since lim (z − 0)
z→∞
If we define,
sin(z) sin(z) = lim z = lim sin(z) = 0. z→∞ z→∞ z z sin(z) z , f (z) = 1,
z,0 , z=0
then f is analytic at z = 0. A simple computation using L’Hospital’s rule yields limz→0 b) We have (see p. 38) ∞ X z2k (−1)k cos(z) = (2k)! k=0 and therefore
sin(z) z
= 1.
∞
1 cos(z) X z2k−1 = = + analytic part. (−1)k z (2k)! z k=0 Thus cosz z has a simple pole at z = 0. c) According to Theorem 1.2 p.103, z = 0 is a removable singularity, since lim (z − 0)
z→∞
If we define,
cos(z) − 1 cos(z) − 1 = lim z = lim cos(z) − 1 = 1 − 1 = 0. z→∞ z→∞ z z cos(z)−1 z , f (z) = 1,
z,0 , z=0
then f is analytic at z = 0. A simple computation using L’Hospital’s rule yields limz→0 e) We know ∞ X zk (−1)k+1 . log(1 + z) = k k=1
cos(z)−1 z
= 0.
So we can write ∞ ∞ ∞ k k−2 k−2 X 1 X log(1 + z) 1 1 X k+1 z k+1 z k+1 z = (−1) = = − + (−1) (−1) k k z 2 k=3 k z2 z2 k=1 k=1
has a pole of order 1 at z = 0 and the singular part is 1z (by Equation 1.7 and and hence f (z) = log(1+z) z2 Definition 1.8 p. 105). 2 z+1 z+1 = 1 + z(z−1) g) First, we simplify zz2+1 . Now partial fraction decomposition of z(z−1) yields −z 1 2 z+1 =− + z(z − 1) z z−1 so f (z) = 1 −
1 2 + z z−1
69
2
2 − 1z (by Equation and hence f (z) = zz2+1 has poles of order 1 at z = 0 and z = 1 and the singular part is z−1 −z 1.7 and Definition 1.8 p. 105). i) We know (see p. 38) ∞ X z2k+1 (−1)k sin(z) = (2k + 1)! k=0 so 2k+1 2k 1 1 ∞ ∞ X X 1 1 1 1 z z k k = =1− (−1) f (z) = z sin = z (−1) + − + ... 2 4 6 z (2k + 1)! (2k + 1)! 3!z 5!z 7!z k=0 k=0
Therefore f (z) = z sin 1z has an essential singularity at z = 0 (it is neither a pole nor a removable singularity by Corollary 1.18c). We have f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] = C by the Casorati-Weierstrass Theorem p. 109. But f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] is either C or C\{a} for one a ∈ C by the Great Picard Theorem (p. 300). In this case, f ({z : 0 < |z| < δ}) = f [ann(0; 0, δ)] = C. Exercise 2. Give the partial fraction expansion of r(z) =
z2 +1 . (z2 +z+1)(z−1)2
Solution. Not available. Exercise 3. Give the details of the derivation of (1.17) from (1.16). Solution. Not available. 1 Exercise 4. Let f (z) = z(z−1)(z−2) ; give the Laurent Expansion of f (z) in each of the following annuli: (a) ann(0; 0, 1); (b) ann(0; 1, 2); (c) ann(0; 2, ∞).
Solution. a) We have 0 < |z| < 1 and using partial fraction decomposition yields
! 1 1 1 −1 1 1 = = + f (z) = z(z − 1)(z − 2) z (z − 1)(z − 2) z z − 1 z − 2 ! ∞ ∞ 1 1 1 1 1 X n 1 X z n = − = z − z 1 − z 2 1 − 2z |z|<1and |z/2|<1 z n=0 2 n=0 2 !n+1 !n+2 ∞ ∞ ∞ X X X n 1 1 n−1 n−1 z z = = z − 1 − 2 2 n=−1 n=0 n=0 =
∞ X 1 − 2−n−2 zn ,
n=−1
(0 < |z| < 1).
b) We have 1 < |z| < 2 and thus 1 f (z) = z(z − 1)(z − 2) !n ∞ ∞ 1 X z n 1 1 X 1 1 1 1 1 1 − = − = − − z z 1 − 1z 2 1 − 2z |1/z|<1and |z/2|<1 z z n=0 z 2 n=0 2 !n+2 X !n+1 !n+1 ∞ ∞ ∞ ∞ X X X 1 1 1 n−1 −n−2 = − − z = zn−1 −z − z 2 2 n=0 n=0 n=0 n=0 !n+2 ∞ −2 X X 1 zn , (1 < |z| < 2). −zn − = 2 n=−∞ n=−1 70
c) We have 2 < |z| < ∞ and thus 1 z(z − 1)(z − 2) 1 1 1 1 1 + = − z z 1 − 1z z 1− ! ∞ ∞ X 1 n+2 X = + − 2n z n=0 n=0
f (z) =
=
−2 X
n=−∞
2 z
(2−n−2 − 1)zn ,
1 z
= |1/z|<1and |2/z|<1
!n+2
=
∞ X n=0
!n !n ∞ ∞ 1 1 X 1 1 X 2 + − z z n=0 z z n=0 z
(2n − 1)z−n−2
(2 < |z| < ∞).
Exercise 5. Show that f (z) = tan z is analytic in C except for simple poles at z = n. Determine the singular part of f at each of these poles.
π 2
+ nπ, for each integer
sin(z) Solution. Define zn = π2 + nπ. We know tan(z) = cos(z) and (cos(z))0 = − sin(z). Hence, tan(z) has a simple 1 at each zn . pole at each zn with residue −1 and with singular part − z−z n
Exercise 6. If f : G → C is analytic except for poles show that the poles of f cannot have a limit point in G. Solution. Not available. Exercise 7. Let f have an isolated singularity at z = a and suppose f (z) . 0. Show that if either (1.19) or (1.20) holds for some s in R then there is an integer m such that (1.19) holds if s > m and (1.20) holds if s < m. Solution. Let limz→a |z − a| s | f (z)| = 0 hold, that is limz→a (z − a) s f (z) = 0 for some s in R. Then it holds obviously for any greater s’s. Hence, there exist an integer n > 0 such that lim(z − a)n f (z) = 0
(5.1)
lim(z − a)n+1 f (z) = 0
(5.2)
z→a
and z→a
(In fact we are free to choose any positive integer greater than s). Now, (z − a)n f (z) thus has a removable singularity at z = a (by Theorem 1.2 p. 103) applied to (5.2)) and its extended value is also zero by (5.2). Thus (z − a)n f (z) has a zero of finite order k at z − a. So (z − a)n f (z) = (z − a)k h(z) where h(z) is analytic at a and h(a) , 0. Hence
Thus,
0, s s−n n s−n+k lim(z − a) f (z) = lim(z − a) (z − a) f (z) = lim(z − a) h(z) = ±∞, z→a z→a z→a h(a) , 0, 0, lim |z − a| | f (z)| = ∞, z→a s
71
s>m s
s>n−k s
where m = n − k. Now let limz→a |z − a| s | f (z)| = ∞ for some s in R. Then there exists an integer n < 0 so that lim |z − a|n | f (z)| = ∞
(5.3)
z→a
(In fact we are free to choose any negative integer less than s). Now (z − a)n f (z) thus has a pole at z = a (by Definition 1.3 p.105 and (5.1)) and (z − a)n f (z) has a pole of finite order l at z = a and so (z − a)n f (z) = (z − a)−l k(z) where k(z) is analytic at a and k(a) , 0 (Proposition 1.4 p.105). Hence
Thus,
0, s s−n+n s−n−l lim(z − a) f (z) = lim(z − a) f (z) = lim(z − a) k(z) = ±∞, z→a z→a z→a k(a) , 0, 0, lim |z − a| s | f (z)| = ∞, z→a
where m = n + l.
s>n+l s
s>m s
Exercise 8. Let f , a, and m be as in Exercise 7. Show: (a) m = 0 iff z = a is a removable singularity and f (a) , 0; (b) m < 0 iff z = a is a removable singularity and f has a zero at z = a of order −m; (c) m > 0 iff r = a is a pole of f of order m. Solution. a) ⇒: Let m = 0. Then by Exercise 7, we have s>0=m 0, 2 lim(z − a) f (z) = . , 0, s = 0 = m z→a
(5.4)
Thus using s = 1, f (z) has a removable singularity (by Theorem 1.2 p. 103) with the extended value equal to f (a) = limz→a f (z) , 0. ⇐: Let z = a be a removable singularity and f (a) = limz→a f (z) , 0. Then limz→a (z − a) f (z) = 0 by Theorem 1.2 p.103. By Exercise 7, we obtain that there is an integer m such that 0, s > m s lim |z − a| | f (z)| = . ∞, s < m z→a
But m = 0, since
0, lim(z − a) f (z) = ∞, z→a s
s>0 , s<0
which follows by (5.4). b) ⇒: Let the algebraic order be defined at z = a and be equal to −m (m > 0). Then by Exercise 7, we have lim f (z) = 0 z→a
lim(z − a) f (z) = 0.
and
z→a
Recall limz→a (z − a) s f (z) = 0 if s > −m so it is true of s = 0 and s = 1. Thus, f (z) has a removable singularity at z = a and also the extended function has the value limz→a f (z) = 0 at z = a (p. 103 Theorem 72
1.2). Now h(z) := (z − a)−m f (z) has a removable singularity at z = a (since limz→a (z − a)−m+1 f (z) = 0) but h(z) has extended value h(a) = lim(z − a)−m f (z) , 0 z→a
m
(by Exercise 7). Therefore f (z) = (z − a) h(z) has a zero of order m at z = a. ⇐: Assume f (z) has a removable singularity at z = a and let z = a be a zero of order m > 0. Then f (z) = (z − a)m h(z) where h(z) is analytic in |z − a| < δ, δ > 0 and h(a) , 0. Then 0, s > −m s s+m lim |z − a| | f (z)| = lim |z − a| |h(z)| = ∞, s < −m z→a z→a by Exercise 7. Hence, the algebraic order of f at z = a is equal to −m (m > 0). c) ⇒: Let the algebraic order be m > 0. Then lim(z − a) s f (z) = 0 z→a
for s > m by Exercise 7, that is lim(z − a)m+1 f (z) = 0 z→a
and lim(z − a)m f (z) , 0 z→a
by Exercise 7. Then for z , a, we have h(z) = (z − a)m f (z) has a removable singularity at z = a with h(a) , 0. Now f (z) = (z − a)−m h(z) where h ∈ A(B(a; δ)) and h(a) , 0. Thus f (z) has a pole of order m at z = a. ⇐: Let f (z) have a pole of order m at z = a. Then f (z) = (z−a)−m g(z) where g(z) ∈ A(B(a; δ)) and g(a) , 0. Thus limz→a (z − a)m f (z) = g(a) , 0 and lim(z − a)m+1 f (z) = lim(z − a)g(z) = 0. z→a
z→a
So s
lim |z − a| | f (z)| = lim |z − a| z→a
s−m
z→a
0, |z − a| | f (z)| = ∞, m
s>m s
by Exercise 7. Therefore, m > 0 is the algebraic order of f at z = a.
Exercise 9. A function f has an essential singularity at z = a iff neither (1.19) nor (1.20) holds for any real number s. Solution. ⇒: We prove this direction by proving the contrapositive (P → Q ⇐⇒ ¬Q ⇒ ¬P). Assume (1.19) or (1.20) hold for some real number s. Then by Exercise 7, there exists an integer m such that (1.19) holds if s > m and (1.20) holds if s < m. Then by Exercise 8, z = a is either a removable singularity or a pole. Hence z = a is not an essential singularity (by Definition 1.3 p.105). ⇐: Also this direction by is being proved by showing the contrapositive. Assume that f has an isolated singularity at z = a which is not an essential singularity. Then either z = a is a removable singularity or a pole (by Definition 1.3 p.105). Again by Problem 7 and 8, in either case there exists an m such that (1.19) or (1.20) holds. (If z = a is removable, then either m = 0 or m < 0; if z = a is a pole, then m > 0 by Exercise 8). Exercise 10. Suppose that f has an essential singularity at z = a. Prove the following strengthened version of the Casorati–Weierstrass Theorem. If c ∈ C and > 0 are given then for each δ > 0 there is a number α, |c − α| < , such that f (z) = α has infinitely many solutions in B(a; δ). 73
Solution. Not available. Exercise 11. Give the Laurent series development of f (z) = exp Solution. Not available.
1 z . Can you generalize this result?
Exercise 12. (a) Let λ ∈ C and show that ( !) ! ∞ X 1 1 1 exp λ z + = a0 + an zn + n 2 z z n=1 for 0 < |z| < ∞, where for n ≥ 0
1 an = π
Z
π
eλ cos t cos nt dt.
0
(b) Similarly, show exp for 0 < |z| < ∞, where
(
1 1 λ z− 2 z 1 bn = π
!)
Z
= b0 +
∞ X
bn zn +
n=1
(−1)n zn
!
π 0
cos(nt − λ sin t) dt.
Solution. Not available. Exercise 13. Let R > 0 and G = {z : |z| > R}; a function f : G → C has a removable singularity, a pole, or an essential singularity at infinity if f (z−1 ) has, respectively, a removable singularity, a pole, or an essential singularity at z = 0. If f has a pole at ∞ then the order of the pole is the order of the pole of f (z−1 ) at z = 0. (a) Prove that an entire function has a removable singularity at infinity iff it is a constant. (b) Prove that an entire function has a pole at infinity of order m iff it is a polynomial of degree m. (c) Characterize those rational functions which have a removable singularity at infinity. (d) Characterize those rational functions which have a pole of order m at infinity. Solution. Not available. Exercise 14. Let G = {z : 0 < |z| < 1} and let f : G → C beR analytic. Suppose that γ is a closed rectifiable curve in G such that n(γ; a) = 0 for all a in C − G. What is γ f ? Why? Solution. Not available.
Exercise 15. Let f be analytic in G = {z : 0 < |z − a| < r} except that there is a sequence of poles {an } in G with an → a. Show that for any ω in C there is a sequence {zn } in G with a = lim zn and ω = lim f (zn ). Solution. Not available. −1 R1 and g(z) = 0 (t − z)−1 dt are Exercise 16. Determine the regions in which the functions f (z) = sin 1z analytic. Do they have any isolated singularities? Do they have any singularities that are not isolated? Solution. Not available. ! Exercise 17. Let f be analytic in the region G =ann(a; 0, R). Show that if G | f (x + iy)|2 dx dy < ∞ then f ! has a removable singularity at z = a. Suppose that p > 0 and G | f (x + iy)| p dx dy < ∞; what can be said about the nature of the singularity at z = a? Solution. Not available. 74
5.2 Residues Exercise 1. Calculate the following integrals: a) Z
∞
0
b)
Z
∞
0
c)
Z
π
Z
π
0
Solution. Note that
Z
∞
0
2 since x4 +xx 2 +1 is even. 2 f (z) = z4 +zz 2 +1 has two
cos x − 1 dx x2
cos 2θ dθ where a2 < 1 1 − 2a cos θ + a2
0
d)
x2 dx x4 + x2 + 1
dθ where a > 1. (a + cos θ)2
x2 dx 1 = x4 + x2 + 1 2
Z
∞ −∞
x2 dx x4 + x2 + 1
Let γR = [−R, R] ∪ CR where CR = Reit , 0 ≤ t ≤ π and R 0 (draw a sketch). π 2π simple poles at a1 = ei 3 , a2 = ei 3 enclosed by γR . By the residual Theorem, we have
2πi(Res( f, a1 ) + Res( f, a2 )) =
Z
f =
Z
R
−R
γR
x2 dx + 4 x + x2 + 1
Z
CR
z4
z2 dz. + z2 + 1
Thus, Z We have Z f (z) dz = CR
=
∞ −∞
x2 dx = − lim 4 R→∞ x + x2 + 1
Z
CR
2 X z2 Res( f, ak ). dz + 2πi z4 + z2 + 1 k=1
Z Z Z z2 |z|2 |z|2 z2 dz ≤ dz |dz| ≤ |dz| ≤ 4 2 4 2 4 2 4 2 CR z + z + 1 CR |z| − |z| − 1 CR |z + z + 1| CR z + z + 1 Z R2 πR3 |dz| = 4 → 0 as R → ∞. 4 2 R − R − 1 CR R − R2 − 1
Z
Next, we calculate Res( f, a1 ). π
Res( f, ei 3 )
=
π
limπ (z − ei 3 )
z→ei 3
z4
2π π z2 1 = ei 3 limπ (z − ei 3 ) 4 2+1 i3 + z2 + 1 z + z z→e 2π
= =
π
ei 3 1 ei 3 = = π 3π 2π 3 i i i 4e 3 + 2e 3 4e 3 + 2 z→ei 3 4z + 2z √ √ 1 1 1 1 1 1 2 + 2 3i 2 + 2 3i = = √ + √ √ 1 1 4(− 2 + 2 3i) + 2 2 3i 4 3i 4 1 1 − √ i. 4 4 3 2π
= ei 3 limπ
75
Finally, we calculate Res( f, a2 ). 2π
Res( f, ei 3 )
2π
lim (z − ei 3 )
=
2π z→ei 3
z4
4π 2π z2 1 = ei 3 lim (z − ei 3 ) 4 2 2π +z +1 z + z2 + 1 z→ei 3 2π
4π
ei 3 1 ei 3 = = 4π 2π 6π 3 2π 4ei 3 + 2 4ei 3 + 2ei 3 z→ei 3 4z + 2z √ √ − 12 + 21 3i − 12 + 21 3i 1 1 = = = √ − √ √ 1 1 4 4(− 2 − 2 3i) + 2 −2 3i 4 3i 1 1 = − − √ i. 4 4 3 4π
= ei 3 lim
So
2 X
! ! 1 1 1 π π 1 1 Res( f, ak ) = 2πi 2πi = 2πi − √ i = − √ i2 = √ . − √ i+ √ − 4 4 4 3 4 3i 2 3 3 3 k=1 Hence,
Z
∞ 0
x4
x2 dx 1 π = √ . 2 +x +1 2 3
eiz −1 z2
and γ = [−R, −r] + γr + [r, R] + γR where γR = Reit , 0 ≤ t ≤ π and γr = re−it , 0 ≤ t ≤ π b) Let f (z) = R and 0 < r < R (A sketch might help to see that γ is a closed and rectifiable curve). Clearly γ f (z) dz = 0 since f is analytic on the region enclosed by γ. Also Z Z −r ix Z R ix Z Z e −1 e −1 f (z) dz = dx + dx + f+ f. x2 x2 γ −R r γr γR We have
Z
So Z
−r
−R
eix − 1 dx + x2
Z
r
R
−r −R
Z
eix − 1 dx = x2
eix − 1 dx x2
Z
R r
e−ix − 1 dx. x2
R
e−ix + e−ix − 2 dx = 2 x2 r Z R cos(x) − 1 dx = 2 x2 r
=
Z
r
−ix −ix R e +e 2 x2
Therefore, Z
cos(x) − 1 1 dx = − 2 2 x
Z
1 cos(x) − 1 dx = lim − R→∞ 2 x2
Z
r
which implies that Z
0
∞
R
Z
γR
eiz − 1 1 dz − 2 2 z
γR
! ! Z iz eiz − 1 e −1 1 dz − lim dz r→0 2 γ z2 z2 r
γr
eiz − 1 dz z2
The first term on the right hand side is zero, since Z iz ! Z Z Z 1 |eiz − 1| e − 1 iz dz ≤ |dz| ≤ 2 |dz| + |e | |dz| z2 |z|2 R γR γR γR γR 2π 1 (πR + πR) = → 0 as R → ∞. ≤ 2 R R 76
−1
dx
The second term on the right hand side is π, since ! Z iz 1 e −1 dz = − 2πi Res( f (z), 0) = −πi2 = π lim 2 r→0 2 z γr ( f (z) can be written as zi +analytic part and therefore Res( f (z), 0) = i). Hence, Z ∞ 1 π 1 cos(x) − 1 dx = − · 0 − · π = − . 2 2 2 2 x 0 c) If a = 0, then Z
0
π
cos 2θ dθ = 1 − 2a cos θ + a2
Z
π
cos 2θ dθ = 0
1 1 [sin 2θ]π0 = [0 − 0] = 0. 2 2
Assume a , 0. We can write Z π Z Z e2θi +e2θi cos 2θ dθ cos 2θ dθ 1 2π 1 2π 2 dθ. = = 2 2 eθi +eθi 2 2 1 − 2a cos θ + a 1 − 2a cos θ + a 1 − 2a 2 + a2 0 0 0 Let z = eiθ . Then dz = iz dθ =⇒ dθ = 1 2
Z
2π 0
1−
e2θi +e2θi 2 eθi +eθi 2a 2
+ a2
dθ
=
1 2
dz iz .
Hence, z2 +
Z
|z|=1
1 = − 4ia
1 z2
Z
|z|=1
Z
z4 + 1 dz 2 2 2 |z|=1 z z − az − a + a z Z 1 z4 + 1 z4 + 1 dz dz = − 2 4ia |z|=1 z2 (z − a) z − 1 z2 z2 − 1+a a z+1 a
1 dz = 1 2 iz 4i 1−a z+ z +a 2
By the Residue Theorem and the fact that |a| < 1, since a2 < 1 Z z4 + 1 dz = 2πi Res f ( f, 0) + Res( f, a) 1 |z|=1 z2 (z − a) z − a where f (z) =
z4 +1 . z2 (z−a)(z− 1a )
It is easily obtained d 2 d z4 + 1 z4 + 1 = lim Res( f, 0) = lim z z→0 dz z→0 dz (z − a) z − 1 z2 (z − a) z − a1 a 1 1 3 2 4 −4z z − az − a z + 1 − (z − 1) 2z − a − a 1 = lim =a+ 2 z→0 a z2 − az − 1a z + 1
and
Res( f, a) = lim(z − a) z→a
So
Z
|z|=1
Finally,
z4 + 1 a4 + 1 a4 + 1 z4 + 1 = lim = = . z→a z2 z − 1 a(a2 − 1) z2 (z − a) z − a1 a2 a − 1a a
! 2a4 4πia3 1 a4 + 1 z4 + 1 dz = 2πi a + + = 2πi 2 =− . 2 a a(a − 1) a(a = 1) 1 − a2 z2 (z − a) z − a1 Z
0
π
! a2 1 cos 2θ dθ 4πia3 = π = − . · − 4ia 1 − 2a cos θ + a2 1 − a2 1 − a2 77
Exercise 2. Verify the following equations: a) Z ∞
0
(x2
b)
Z
π dx = 3, 2 2 +a ) 4a ∞
0
c)
Z
d)
∞ 0
Z
(log x)3 dx = 0; 1 + x2
π(a + 1)e−a cos ax dx = , 4 (1 + x2 )2
π/2 0
Z
∞ 0
f)
∞
dx π = ; 1 + x2 2
0
h)
Z
∞ −∞
if a > 0;
π log x dx = − ; 4 (1 + x2 )2
Z Z
if a > 0;
π dθ = , 1 2 a + sin θ 2 [a(a + 1)] 2
e)
g)
a > 0;
eax π dx = , x 1+e sin aπ
2π
log sin2 2θ dθ = 4 0
Z
if
0 < a < 1;
4π 0
log sin θ dθ = −4π log 2.
1 it Solution. a) Let f (z) = (z2 +a 2 )2 and γR = [−R, R] + C R where C R = Re , 0 ≤ t ≤ π. So f has z1 = ai as a pole of order 2 enclosed by γR . So we have by the Residue Theorem Z f (z) dz = 2πi Res( f, z1 ) γR
where Res( f, z1 )
= =
# # " " d d 1 1 = lim (z − ai)2 z→ai dz z→ai dz (z + ai)2 (z − ai)2 (z + ai)2 −2 2 1 lim =− = 3 . 3 3 z→ai (z + ai) (2ai) 4a i lim
Also 2πi
π 1 = = 4a3 i 2a3
Z
R
−R
dx + (x2 + a2 )2
Z
f (z) dz
CR
where the latter integral on the right hand side is zero since Z Z Z Z 1 1 1 1 dz ≤ |dz| ≤ |dz| ≤ 2 |dz| 2 2 2 2 22 2 2 2 (R − a2 )2 CR CR |z + a | CR (|z| − |a| ) CR (z + a ) πR → 0 as R → ∞. = (R2 − a2 )2 78
Hence,
Z
∞
0
eiaz c) Let f (z) = (1+z 2 )2 enclosed by γR . By
1 dx = (x2 + a2 )2 2
Z
∞ −∞
1 dx = lim (x2 + a2 )2 2 R→∞
Z
R
−R
1 π π dx = = . (x2 + a2 )2 2 2a3 4a3
it
and γR = [−R, R] + CR where CR = Re , 0 ≤ t ≤ π. So f has z1 = i as a pole of order 2 the Residue Theorem, we have Z f (z) dz = 2π Res( f, z1 ) γR
where Res( f, z1 )
# " " iaz # d d eiaz e 2 = lim = lim (z − i) 2 2 z→i dz z→i dz (z + i)2 (z − i) (z + i) " # # " 2 iaz iaz (z + i)iaeiaz − 2eiaz (z + i) iae − 2(z + i)e = lim = lim z→i z→i (z + i)4 (z + i)3 2
Also 2πi
2
2i2 aei a − 2ei (2i)3
=
a
=
−2ae−a − 2e−a (a + 1)e−a = . −8i 4i
(a + 1)e−a π(a + 1)e−a = = 4i 2
Z
R
f (x) dx +
−R
Z
f (z) dz
CR
where the latter integral on the right hand side is zero since Z Z Z Z eiaz |eiaz | 1 ≤ f (z) dz = dz |dz| ≤ |dz| 2 2 2 2 2 2 CR (1 + z ) CR |1 + z | CR CR (|z| − 1) 1 = Rπ → 0 as R → ∞, (R2 − 1)2 since |eiaz | = e−Im(az) ≤ 1 since Im(az) ≥ 0 (a > 0 by assumption). Hence Z Z R Z ∞ cos ax 1 ∞ cos ax 1 1 π(a + 1)e−a π(a + 1)e−a cos ax lim = dx = dx = dx = 2 2 2 2 2 2 2 −∞ (1 + x ) 2 R→∞ −R (1 + x ) 2 2 4 0 (1 + x ) provided a > 0. g) First of all substitute u = e x , so Z ∞
eax dx = 1 + ex
Z
∞
ua−1 du = 2 1+u
Z
∞
−∞
Now, let u = t2 , then
Hence,
Z
∞ 0
Z
∞
−∞
0
0
ua du = 1+u u
Z
t2a−2 t dt = 2 1 + t2
eax dx = 2 1 + ex
2a−1
Z
∞ 0
∞
0
Z
ua−1 du. 1+u
∞ 0
t2a−1 dt. 1 + t2
x2a−1 dx. 1 + x2
Let f (z) = z1+z2 and γ = [−R, −r] + γr + [r, R] + γR (0 < r < R). Define log z on G = {z ∈ C : z , 0, − π2 < arg(z) < 3π 2 } with log(z) = log |z| + iθ, θ = arg(z). f (z) has a simple pole i enclosed by γ. By the Residue 79
Theorem Z
= 2πRes( f, i) = 2πi lim(z − i)
f (z) dz
z→i
γ
= πi Also
Z
f (z) dz =
γ
2a−1
Z
Z
IV)
−r
π
= πe
Z
Z
−r
π
= πe(2a−1)(0+i 2 ) = πeaπi e−i 2 = −πieaπi .
(2a−1) log i
Z
R
f (x) dx + f (x) dx + f (z) dz + f (z) dz . |−R {z } |γr {z } |r {z } |γR {z } I
I)
i2a−1 z2a−1 = 2πi (z − i)(z + i) 2i
Z
III
II
IV
Z R Z R (2a−1)(log |−x|+iπ) (−x)2a−1 e f (−x) dx = dx = dx 2 1 + x 1 + x2 r r r Z R (2a−1) log x Z R (2a−1) log x e e 2aπi dx = −e dx = e(2a−1)iπ 2 1 + x 1 + x2 r r Z R 2a−1 x = −e2aπi dx. 1 + x2 r
f (x) dx
=
−R
R
Z Z Z z2a−1 |z|2a−1 R2a−1 πR2a f (z) dz = ≤ dz |dz| ≤ πR = → 0 as R → ∞ 2 γR 1 + z2 R2 − 1 R2 − 1 γR γR |1 + z |
since a < 1. II) Similar to IV)
Z πr2a f (z) dz ≤ 2 → 0 as r → 0 r −1 γr
since a > 0. Hence, if r → 0 and R → ∞, we get Z Z aπi 2aπi −πie = f (z) dz = −e γ
∞
f (x) dx + 0 +
0
Z
0
Thus, aπi
−πie and so
Z
∞
f (x) dx =
0
Therefore,
Z
∞
−∞
∞
2aπi
= (1 − e
)
Z
2aπi
f (x) dx + 0 = (1 − e
∞
)
Z
∞
f (x) dx.
0
f (x) dx
0
−iπ iπ iπ π −πieaπi = = = = . 1 − e2aπi e−aπi − eaπi eaπi − e−aπi 2i sin(aπ) 2 sin(aπ) eax dx = 2 1 + ex
Z
∞ 0
π π x2a−1 dx = 2 = 2 sin(aπ) sin(aπ) 1 + x2
provided 0 < a < 1. Exercise 3. Find all possible values of Solution. Not available.
R
γ
exp z−1 dz where γ is any closed curve not passing through z = 0.
80
Exercise 4. Suppose that f has a simple pole at z = a and let g be analytic in an open set containing a. Show that Res( f g; a) = g(a)Res( f ; a). Solution. Not available. Exercise 5. Use Exercise 4 to show that if G is a region and f is analytic in G except for simple poles at a1 , . . . , an ; and if g is analytic in G then 1 2πi
Z
fg = γ
n X
n(γ; ak )g(ak )Res( f ; ak )
k=1
for any closed rectifiable curve γ not passing through a1 , . . . , an such that γ ≈ 0 in G. Solution. Not available. Exercise 6. Let γ be Rthe rectangular path [n + 21 + ni, −n − 12 + ni, −n − 21 − ni, nR + 12 − ni, n + i + ni] and evaluate the integral γ π(z + a)−2 cot πz dz for a , an integer. Show that limn→∞ γ π(z + a)−2 cot πz dz = 0 and, by using the first part, deduce that ∞ X π2 1 = 2 sin πa n=−∞ (a + n)2
(Hint: Use the fact that for z = x + iy, | cos z|2 = cos2 x + sinh2 y and | sin z|2 = sin2 x + sinh2 y to show that | cot πz| ≤ 2 for z on γ if n is sufficiently large.) Solution. Not available. Exercise 7. Use Exercise 6 to deduce that ∞
1 π2 X = 8 (2n + 1)2 n=0 Solution. Not available. Exercise 8. Let γ be the polygonal path defined in Exercise 6 and evaluate an integer. Show that limn→∞ π(z2 − a2 )−1 cot πz dz = 0, and consequently π cot πa =
∞
1 X 2a + a n=1 a2 − n2
for a , an integer. Solution. Not available. Exercise 9. Use methods similar to those of Exercises 6 and 8 to show that ∞
1 X 2(−1)n a π = + sin πa a n=1 a2 − n2 for a , an integer. Solution. Not available. 81
R
γ
π(z2 − a2 )−1 cot πz dz for a ,
Exercise 10. Let γ be the circle |z| = 1 and let m and n be non-negative integers. Show that (±1) p (n+2p)! Z 2 (z ± 1)m dz 1 p!(n+p)! , if m = 2p + n, p ≥ 0 = m+n+1 0, 2πi γ z otherwise Solution. Not available.
Exercise 11. In Exercise 1.12, consider an and bn as functions of the parameter λ and use Exercise 10 to compute power series expansions for an (λ) and bn (λ). (bn (λ) is called a Bessel function.) Solution. Not available. Exercise 12. Let f be analytic in the plane except for isolated singularities at a1 , a2 , . . . , am . Show that Res( f ; ∞) = −
∞ X
Res( f, ak ).
k=1
R 1 f when (Res( f ; ∞) is defined as the residue of −z−2 f (z−1 ) at z = 0. Equivalently, Res( f ; ∞) = − 2πi γ(t) = Reit , 0 ≤ t ≤ 2π, for sufficiently large R.) What can you say if f has infinitely many isolated singularities? Solution. Not available. it Exercise 13. Let f be an R entire function−1and let a, b ∈ C such that |a| < R and |b| < R. If γ(t) = Re , 0 ≤ t ≤ 2pi, evaluate γ [(z − a)(z − b)] f (z) dz. Use this result to give another proof of Liouville’s Theorem.
Solution. Not available.
5.3 The Argument Principle Exercise 1. Prove Theorem 3.6. Solution. Not available. ¯ 1) and satisfies | f (z)| < 1 for |z| = l. Find the number of solutions Exercise 2. Suppose f is analytic on B(0; (counting multiplicities) of the equation f (z) = zn where n is an integer larger than or equal to 1. Solution. Not available. ¯ R) with f (0) = 0, f 0 (0) , 0 and f (z) , 0 for 0 < |z| ≤ R. Put Exercise 3. Let f be analytic in B(0; ρ = min{| f (z)| : |z| = R} > 0. Define g : B(0; ρ) → C by Z 1 z f 0 (z) g(ω) = dz 2πi γ f (z) − ω where γ is the circle |z| = R. Show that g is analytic and discuss the properties of g. Solution. Not available. Exercise 4. If f is meromorphic on G and f˜ : G → C∞ is defined by f˜(z) = ∞ when z is a pole of f and f˜(z) = f (z) otherwise, show that f˜ is continuous. 82
Solution. Not available. Exercise 5. Let f be meromorphic on the region G and not constant; show that neither the poles nor the zeros of f have a limit point in G. Solution. Not available. Exercise 6. Let G be a region and let H(G) denote the set of all analytic functions on G. (The letter “H” stands for holomorphic. Some authors call a differentiable function holomorphic and call functions analytic if they have a power series expansion about each point of their domain. Others reserve the term “analytic” for what many call the complete analytic function, which we will not describe here.) Show that H(G) is an integral domain; that is, H(G) is a commutative ring with no zero divisors. Show that M(G), the meromorphic functions on G, is a field. We have said that analytic functions are like polynomials; similarly, meromorphic functions are analogues of rational functions. The question arises, is every meromorphic function on G the quotient of two analytic functions on G? Alternately, is M(G) the quotient field of H(G)? The answer is yes but some additional theory will be required before this answer can be proved. Solution. Not available. Exercise 7. State and prove a more general version of Rouche’s Theorem for curves other than circles in G. Solution. Not available. Exercise 8. Is a non-constant meromorphic function on a region G an open mapping of G into C? Is it an open mapping of G into C∞ ? Solution. Not available. Exercise 9. Let λ > 1 and show that the equation λ − z − e−z = 0 has exactly one solution in the half plane {z : Re z > 0}. Show that this solution must be real. What happens to the solution as λ → 1? Solution. Not available. ¯ 1). If | f (z)| < 1 for |z| = 1, show that there is Exercise 10. Let f be analytic in a neighborhood of D = B(0; a unique z with |z| < 1 and f (z) = z. If | f (z)| ≤ 1 for |z| = 1, what can you say? ¯ 1) since f is Solution. Let g(z) = f (z) − z and h(z) = z. Clearly g, h are analytic in a neighborhood of B(0, analytic. Note that g, h have no poles. Let γ = {z : |z| = 1}. We have |g(z) + h(z)| = | f (z) − z + z| = | f (z)| < 1 = |z| = |h(z)| ≤ |g(z)| + |h(z)| on γ since | f (z)| < 1 for |z| = 1. By Roché’s Theorem, we get Zg − Pg = Zh − Ph . But Pg = Ph = 0, so Zg = Zh
(inside the unit circle).
Since h(z) = z has only one zero (z = 0) inside the unit circle, we get that g(z) = f (z) − z has exactly one zero inside the unit circle. Hence, there is a unique z such that f (z) − z = 0 ⇐⇒ f (z) = z with |z| < 1. If we choose f (z) = z, then | f (z)| ≤ 1 for |z| = 1 and therefore we have infinitely many fixed points. If we choose f (z) = 1, then | f (z)| ≤ 1 for |z| = 1 and therefore we do not have any fixed points. 83
Chapter 6
The Maximum Modulus Theorem 6.1 The Maximum Principle Exercise 1. Prove the following Minimum Principle. If f is a non-constant analytic function on a bounded open set G and is continuous on G− , then either f has a zero in G or | f | assumes its minimum value on ∂G. (See Exercise IV. 3.6.) ¯ we have | f | ∈ C(G). ¯ Hence ∃ a ∈ G¯ such that | f (a)| ≤ | f (z)| ∀z ∈ G. ¯ Solution. Since f ∈ C(G) If a ∈ ∂G, then | f | assumes its minimum value on ∂G and we are done. Otherwise, if a < ∂G, then a ∈ G S and we can write G = Ai where Ai are the components of G, that is a ∈ Ai for some i. But each Ai is a region, so we can use Exercise IV 3.6 which yields either f (a) = 0 or f is constant. But f was assumed to be non-constant, so f has to have a zero in G. Therefore, either f has a zero in G or | f | assumes its minimum value on ∂G. Exercise 2. Let G be a bounded region and suppose f is continuous on G− and analytic on G. Show that if there is a constant c ≥ 0 such that | f (z)| = c for all z on the boundary of G then either f is a constant function or f has a zero in G. Solution. Assume there is a constant c ≥ 0 such that | f (z)| = c for all z ∈ ∂G. According to the Maximum Modulus Theorem (Version 2), we get ¯ = max{| f (z)| : z ∈ ∂G} = c. max{| f (z)| : z ∈ G} So ¯ ∀z ∈ G.
| f (z)| ≤ c,
(6.1)
¯ and hence there exists an a ∈ G such that Since f ∈ C(G) which implies | f | ∈ C(G) | f (a)| ≤ | f (z)| ≤ c, (6.1) then by Exercise IV 3.6 either f is constant of f has a zero in G. Exercise 3. (a) Let f be entire and non-constant. For any positive real number c show that the closure of {z : | f (z)| < c} is the set {z : | f (z)| ≤ c}. (b) Let p be a polynomial and show that each component of {z : |p(z)| < c} contains a zero of p. (Hint: Use 84
Exercise 2.) (c) If p is a polynomial and c > 0 show that {z : |p(z)| = c} is the union of a finite number of closed paths. Discuss the behavior of these paths as c → ∞. Solution. Not available. Exercise 4. Let 0 < r < R and put A = {z : r ≤ |z| ≤ R}. Show that there is a positive number > 0 such that for each polynomial p, sup{|p(z) − z−1 | : z ∈ A} ≥ This says that z−1 is not the uniform limit of polynomials on A. Solution. Not available. ¯ R) with | f (z)| ≤ M for |z| ≤ R and | f (0)| = a > 0. Show that the Exercise 5. Let f be analytic on B(0; number of zeros of f in B(0; 13 R) is less than or equal to log1 2 log Ma . Hint: If z1 , . . . , zn are the zeros of f in B(0; 13 R); consider the function n !−1 Y z , 1− g(z) = f (z) z k
k=1
and note that g(0) = f (0). (Notation: Solution. Not available.
Qn
k=1
ak = a1 a2 . . . an .)
¯ R) with | f (z)| = |g(z)| for |z| = R. Show that if Exercise 6. Suppose that both f and g are analytic on B(0; neither f nor g vanishes in B(0; R) then there is a constant λ, |λ| = 1, such that f = λg. Solution. Not available. Exercise 7. Let f be analytic in the disk B(0; R) and for 0 ≤ r < R define A(r) = max{Re f (z) : |z| = r}. Show that unless f is a constant, A(r) is a strictly increasing function of r. Solution. Not available. Exercise 8. Suppose G is a region, f : G → C is analytic, and M is a constant such that whenever z is on ∂∞G and {zn } is a sequence in G with z = lim zn we have lim sup | f (zn )| ≤ M. Show that | f (z)| ≤ M, for each z in G. Solution. We need to show lim sup | f (z)| ≤ M z→a
∀u ∈ ∂∞G.
Then we can use the Maximum Modulus Theorem (Version 3). Instead of showing that lim sup | f (zn )| ≤ M ⇒ lim sup | f (z)| ≤ M, z→a
we show the contrapositive, that is lim sup | f (z)| > M ⇒ lim sup | f (zn )| > M. z→a
So assume lim supz→a | f (z)| > M. But this implies lim supz→a | f (zn )| > M since zn → z as z → ∞, that is zn gets arbitrarily close to z and since f is analytic, that is continuous, we have f (zn ) gets arbitrarily close to f (z). 85
6.2 Schwarz’s Lemma Exercise 1. Suppose | f (z)| ≤ 1 for |z| < 1 and f is a non-constant analytic function. By considering the function g : D → D defined by f (z) − a g(z) = 1 − a¯ f (z) where a = f (0), prove that
| f (0)| + |z| | f (0)| − |z| ≤ | f (z)| ≤ 1 + | f (0)| |z| 1 − | f (0)| |z|
for |z| < 1.
Solution. We claim that |g(z)| ≤ 1 and g(0) = 0. It is easy to check that g(0) = 0 since g(0) =
f (0) − a a−a = = 0. 1 − a¯ f (0) 1 − a¯ a
We also have |g(z)| =
| f (z) − a| ≤1 |1 − a¯ f (z)|
since | f (z)| ≤ 1. (Can you see it?) Now, apply Schwarz Lemma. ⇐⇒ ⇒ ⇒ ⇒
⇒
|g(z)| ≤ |z| for |z| < 1 | f (z) − a| ≤ |z|· |1 − a¯ f (z)|
|| f (z)| − |a|| ≤ | f (z) − a| ≤ |z|· |1 − a¯ f (z)| ≤ |z| + |z|· |a|· | f (z)| || f (z)| − |a|| ≤ |z| + |z|· |a|· | f (z)| −|z| − |z|· |a|· | f (z)| ≤ | f (z)| − |a| ≤ |z| + |z|· |a|· | f (z)| |a| − |z| |a| + |z| ≤ | f (z)| ≤ . 1 + |a| |z| 1 − |a| |z|
Set a = f (0) to obtain the desired result. Exercise 2. Does there exist an analytic function f : D → D with f ( 21 ) =
3 4
and f 0 ( 21 ) = 32 ?
Solution. The answer is no. Assume there exist an analytic function f : D → D with f ( 21 ) = 34 . According to the definition, we have Eαa = { f ∈ A(D) : | f (z)| ≤ 1, f (a) = α}. In our case, we have a =
1 2
and α = 34 . We know,
0 1 − |α|2 f (a) ≤ 1 − |a|2
and therefore we must have
But f 0 ( 12 ) =
2 3
2 ! 3 0 1 1 − 4 ≤ f 2 = 2 1 − 12
7 16 3 4
= 0.6¯ which is not possible.
86
=
7 3 7 ¯ = = 0.583. 16 4 12
Exercise 3. Suppose f : D → C satisfies Re f (z) ≥ 0 for all z in D and suppose that f is analytic. (a) Show that Re f (z) > 0 for all z in D. (b) By using an appropriate Möbius transformation, apply Schwarz’s Lemma to prove that if f (0) = 1 then | f (z)| ≤
1 + |z| 1 − |z|
| f (z)| ≥
1 − |z| 1 + |z|
for |z| < 1. What can be said if f (0) , 1? (c) Show that f (0) = 1, f also satisfies
(Hint: Use part (a)). Solution. Not available.
¯ R) Exercise 4. Prove Caratheodory’s Inequality whose statement is as follows: Let f be analytic on B(0; and let M(r) = max{| f (z)| : |z| = r}, A(r) = max{Re f (z) : |z| = r}; then for 0 < r < R, if A(r) ≥ 0, M(r) ≤
R+r [A(R) + | f (0)|] R−r
(Hint: First consider the case where f (0) = 0 and examine the function g(z) = f (Rz)[2A(R) + f (Rz)]−1 for |z| < 1.) Solution. Not available. Exercise 5. Let f be analytic in D = {z : |z| < 1} and suppose that | f (z)| ≤ M for all z in D. (a) If f (zk ) = 0 for 1 ≤ k ≤ n show that n Y |z − zk | | f (z)| ≤ M |1 − z¯k z| k=1 for |z| < 1. (b) If f (zk ) = 0 for 1 ≤ k ≤ n, each zk , 0, and f (0) = Meia (z1 z2 , . . . , zn ), find a formula for f . Solution. Not available. ¯ 1) and | f (z)| = 1 where |z| = 1. Find a Exercise 6. Suppose f is analytic in some region containing B(0; formula for f . (Hint: First consider the case where f has no zeros in B(0; 1).) Solution. Not available. ¯ 1) and | f (z)| = 1 when |z| = 1. Suppose that Exercise 7. Suppose f is analytic in a region containing B(0; f has a zero at z = 41 (1 + i) and a double zero at z = 12 . Can f (0) = 21 ? Solution. Not available. Exercise 8. Is there an analytic function f on B(0; 1) such that | f (z)| < 1 for |z| < 1, f (0) = f 0 (0) = 34 ? If so, find such an f . Is it unique? Solution. Not available.
87
1 2,
and
6.3 Convex functions and Hadamard’s Three Circles Theorem Exercise 1. Let f : [a, b] → R and suppose that f (x) > 0 for all x and that f has a continuous second derivative. Show that f is logarithmically convex iff f 00 (x) f (x) − [ f 0 (x)]2 ≥ 0 for all x. Solution. Not available. Exercise 2. Show that if f : (a, b) → R is convex then f is continuous. Does this remain true if f is defined on the closed interval [a, b]? Solution. Since f is convex on (a, b) and if a < s < t < u < b, we have f (u) − f (s) f (u) − f (t) f (t) − f (s) ≤ ≤ t−s u−s u−t
(6.2)
t f (u) − s f (u) + u f (s) − t f (s) − u f (t) + s f (t) ≥ 0.
(6.3)
u−t ∈ (0, 1). Since f is convex on which we show now. Let a < s < t < u < b, t = λs + (1 − λ)u with λ = u−s (a, b), we get u−t t−s f (t) ≤ f (s) + f (u) u−s u−s which implies (t − s) f (u) + (u − t) f (s) − (u − s) f (t) ≥ 0
which is equivalent to
From (6.3), we get (rearrange terms and add/subtract a term)
⇐⇒ ⇐⇒
t f (u) − t f (s) − s f (u) + s f (s) − u f (t) − u f (s) + s f (t) − s f (s) ≥ 0
(t − s)( f (u) − f (s)) − (u − s)( f (t) − f (s)) ≥ 0 f (t) − f (s) f (u) − f (s) ≤ . t−s u−s
Similar, from (6.3), we get
⇐⇒ ⇐⇒
u f (u) − s f (u) − u f (t) + s f (t) − u f (u) + u f (s) + t f (u) − t f (s) ≥ 0 (u − s)( f (u) − f (t)) − (u − t)( f (u) − f (s)) ≥ 0 f (u) − f (s) f (u) − f (t) ≤ . u−s u−t
Thus, we have proved (6.2). Given x ∈ (a, b) choose δ > 0 such that [x − δ, x + δ] ⊂ (a, b). Claim: f (x) − f (x − δ) f (z) − f (x) f (x + δ) − f (x) ≤ ≤ δ z−x δ
∀z ∈ (x − δ, x + δ).
(6.4)
If the claim is true, (6.4) is equivalent to f (x + δ) − f (x) f (x) − f (x − δ) (z − x) ≤ f (z) − f (x) ≤ (z − x). δ δ f (x) Taking the limit z → x, we have that f (x)−δf (x−δ) (z − x) → 0 and f (x+δ)− (z − x) → 0. Thus f (z) − f (x) → 0, δ that is |z − x| < δ ⇒ | f (z) − f (x)| < which shows that f is continuous.
88
Thus it remains to show the claim. The proof uses (6.2). First, consider a point z ∈ (x − δ, x) and apply the second inequality in (6.2) gives f (x) − f (x − δ) f (z) − f (x) ≤ δ z−x which gives the first inequality in (6.4). Applying the outer inequality in (6.2) to the three points z < x < x + δ gives f (x) − f (z) f (x + δ) − f (x) f (z) − f (x) f (x + δ) − f (x) ≤ ⇐⇒ ≤ x−z x+δ−x z−x δ which gives the second inequality in (6.4). Now, consider the case x < z < x + δ. Then, the first inequality in (6.4) follows from the outer inequality in (6.2) applied to the three points x − δ, x, z. f (x) − f (x − δ) f (z) − f (x) f (x) − f (x − δ) f (z) − f (x) ≤ ⇐⇒ ≤ . x−x+δ z−x δ z−x The second inequality in (6.4) follows from the first inequality in (6.2) applied to x, z, x + δ. f (z) − f (x) f (x + δ) − f (x) f (z) − f (x) f (x + δ) − f (x) ≤ ⇐⇒ ≤ . z−x x+δ−x z−x δ Thus, we have proved the claim. The statement is not true if f is defined on the closed interval [a, b]! Here is a counterexample: Define f : [a, b] → R by 0, x ∈ [a, b) . f (x) = 1, x = b
Clearly, f is convex on [a, b], but f is not continuous at x = b (but the function is continuous on (a, b)).
Exercise 3. Show that a function f : [a, b] → R is convex iff any of the following equivalent conditions is satisfied: f (u) u 1 (a) a ≤ x < u < y ≤ b gives det f (x) x 1 ≥ 0; f (y) y 1 f (y)− f (x) f (u)− f (x) (b) a ≤ x < u < y ≤ b gives u−x ≤ y−x ; f (x) f (u) (c) a ≤ x < u < y ≤ b gives f (u)− ≤ f (y)− u−x y−u ; Interpret these conditions geometrically.
Solution. Not available. Exercise 4. Supply the details in the proof of Hadamard’s Three Circle Theorem. Solution. Not available. Exercise 5. Give necessary and sufficient conditions on the function f such that equality occurs in the conclusion of Hadamard’s Three Circle Theorem. Solution. Not available.
89
¯ R) and not constant then Exercise 6. Prove Hardy’s Theorem: If f is analytic on B(0; Z 2π 1 | f (reiθ )| dθ I(r) = 2π 0 is strictly increasing and log I(r) is a convex function of log r. Hint: If 0 < r1 < r < r2 find a continuous function ϕ : [0, 2π] → C such that ϕ(θ) f (reiθ ) = | f (reiθ )| and consider the function F(z) = R 2π f (zeiθ )ϕ(θ) dθ. (Note that r is fixed, so ϕ may depend on r.) 0
Solution. Not available.
Exercise 7. Let f be analytic in ann(0; R1 , R2 ) and not identically zero; define Z 2π 1 | f (reiθ )|2 dθ. I2 (r) = 2π 0 Show that log I2 (r) is a convex function of log r, R1 < r < R2 . Solution. Not available.
6.4 The Phragmén-Lindelöf Theorem Exercise 1. In the statement of the Phragmén-Lindelöf Theorem, the requirement that G be simply connected is not necessary. Extend Theorem 4.1 to regions G with the property that for each z in ∂∞G there is a sphere V in C∞ centered at z such that V ∩ G is simply connected. Give some examples of regions that are not simply connected but have this property and some which don’t. Solution. Not available. Exercise 2. In Theorem 4.1 suppose there are bounded analytic functions ϕ1 , ϕ2 , . . . , ϕn on G that never vanish and ∂∞G = A ∪ B1 ∪ . . . ∪ Bn such that condition (a) is satisfied and condition (b) is also satisfied for each ϕk and Bk . Prove that | f (z)| ≤ M for all z in G. Solution. We have ϕ1 , . . . , ϕn ∈ A(G), bounded and ϕk . 0 by assumption. Hence, |ϕk (z)| ≤ κk ∀ z in G for each k = 1, . . . , n. Further a) for every a in A, lim supz→a | f (z)| ≤ M and b) for every b in Bk and ηk > 0, lim supz→b | f (z)||ϕk (z)|ηk ≤ M ∀ k = 1, . . . , n. Also because G is simply connected, there is an analytic branch of log ϕk (z) on G for each k = 1, . . . , n (Corollary IV 6.17). Hence, gk (z) = exp{ηk log ϕk (z)} is an analytic branch of ϕk (z)ηk for ηk > 0 and |gk (z)| = |ϕk (z)|ηk ∀ k = 1, . . . , n. Define F : G → C by F(z) = f (z)
n Y
gk (z)κk−ηk ;
k=1
then F is analytic on G and |F(z)| ≤ | f (z)| Hence,
n Y k=1
|ϕk (z)|ηk κk−ηk
≤
|ϕk (z)|≤κk ∀k
| f (z)|
n Y k=1
M, lim sup | f (z)| = Mκ−ηk , (a),(b) z→a k 90
κkηk κk−ηk |ϕk (z)| = | f (z)| ∀z ∈ G.
a∈A a ∈ Bk
∀k
.
By the Maximum Modulus Theorem III, we have |F(z)| ≤ max{M, Mκ1−η1 , . . . , Mκn−ηn } ∀z ∈ G. Thus, | f (z)| = |F(z)|
n Y k=1
|gk (z)|−1 κkηk ≤ +
η n Y κk k max{M, Mκ−η1 , . . . , Mκ−ηn } n 1 ϕk (z) k=1
for all z in G and for all ηk > 0. Letting ηk → 0 ∀k = 1, . . . , n gives that | f (z)| ≤ M for all z in G. Exercise 3. Let G = {z : |Im z| < 12 π} and suppose f : G → C is analytic and lim supz→w | f (z)| ≤ M for w in ∂G. Also, suppose A < ∞ and a < 1 can be found such that | f (z)| < exp[A exp(a|Re z|)] for all z in G. Show that | f (z)| ≤ M for all z in G. Examine exp(exp z) to see that this is the best possible growth condition. Can we take a = 1 above? Solution. Not available. Exercise 4. Let f : G → C be analytic and suppose M is a constant such that lim sup | f (zn )| ≤ M for each sequence {zn } in G which converges to a point in ∂∞G. Show that | f (z)| ≤ M. (See Exercise 1.8). Solution. Claim: Let M be a constant such that lim supn→∞ | f (zn )| ≤ M for each sequence {zn } in G with z = lim zn , n→∞
∀z ∈ ∂∞G
(6.5)
implies lim sup | f (z)| ≤ M
for all
z→a
a ∈ ∂∞G.
(6.6)
If G would be a region together with the claim would give | f (z)| ≤ M ∀z ∈ G by the Maximum Modulus Theorem III. However, G is assumed to be an open set. But every open set can be written as a union of components (components are open and connected). So each component is a region. Therefore, we can apply the Maximum Modulus Theorem III to each component once we have shown the claim and additionally we have to show that the boundary of G is the same as taking the boundaries of the components: Clearly, since the components of G add up to G, we have that the boundary of the components must include the boundary of G. So it remains to show that the boundary of the component does not include more than the boundary of G. Assume it would be the case. Then there needs to be a boundary of a component lying inside G (since every component is lying inside G). But this would imply that it must be connected to another component of G contradicting the fact a component is a maximally connected subset. Hence, the boundary of G equals the boundary of all its components. It remains to show the claim which we will do by showing the contrapositive. Proof of the claim: Assume not (6.6), that is ∀M > 0 ∃a ∈ ∂∞G : lim sup | f (z)| > M z→a
which is equivalent to ∀M > 0 ∃a ∈ ∂∞G : lim+ sup{| f (z)| : z ∈ G ∩ B(a; r)} > M r→0
91
by definition. Then, clearly ∃δ > 0 such that limr→0+ sup{| f (z)| : z ∈ G ∩ B(a; r)} > M + δ. Let {zn } be a sequence with limn→∞ = a and let rn = 2|zn − a|. Obviously, we have limn→∞ rn = 0. Next, consider the sequence αn = sup{| f (z)| : z ∈ F ∩ B(a; rn )}. Then limn→∞ αn > M + δ. In addition, {αn } is a nonincreasing sequence, therefore limn→∞ αn > M + δ implies αn > M + δ ∀n. Thus, ∃ yn ∈ G ∩ B(a; rn ) such that | f (yn )| > M + δ. Taking lim sup in this inequality yields lim sup | f (yn )| ≥ M + δ, n→∞
that is lim sup | f (yn )| ≥ M n→∞
where limn→∞ yn = a, which gives “not (6.5)”. Hence (6.5) implies (6.6). Exercise 5. Let f : G → C be analytic and suppose that G is bounded. Fix z0 in ∂G and suppose that lim supz→w | f (z)| ≤ M for w in ∂G, w , z0 . Show that if limz→z0 |z − z0 | | f (z)| = 0 for every > 0 then | f (z)| ≤ M for every z in ∂G. (Hint: If a < G, consider ϕ(z) = (z − z0 )(z − a)−1 .) Solution. Not available. Exercise 6. Let G = {z : Re z > 0} and let f : G → C be an analytic function with lim supz→w | f (z)| ≤ M for w in ∂G, and also suppose that for every > 0, lim sup{exp(−/r)| f (reiθ )| : |θ| <
r→∞
1 π} = 0. 2
Show that | f (z)| ≤ M for all z in G. Solution. Not available. Exercise 7. Let G = {z : Re z > 0} and let f : G → C be analytic such that f (1) = 0 and such that lim supz→w | f (z)| ≤ M for w in ∂G. Also, suppose that for every δ, 0 < δ < 1, there is a constant P such that | f (z)| ≤ P exp(|z|1−δ ). Prove that | f (z)| ≤ M Hint: Consider f (z)
1+z 1−z
"
(1 − x)2 + y2 (1 + x)2 + y2
.
Solution. Not available.
92
# 12
.
Chapter 7
Compactness and Convergence in the Space of Analytic Functions 7.1 The space of continuous functions C(G, Ω) Exercise 1. Prove Lemma 1.5 (Hint: Study the function f (t) =
t 1+t
for t > −1.)
d(s,t) Solution. To show that µ : S × S → R, µ(s, t) = 1+d(s,t) is a metric we proof that µ satisfies the conditions of a metric function. d(s,t) is well defined in the nonnegative real numbers because d is a metric. For s, t ∈ S the value µ(s, t) = 1+d(s,t) Also µ(s, t) = 0 if and only if d(s, t) = 0 and since d is a metric it follows that s = t. The metric d is symmetric and therefore µ is also. It remains to show the triangle inequality. Let therefore s, t, u ∈ S . The metric d satisfies the triangle inequality
d(s, u) ≤ d(s, t) + d(t, u).
(7.1)
t The function f (t) = 1+t is a strictly increasing, concave function on the interval [0, ∞). Thus if d(s, u) ≤ max{d(s, t), d(t, u)}, then also µ(s, u) ≤ max{µ(s, t), µ(t, u)} and by nonnegativity of µ also
µ(s, u) ≤ µ(s, t) + µ(t, u). If instead max{d(s, t), d(t, u)} < d(s, u) then ) ( 1 1 1 . < min , 1 + d(s, u) 1 + d(s, t) 1 + d(t, u) Together with equation (7.1) we have d(s, t) + d(t, u) d(s, u) ≤ 1 + d(s, u) 1 + d(s, u) d(s, t) d(t, u) d(s, t) d(t, u) = + ≤ + 1 + d(s, u) 1 + d(s, u) 1 + d(s, t) 1 + d(t, u) = µ(s, t) + µ(t, u),
µ(s, u) =
which gives the triangle inequality. To show that the metrics d and µ are equivalent on S , let O be on open set in (S , d) and let x ∈ O. There is 93
ε , then Bµ (x, δ) ⊂ Bd (x, ε). Similarly, if ε ∈ (0, 1) such that Bd (x, ε) ⊂ O. Choose δ positive such that δ < 1+ε δ O is an open set in (S , µ) and δ is such that x ∈ Bµ (x, δ) ⊂ O then choose ε > 0 with ε ≤ 1−δ which implies Bd (x, ε) ⊂ Bµ (x, δ). Since this can be done for any element x ∈ O, open sets in (S , d) are open in (S , µ) and vice-versa. The fact that exactly the open sets in (S , d) are open in (S , µ) leads to the statement about Cauchy sequences. Assume that {xn }n is a Cauchy sequence in (S , d). To see that it is Cauchy in (S , µ), let ε > 0 be arbitrary but fixed. By the above there is a δ small enough that Bd (x, δ) ⊂ Bµ (x, ε) and δ depends only on ε, but not on x ∈ S . Since {xn }n is Cauchy in (S , d) there is N ∈ N such that d(xm , xn ) < δ whenever m, n ≥ N and therefore also µ(xm , xn ) < ε for m, n ≥ N. The opposite statement holds with a similar argument.
Exercise 2. Find the sets Kn obtained in Proposition 1.2 for each of the following choices of G: (a) G is an open disk; (b) G is an open annulus; (c) G is the plane with n pairwise disjoint closed disks removed; (d) G is an infinite strip; (e) G = C − Z. Solution.
a) G is an open disk,
Let a ∈ C and r > 0 such that G = B(a, r). Set 1 Kn = B a, r − n
!
where a ball of negative radius is to be understood as the empty set. Depending on the size of r, a S finite number of Kn may be empty, not violating the condition Kn ⊂ int Kn+1 , and obviously G = ∞ n=1 .
b) G is an open annulus,
Let a ∈ C be the center of the annulus with radii r, R, 0 ≤ r < R < ∞ so that G = ann(a, r, R). Define ! 1 1 Kn = ann a, r + , R − , n n again with the interpretation that an annulus with inner radius larger than the outer radius is the empty set. d) G = {z ∈ C : |Imz| < 1},
For this open set G define
) ( 1 . Kn = {z = x + iy ∈ C : x, y ∈ R, |x| ≤ n} ∩ z = x + iy ∈ C : x, y ∈ R|y| ≤ 1 − n S Although G is unbounded each Kn is bounded and closed and hence compact and G = ∞ n=1 Kn .
e) G = C − Z.
Here we can define n !C [ 1 Kn = {z ∈ C : |z| ≤ n} ∩ B ( j, , n j=−n
an intersection of closed set, one of which is bounded, hence Kn is compact. Also Kn ⊂ int Kn+1 by S definition and G = ∞ n=1 .
Exercise 3. Supply the omitted details in the proof of Proposition 1.18. 94
Solution. Not available. Exercise 4. Let F be a subset of a metric space (X, d) such that F − is compact. Show that F is totally bounded. Solution. Not available. Exercise 5. Suppose { fn } is a sequence in C(G, Ω) which converges to f and {zn } is a sequence in G which converges to a point z in G. Show lim fn (zn ) = f (z). Solution. Not available. Exercise 6. (Dini’s Theorem) Consider C(G, R) and suppose that { fn } is a sequence in C(G, R) which is monotonically increasing (i.e., fn (z) ≤ fn+1 (z) for all z in G) and lim fn (z) = f (z) for all z in G where f ∈ C(G, R). Show that fn → f . S Solution. By Proposition 1.2 we can find a sequence {Kn } of compact subsets of G such that ∞ n=1 Kn . Thus, WLOG we can assume that G is compact. Define gn = f − fn ∀n. Clearly gn is continuous ∀n, since f and fn are continuous by assumption. Since fn is monotonically increasing, gn is monotonically decreasing. Since fn converges to f , gn converges pointwise to zero, that is ∀ > 0 ∃N ∈ N ∀x ∈ G : 0 ≤ gN (x) <
. 2
Since gN is continuous, there is an open neighborhood B(x; r) around x (with r < ∞). For any y ∈ B(x; r) we have gN (y) < , that is ∀ > 0 ∀x ∈ G ∃N ∈ N ∃B(x; r) : ∀y ∈ B(x; r) : gN (y) < . Since gN is monotonically decreasing, we surely get gn (y) < for every n ≥ N and for every y ∈ B(x; r). These open neighborhoods will cover G. Since G is compact, we can find finitely many x1 , . . . , xk such that B(xi ; r) cover G for every value of . Note that the values of N can change the chosen center x of each those neighborhoods, say Ni . Therefore, let M = maxi Ni . Because of the monotonicity of gn , we have that gn (y) < ∀ n ≥ M in every open neighborhood B(xi ; r), that is, for every value in G. Hence, ∀ > 0 ∃M ∈ N, ∀n ≥ M ∀x ∈ G : | fn (x) − f (x)| < , which shows fn → f . Exercise 7. Let { fn } ⊂ C(G, Ω) and suppose that { fn } is equicontinuous at each point of G. If f ∈ C(G, Ω) and f (z) = lim fn (z) for each z then show that fn → f . S Solution. By Proposition 1.2 we can find a sequence {Kn } of compact subsets of G such that ∞ n=1 Kn . Thus, WLOG we can assume that G is compact. Since fn is equicontiuous at each point of G, we have that ∀ > 0 ∃δ > 0 : d( fn (x), fn (y)) <
whenever |x − y| < δ. 3
(7.2)
Taking the limit as n → ∞, we also get d( f (x), f (y)) <
whenever |x − y| < δ. 3
(7.3)
by the pointwise convergence ( f (z) = lim fn (z) ∀z ∈ G by assumption). As in the previous exercise, we can find finitely many points y1 , . . . , yk ∈ G such that the open neighborhoods B(yi ; δ) cover G, since G is 95
compact. To be more precise: For every point x in G, we can find an open neighborhood B(yi ; δ) for some i = 1, . . . , k. Since fn → f pointwise, we have d( fn (yi ), f (yi )) <
3
∀n ≥ Ni .
Set N = maxi Ni , then we have d( fn (yi ), f (yi )) <
3
∀n ≥ N
∀i = 1, . . . , k.
Finally, for an arbitrary x ∈ G and n ≥ N we have d( fn (x), f (x)) ≤ d( fn (x), fn (yi )) + d( fn (yi ), f (yi )) + d( f (yi ), f (x))
(7.4)
< + + = (7.2),(7.3),(7.4) 3 3 3
where we choose yi such that |x − yi | < δ. This shows fn → f . P Pn n k Exercise 8. (a) Let f be analytic on B(0; R) and let f (z) = ∞ n=0 an z for |z| < R. If fn (z) = k=0 ak z , show that fn → f in C(G; C). P n (b) Let G = ann(0; 0, R) and let f be analytic on G with Laurent series development f (z) = ∞ n=−∞ an z . Put Pn k fn (z) = k=−∞ ak z and show that fn → f in C(G; C).
Solution. a) Let G = B(0; R). This result is proved with the Arzela-Ascoli Theorem. With the given function f define an analytic function ∞ X |ak |zk . (7.5) g : G → C, g(z) := k=0
Let z ∈ G and define r := we have
|z|+R 2 (<
R) then z ∈ B(0; r). Let F = { fn }n . Then for this fixed z and for any n ∈ N
n n X X |ak ||z|k = g(|z|) < ∞. | fn (z)| = | ak zk | ≤ k=0
k=0
Thus for a given z ∈ G, { fn (z) : fn ∈ F } is bounded and therefore has compact closure. Next, we want to show equicontinuity at each point of G. Fix z0 ∈ G and let ε be positive. Let r > 0 be such that B(z0 ; r) ⊂ G. With the function g(z) defined in equation (7.5) and the computations from above we have for all w ∈ ∂B(z0 ; r) and for all n ∈ N | fn (w)| ≤ g(|w|). The boundary of the disk with radius r is compact and so is g(∂B(z0 ; r)). Thus there is a positive number M such that |g(∂B(z 0 , r))| ≤ M. εr Choose δ ∈ 0, min{ 2r , 2M } and let |z − z0 | < δ then as in the proof of Montel’s theorem the following estimate holds I fn (w)(z − z0 ) 1 dw| | fn (z) − fn (z0 )| = | 2πi |w−z0 |=r (w − z)(w − z0 ) 2M ≤ δ ≤ ε. r By Arzela-Ascoli the sequence { fn } is normal in C(G) so there is a subsequence { fnk } that converges in C(G). Since f is analytic and { fn } is the sequence of partial sums of f , { fn } itself is an analytic Cauchy sequence in C(G). Hence every subsequence converges to the same limit, which shows that fn → f in C(G). As a last step recall that H(G) is complete as a subspace of C(G) (Corollary 2.3) and { fn } ⊂ H(G) thus also the limit function f ∈ H(G). 96
7.2 Spaces of analytic functions Exercise 1. Let f, f1 , f2 , . . . be elements of H(G) and show that fn → f iff for each closed rectifiable curve γ in G, fn (z) → f (z) uniformly for z in {γ}. Solution. Not available. Exercise 2. Let G be a region, letRa ∈ R, and suppose that f : [a, ∞] × G → C is a continuous func∞ tion. Define the integral F(z) = a f (t, z) dt to be uniformly convergent on compact subsets of G if Rb limb→∞ a f (t, z) dt exists uniformly for z in any compact subset of G. Suppose that this integral does converge uniformly on compact subsets of G and that for each t in (a, ∞), f (t, · ) is analytic on G. Prove that F is analytic and Z ∞ k ∂ f (t, z) F (k) (z) = dt ∂zk a Solution. Not available. Exercise 3. The proof of Montel’s Theorem can be broken up into the following sequence of definitions and propositions: (a) Definition. A set F ⊂ C(G, C) is locally Lipschitz if for each a in G there are constants M and r > 0 such that | f (z) − f (a)| ≤ M|z − a| for all f in F and |z − a| < r. (b) If F ∈ C(G, C) is locally Lipschitz then F is equicontinuous at each point of G. (c) If F ⊂ H(G) is locally bounded then F is locally Lipschitz. Solution. Not available. Exercise 4. Prove Vitali’s Theorem: If G is a region and { fn } ⊂ H(G) is locally bounded and f ∈ H(G) that has the property that A = {z ∈ G : lim fn (z) = f (z)} has a limit point in G then fn → f . Solution. The sequence { fn } is a locally bounded sequence in H(G) and by Montel’s Theorem then { fn } is normal, so there is a subsequence { fnk } ⊂ { fn } that converges to f in H(G). Suppose that fn 9 f in H(G) then there must be a compact set K ⊂ G the convergence is not uniform on K. In other words there must be an ε > 0 so that for all n ∈ N there is zn ∈ K with | fn (zn ) − f (zn )| ≥ ε. By compactness of K extract a subsequence of {zn }, say {znm } that converges to a point z0 ∈ K. The sequence { fn } is locally bounded, so in particular the subsequence { fnm } with indices corresponding to {znm } is, and again by Montel’s Theorem now applied to the subsequence and again the completeness of H(G) there is an analytic function g such that fnm → g in H(G). On the set A of points of pointwise convergence fn (z) → f and fn (z) → g. Now G is a region and A has a limit point in G so by the Identity Theorem (Chapter 4, Corollary 3.8) already f = g on G which gives a contradiction on the set K and the point z0 because | fnm (znm ) − f (znm )| → |g(z0 ) − f (z0 )| ≥ ε. Hence we can conclude that fn → f in H(G). Exercise 5. Show that for a set F ⊂ H(G) the following are equivalent conditions: (a) F is normal; (b) For every > 0 there is a number c > 0 such that {c f : f ∈ F } ⊂ B(0; ) (here B(0; ) is the ball in H(G) with center at 0 and radius ). Solution. Not available. Exercise 6. Show that if F ⊂ H(G) is normal then F 0 = { f 0 : f ∈ F } is also normal. Is the converse true? Can you add something to the hypothesis that F 0 is normal to insure that F is normal? 97
Solution. Not available. Exercise 7. Suppose F is normal in H(G) and Ω is open in C such that f (G) ⊂ Ω for every f in F . Show that if g is analytic on Ω and is bounded on bounded sets then {g ◦ f : f ∈ F } is normal. Solution. Not available. Exercise 8. Let D = {z : |z| √ < 1} and show that F ⊂ H(D) is normal iff there is a sequence {Mn } of positive P n constants such that lim sup n Mn ≤ 1 and if f (z) = ∞ 0 an z is in F then |an | ≤ Mn for all n.
Solution. Not available.
2πit Exercise 9. Let D = B(0; R 1) and for 0 < r < 1 let γr (t) = re , 0 ≤ t ≤ 1. Show that a sequence { fn } in H(D) converges to f iff γ | f (r) − fn (z)| |dz| → 0 as n → ∞ for each r, 0 < r < 1. r
Solution. ⇒: Assume limn→∞ fn (z) = f (z) ∀z ∈ D. Then Z Z | f (z) − fn (z)| |dz| ≤ sup | f (z) − fn (z)| z∈γr
γr
= γr =re2πit ,0≤t≤1
γr
|dz|
sup | f (z) − fn (z)|· 2πr, z∈γr
∀0 < r < 1.
But supz∈γr | f (z) − fn (z)| → as n → ∞ since limn→∞ fn (z) = f (z) ∀ z ∈ D. Therefore, Z | f (z) − fn (z)| |dz| → 0 γr
as n → ∞ for R each r, 0 < r < 1. ⇐: Assume γ | f (z) − fn (z)| |dz| → 0 as n → ∞ for each r, 0 < r < 1. Let K be an arbitrary compact subset r of D = B(0; 1) and choose 0 < r < 1 such that K ⊂ B(0; r) and denote the shortest distance between an arbitrary point in K and an arbitrary point on ∂B by δ > 0. By the assumption, we have Z | f (z) − fn (z)| |dz| < 2πδ. (7.6) ∂B
Let a ∈ K, then by Cauchy’s Integral formula, we have Z f (z) − fn (z) 1 dz. f (a) − fn (a) = 2πi ∂B z−a Thus, | f (a) − fn (a)| ≤
1 2π
Z
∂B
| f (z) − fn (z)| 1 |dz| ≤ |z − a| 2πδ
Z
2πδ | f (z) − fn (z)| |dz| < = . (7.6) 2πδ ∂B
Hence, fn (a) → f (z) uniformly ∀a ∈ K. Since it converges uniformly on all compact subsets of D, Proposition 1.10 b) yields that { fn } converges to f ∀ z ∈ D. Exercise 10. Let { fn } ⊂ H(G) be a sequence of one-one functions which converge to f . If G is a region, show that either f is one-one or f is a constant function. Solution. Assume { fn } ⊂ H(G) is a sequence of one-one functions which converge to f where G is a region (open and connected). Suppose f is not the constant function. Then, we have to show that f is one-one. Choose an arbitrary point z0 ∈ G and define the sequence gn (z) = fn (z) − fn (z0 ). Clearly, the sequence {gn } converges to g(z) = f (z) − f (z0 ) on the open connected set G\{z0 }, which is again a region. In addition, we 98
have that gn never vanishes on G\{z0 }, since fn are assumed to be one-one functions. Therefore, the sequence {gn } satisfies the conditions of the Corollary 2.6 resulting from Hurwitz’s Theorem, where the region is G\{z0 }. Thus, either g ≡ 0 or g never vanishes (g has no zeros). Since g(z) = f (z) − f (z0 ) is not identically zero on G\{z0 }, we must have g(z) = f (z) − f (z0 ) has no zeros in G\{z0 }. This implies ∀z ∈ G\{z0 }.
f (z) , f (z0 )
Since z0 was an arbitrary point in G, we have shown that f is one-one on G. Exercise 11. Suppose that { fn } is a sequence in H(G), f is a non-constant function, and fn → f in H(G). Let a ∈ G and α = f (a); show that there is a sequence {an } in G such that: (i) a = lim an ; (ii) fn (an ) = α for sufficiently large n. Solution. Not available. Exercise 12. Show that lim tan nz = −i uniformly for z in any compact subset of G = {z : Im z > 0}. Solution. Not available. ¯ R) then Exercise 13. (a) Show that if f is analytic on an open set containing the disk B(a; Z 2π Z R 1 | f (a + reiθ )|2 r dr dθ. | f (a)|2 ≤ πR2 0 0 (b) Let G ! be a region and let M be a fixed positive constant. Let F be the family of all functions f in H(G) such that G | f (z)|2 dx dy ≤ M. Show that F is normal. ¯ R) ⊂ G and f : G → C be analytic. Let γ(t) = a + reit for 0 ≤ t ≤ 2π, then by Solution. a) Let B(a; Cauchy’s Integral Formula we have Z Z 2π g(w) 1 1 f (a + reit ) dt. dw = g(a) = 2πi γ w − a 2π 0
Thus, letting g(z) = f (z)2 , we obtain f 2 (a) =
1 2π
Z
Z
2π
2π
h
0
f (a + reit )
i2
dt.
Multiply by r, we get f 2 (a)r =
1 2π
0
h
i2 f (a + reiθ ) r dθ
(7.7)
and then integrate with respect to r yields Z R Z 2π h Z 2π Z R h Z R i2 i2 1 1 R2 f (a + reiθ ) r dθ dr = f (a + reiθ ) r dr dθ = f 2 (a)r dr = f 2 (a) 2 (7.7) 0 2π 0 2π 0 0 0 which implies 1 f (a) = πR2 2
So | f (a)|2 ≤
1 πR2
Z
2π 0
Z
Z
0
2π 0
R
Z
h
R
0
99
i2 f (a + reiθ ) r dr dθ.
2 f (a + reiθ ) r dr dθ.
b) Let G be a region and M be a fixed constant. Let F be the family of all functions f in H(G) such that " | f (z)|2 dx dy ≤ M. (7.8) G
To show that F is normal. According to Montel’s Theorem, it suffices to show that F is locally bounded, that is ∃ r > 0 such that sup{| f (z)| : |z − a| < r, f ∈ F } < ∞ (p. 153). We will prove this fact by contradiction: Assume F is not locally bounded. Then there is a compact set K ⊂ G such that sup{| f (z)| : z ∈ K, f ∈ F } = ∞. That is, there is a sequence { fn } in F such that sup{| fn (z)| : z ∈ K} ≥ n. Therefore, ∃ zn ∈ K such that | fn (zn )| ≥
n . 2
(7.9)
Every sequence has a convergence subsequence, say {znk }, with znk → z0 ∈ K since K is compact. WLOG, we write {znk } = {zn }. Clearly z0 ∈ G since z0 ∈ K (K ⊂ G), hence ∃ R > 0 such that B(z0 ; R) ⊂ G (since G is open). If we pick n large enough, we can get zn ∈ B(z0 ; R2 ) (a picture might help). In addition, we can ˆ ⊂ B(z0 ; R ) because B(z0 ; R ) is open. Apply part a) to the ball B(zn ; R) ˆ to find an Rˆ > 0 such that B(zn ; R) 2 2 obtain: Z 2π Z Rˆ " 1 1 1 f (z + reiθ ) 2 r dr dθ ≤ M < ∞. | fn (zn )|2 ≤ | fn (z)|2 dx dy ≤ n n 2 ˆ πRˆ 2 0 π R π Rˆ 3 G 0 (7.8) ˆ B(z ;R)⊂G n
So
| fn (zn )| ≤
r
M1 < ∞. π Rˆ
But by (7.9) we have
n . 2 This gives a contradiction if we let n → ∞. Thus, F is locally bounded and therefore normal. | fn (zn )| ≥
7.3 Spaces of meromorphic functions Exercise 1. Prove Proposition 3.3. Solution. a) Let a ∈ C and r > 0. To show: There is a number δ > 0 such that B∞ (a; δ) ⊂ B(a; r). Let R > 0 such that B(a; r) ⊂ B(0; R) and B∞ (a; δ) ⊂ B(0; R). Claim: Pick 2r δ= √ p 1 + R2 1 + |a|2 to obtain B∞ (a; δ) ⊂ B(a; R). Proof of the claim: Let z ∈ B∞ (a; δ). This is equivalent to
⇐⇒ ⇐⇒
2|z − a| <δ p 1 + |z|2 1 + |a|2 2|z − a| 2r < √ p p p 2 2 2 1 + |z| 1 + |a| 1 + R 1 + |a|2 p r 1 + |z|2 |z − a| < √ 2 1+R p
100
which implies (z ∈ B∞ (a; δ) ⇒ z ∈ B(0; R), that is |z| < R) |z − a| < √ ⇐⇒ So z ∈ B(a; r). Therefore
r 1 + R2
√ 1 + R2
|z − a| < r. B∞ (a; δ) ⊂ B(a; r).
b) Let δ > 0 and a ∈ C. To show: There is a number r > 0 such that B(a; r) ⊂ B∞ (a; δ). Claim: Pick r < 2δ to obtain B(a; r) ⊂ B∞ (a; δ). Proof of the claim: Let z ∈ B(a; r). This is equivalent to ⇒
r< 2δ
⇐⇒ 1≤
√ ⇒√ 1+|z|2
1+|a|2
|z − a| < r δ |z − a| < 2 δ |z − a| < · 1 2 p δp 1 + |z|2 1 + |a|2 |z − a| < 2 2|z − a| <δ p 1 + |z|2 1 + |a|2 d(z, a) < δ.
⇐⇒
p
⇐⇒ So z ∈ B∞ (a; δ). Therefore,
B(a; r) ⊂ B∞ (a; δ).
c) Let δ > 0. To show: There is a compact q set K ⊂ C such that C∞ − K ⊂ B∞ (∞; δ). Claim: Choose K = B(0; r) where r >
4 δ2
− 1 to obtain
C∞ − K ⊂ B∞ (∞; δ). (Clearly K is compact and r > 0 since δ ≤ 2.)
101
Proof of the claim: C∞ − K
C∞ − B(0; r)
= = =
r>
q⊂
4 δ2
−1
=
=
=
{C − B(0; r)} ∪ {∞} {C − {z ∈ C : |z| ≤ r}} ∪ {∞} r 4 C − z ∈ C : |z| ≤ − 1 ∪ {∞} 2 δ r 4 − 1 ∪ {∞} z ∈ C : |z| > 2 δ r 4 z∈C: − 1 < |z| ∪ {∞} 2 δ 2 < δ ∪ {∞} z∈C: p 2 1 + |z| {z ∈ C : d(∞, z) < δ} ∪ {∞}
=
B∞ (∞; δ).
= Therefore,
C∞ − K ⊂ B∞ (∞; δ).
d) Let K ⊂ C where K is compact. To show: There is a number δ > 0 such that B∞ (∞; δ) ⊂ C∞ − K. Let B(0; r) such that B(0; r) ⊃ K where r > 0. Clearly C∞ − B(0; r) ⊂ C∞ − K. So it suffices to show: There is a number δ > 0 such that B∞ (∞; δ) ⊂ C∞ − B(0; r) for a given r > 0. Claim: Choose δ ≤
√2 r2 +1
to obtain B∞ (∞; δ) ⊂ C∞ − B(0; r)
and hence B∞ (∞; δ) ⊂ C∞ − K. Proof of the claim: B∞ (∞; δ)
= =
⊂
δ≤ √ 2
r2 +1
= = = =
{z ∈ C : d(∞, z) < δ} ∪ {∞} 2 < δ ∪ {∞} z∈C: p 2 1 + |z| 2 2 < z ∈ C : p √ ∪ {∞} 2 2 r + 1 1 + |z| {z ∈ C : r ≤ |z|}} ∪ {∞} {C ∪ {z ∈ C : |z| < r} ∪ {∞} C∞ − {z ∈ C : |z| < r} C∞ − B(0; r). 102
Therefore, B∞ (∞; δ) ⊂ C∞ − B(0; r). Exercise 2. Show that if F ⊂ M(G) is a normal family in C(G, C∞ ) then µ(F ) is locally bounded. Solution. Not available.
7.4 The Riemann Mapping Theorem Exercise 1. Let G and Ω be open sets in the plane and let f : G → Ω be a continuous function which is one-one, onto, and such that f −1 : Ω → G is also continuous (a homeomorphism). Suppose {zn } is a sequence in G which converges to a point z in ∂G; also suppose that w = lim f (zn ) exists. Prove that w ∈ ∂Ω. Solution. Not available. Exercise 2. (a) Let G be a region, let a ∈ G and suppose that f : (G − {a}) → C is an analytic function such that f (G − {a}) = Ω is bounded. Show that f has a removable singularity at z = a. If f is one-one, show that f (a) ∈ ∂Ω. (b) Show that there is no one-one analytic function which maps G = {z : 0 < |z| < 1} onto an annulus Ω = {z : r < |z| < R} where r > 0. Solution. Not available. Exercise 3. Let G be a simply connected region which is not the whole plane and suppose that z¯ ∈ G whenever z ∈ G. Let a ∈ G ∩ R and suppose that f : G → D = {z : |z| < 1} is a one-one analytic function with f (a) = 0, f 0 (a) > 0 and f (G) = D. Let G+ = {z ∈ G : Im z > 0}. Show that f (G+ ) must lie entirely above or entirely below the real axis. Solution. Not available. Exercise 4. Find an analytic function f which maps {z : |z| < 1, Re z > 0} onto B(0; 1) in a one-one fashion. Solution. Not available. Exercise 5. Let f be analytic on G = {z : Re z > 0}, one-one, with Re f (z) > 0 for all z in G, and f (a) = a for some real number a. Show that | f 0 (a)| ≤ 1. Solution. Clearly G is a simply connected region (not the whole plane). Let a ∈ G. Then, there is a unique analytic function g : G → D having the properties a) g(a) = 0 and g0 (a) > 0 b) g is 1-1 c) g(G) = D 1−1
(by the Riemann Mapping Theorem). Since, we have f : G −−−−−−→ G (since Re f (z) > 0) and g : 1−1
1−1
−−−−−−→ D so g−1 : D −−−−−−→ G we can define h(z) = g( f (g−1 (z))). Clearly h(z) is analytic and oneonto onto one, since f and g are analytic and one-one, and we have h(D) = D by construction. We have h(0) = g( f (g−1 (0)))
= g−1 (0)=a
103
g( f (a)) = g(a) = 0. f (a)=a
We also have |h(z)| ≤ 1
∀z ∈ D
since h : D → D. Thus, the hypothesis of Schwarz’s Lemma are satisfied, and hence, we get |h0 (0)| ≤ 1. We have h0 (z) =
h
i0 0 g( f (g−1 (z))) = g0 ( f (g−1 (z))) f 0 (g−1 (z)) g−1 (z)
= g0 ( f (g−1 (z))) f 0 (g−1 (z))
1 g0 (g−1 (z))
where the last step follows from Proposition 2.20 provided g0 (g−1 (z)) , 0. So h0 (0)
= g0 ( f (g−1 (0))) f 0 (g−1 (0)) = g0 ( f (a)) f 0 (a)
(g0 (a) > 0 by assumption). Therefore
1 g0 (g−1 (0))
1 1 = g0 (a) f 0 (a) 0 = f 0 (a) g0 (a) g (a)
|h0 (0)| = | f 0 (a)| ≤ 1.
Thus, we have shown that | f 0 (a)| ≤ 1. Exercise 6. Let G1 and G2 be simply connected regions neither of which is the whole plane. Let f be a one-one analytic mapping of G1 onto G2 . Let a ∈ G1 and put α = f (a). Prove that for any one-one analytic map h of G1 into G2 with h(a) = α it follows that |h0 (a)| ≤ | f 0 (a)|. Suppose h is not assumed to be one-one; what can be said? 1−1
Solution. Define the function F(z) = f −1 (h(z)). This is well-defined since f : G1 −−−−−−→ G2 and h : onto
1−1
G1 −−−−−−→ G2 . Clearly F(z) is analytic since f and h are analytic, F(z) is one-one and F : G1 → G1 by construction. We have F(a) = f −1 (h(a)) = f −1 (α) = a. Thus by Exercise 5, we get |F 0 (a)| ≤ 1. We have 0 F 0 (z) = f −1 (h(z)) = f 0 (h(z))h0 (z) =
1 f 0 ( f −1 (h(z)))
h0 (z)
where the last step follows by Proposition 2.20 provided f 0 ( f −1 (h(z))) , 0. So F 0 (a) =
1 f 0 ( f −1 (h(a)))
h0 (a) =
h0 (a) f 0 ( f −1 (α))
=
which is well-defines since f 0 (a) , 0 by assumption (f is one-one). Since |F 0 (a)| =
|h0 (a)| ≤1 | f 0 (a)|
104
h0 (a) f 0 (a)
gives |h0 (a)| ≤ | f 0 (a)|. If h is not one-one, then we get the same result, since I do not need the assumption that F(z) has to be one-one for exercise 5. Exercise 7. Let G be a simply connected region and suppose that G is not the whole plane. Let ∆ = {ξ : |ξ| < 1} and suppose that f is an analytic, one-one map of G onto ∆ with f (a) = 0 and f 0 (a) > 0 for some point a in G. Let g be any other analytic, one-one map of G onto ∆ and express g in terms of f . Solution. Let a ∈ G such that f (a) = 0 and f 0 (a) > 0. Since g is one-one mapping from G onto ∆, we clearly have g(a) = α, where α ∈ ∆ (α does not have to be zero). Next, define g1 (z) =
z−α . 1 − αz ¯
Clearly, g1 is a Möbius transformation mapping ∆ to ∆ and in addition g1 is one-one and analytic. Now define g2 (z) = g1 (g(z)). Then, g2 is one-one and analytic and maps G onto ∆. In addition, we have g2 (a) = g1 (g(a)) = g1 (α) = 0.
(7.10)
But g02 (a) need not necessarily satisfy g02 (a) > 0. However, g02 (a) , 0 ∀z ∈ G since g2 is one-one. Therefore, define the rotation g3 (z) = eiθ z for θ ∈ [0, 2π). Clearly g3 maps ∆ onto ∆. g3 is one-one and analytic. Finally, let h(z) = g3 (g2 (z)). Also h is one-one, analytic and maps G onto ∆. We have h(a) = g3 (g2 (a))
= g3 (0) = 0. (7.10)
And, h0 (z) = (g3 (g2 (z)))0 = (g3 (g1 (g(z))))0 = g03 (g1 (g(z)))g01 (g(z))g0 (z) (1 − αg(z))· ¯ 1 − (g(z) − α)· (−α) ¯ 0 g (z). = eiθ 2 (1 − αg(z)) ¯ So, h0 (a) = eiθ
2 1 (1 − αα) ¯ −0 0 iθ 1 − |α| g (a) = e g0 (a) = eiθ g0 (a). 2 2 2 (1 − αα) ¯ (1 − |α| ) 1 − |α|2
Clearly, we can choose a θ ∈ [0, 2π) such that h0 (a) > 0 (depending on the sign of g0 (a). Recall g0 (a) , 0 1 0 iθ since g is one-one and α ∈ ∆, so 1−|α| = −1 or 2 > 0). For example: If g (a) < 0, then pick θ = π, so e θ = −Arg(g0 (a)) will work, too. By uniqueness of the map f ( f satisfies the properties of the Riemann Mapping Theorem), we have f (z) = h(z) 105
since h satisfies the properties of the Riemann Mapping Theorem, too. Thus, f (z) = h(z) = g3 (g2 (z)) = eiθ g2 (z) = eiθ g1 (g(z)) = eiθ
g(z) − α . 1 − αg(z) ¯
So
⇐⇒ ⇐⇒ ⇐⇒
1 − αg(z) ¯ = eiθ g(z) − αeiθ
f (z) − αg(z) ¯ f (z) = eiθ g(z) − αeiθ iθ (e + α¯ f (z))g(z) = f (z) + αeiθ f (z) + αeiθ g(z) = iθ . e + α¯ f (z)
Exercise 8. Let r1 , r2 , R1 , R2 , be positive numbers such that R1 /r1 = R2 /r2 ; show that ann(0; r1 , R1 ) and ann(0; r2 , R2 ) are conformally equivalent. Solution. The function f : C → C; f (z) = rr21 z is analytic and it maps the annulus ann(0; r1 , R1 ) bijectively into ann(0; r2 , R2 ). The inner boundary {z : |z| = r1 } of the domain is mapped to the inner boundary of ann(0; r2 , R2 ). For the outer boundary note that by assumption R1 = Rr22r1 , so if |z| = R1 then | f (z)| = rR2 r11 = R2 . The inverse function is f −1 (z) = rr12 z, an analytic and non-constant function also. Therefore the result is proved. Exercise 9. Show that there is an analytic function f defined on G =ann(0; 0, 1) such that f 0 never vanishes and f (G) = B(0; 1). Solution. Not available.
7.5 The Weierstrass Factorization Theorem Q (1 + zn ) converges absolutely iff (1 + |zn |) converges. Q P Solution. Assume Re(zn ) > −1. Then the product (1 + zn ) converges absolutely iff ∞ k=1 zn converges P P∞ absolutely (by Corollary 5.6 p. 166 applied to wn = 1 + zn ). k=1 zn converges absolutely iff ∞ n=1 |zn | converges (by definition). P P Claim: ∞ |z | converges iff ∞ + |zn |) converges. n=1 log(1 P∞ n=1 n Q Then n=1 log(1+|zn |) converges iff (1+|zn |) converges (by Proposition 5.2 p.165 applied to wn = 1+|zn |). Q Q Thus, ones we have proved the claim, we obtain: (1 + zn ) converges absolutely iff (1 + |zn |) converges. Proof of the claim: Let xn = |zn | for convenience. Clearly xn ≥ 0 ∀n ∈ N. P P ⇒: Assume |zn | converges, that is xn converges. Therefore, we have xn → 0. So given > 0, we have Exercise 1. Show that
Q
0 ≤ log(1 + |xn |) ≤ (1 + )xn
for sufficiently large n where the last inequality follows since limz→0 log(1+z) = 1 by Exercise 2. By the z P P comparison test, log(1 + xn ) converges, that is log(1 + |zn |) converges. P P ⇐: Assume log(1 + |zn |) converges, that is log(1 + xn ) converges. Therefore, we have log(1 + xn ) → 0 and therefore xn → 0. So given > 0, we have (1 − )xn ≤ log(1 + xn ) for sufficiently large n where the last inequality follows since limz→0 P P comparison test, xn converges, that is |zn | converges. P∞ P∞ Therefore, n=1 |zn | converges iff n=1 log(1 + |zn |) converges. 106
log(1+z) z
= 1 by Exercise 2. By the
= 1. Exercise 2. Prove that limz→0 log(1+z) z Solution. First note that, f (z) =
log(1 + z) z
has a removable singularity at z = 0, since lim z f (z) = lim z z→0
z→0
log(1 + z) = lim log(1 + z) = log(1) = 0 z→0 z
(by Theorem 1.2 Chapter 5 p. 103). We know ∞ X zk (−1)k+1 p.165 k k=1
log(1 + z) = if |z| < 1. Therefore, f (z) =
log(1 + z) = z
P∞
k=1 (−1)
z
k+1 zk k
=
∞ ∞ ∞ X X X zk−1 zk−1 zk =1+ =1+ (−1)k+1 (−1)k+2 (−1)k k k k+1 k=1 k=2 k=1
where the sum is an analytic function. Hence, ∞ ∞ ∞ k X X X (limz→0 z)k k z =1+ (−1)k 0=1 lim f (z) = lim 1 + (−1) = 1 + z→0 z→0 k+1 k+1 k=1 k=1 k=1
and therefore
log(1 + z) = 1. z Exercise 3. Let f and g be analytic functions on a region G and show that there are analytic functions f1 , g1 , and h on G such that f (z) = h(z) f1 (z) and g(z) = h(z)g1 (z) for all z in G; and f1 and g1 have no common zeros. limz→0
Solution. Let f and g be analytic functions on the region G. Let {a j } be the zeros of f with multiplicity n j . Let {b j } be the zeros of g with multiplicity n˜ j . If f and g have no common zeros, let h = 1. Then f1 = f and g1 = g. Clearly f1 , g1 and h are analytic on G such that f (z) = h(z) f1 (z) and g(z) = h(z)g1 (z) for all z in G and f1 and g1 have no common zeros. Otherwise let {z j } be the set of common zeros of f and g with multiplicity m j = min(n j , n˜ j ). Then {z j } form a sequence of distinct points in G with no limit points on G (since the zeros are isolated). Then there is an analytic function h defined on G whose only zeros are at the points z j with multiplicity m j (by Theorem 5.15 p. 170). Let f1 = hf and g1 = gh , then clearly f1 and g1 are analytic functions on G such that f (z) = h(z) f1 (z) and g(z) = h(z)g1 (z) for all z in G. (since f1 and g1 have removable singularities at the z j ’s by construction; the multiplicity of z j ’s in f and g is greater or equal than the multiplicity of z j ’s in h). In addition, we have that f1 and g1 have no common zeros by construction. Exercise 4. (a) Let 0 < |a| < 1 and |z| ≤ r < 1; show that a + |a|z ≤ 1 + r (1 − a¯ z)a 1 − r
(b) Let {an } be a sequence of complex numbers with 0 < |an | < 1 and product ! ∞ Y |an | an − z B(z) = an 1 − a¯ n z n=1 107
P (1 − |an |) < ∞. Show that the infinite
converges in H(B(0; 1)) and that |B(z)| ≤ 1. What are the zeros of B? (B(z) is called a Blaschke Product.) P (c) Find a sequence {an } in B(0; 1) such that (1 − |an |) < ∞ and every number eiθ is a limit point of {an }. Solution. Not available.
Exercise 5. Discuss the convergence of the infinite product Solution. Not available.
Q∞
1 n=1 n p
for p > 0.
i Qh Q Exercise 6. Discuss the convergence of the infinite products 1 + ni and 1 + ni . Q i Solution. First consider ∞ n=1 1 + n . We claim that this product does not converge. q Q Let zn := 1 + ni = rn eiθn with rn = 1 + n12 and θ = arctan 1n . If there is z = reiθ such that ∞ n=1 zn = z, Q∞ P∞ then n=1 rn = r and n=1 θn = θ. Since arctan(x) ≥ 12 for all values x ∈ (0, 1] the estimate x ! ∞ ∞ ∞ X X 1X1 1 ≥ θn = arctan n 2 n n n=1 n=1 1
is valid. The right hand side does not converge, therefore the angles do not converge and there is no z ∈ C with the desired properties (note that rn > 1 for all n, thus z , 0). q The second infinite product is a product of real numbers and zn = |1 + ni | = 1 + n12 . The estimate 2 1 + n12 ≤ 1 + n12 implies that 1 ≤ zn ≤ 1 + n12 . Thus for any integer N ∈ N N Y n=1
1≤
N Y n=1
zn ≤
N Y n=1
1+
1 n2
!
where the last term is absolutely convergent by an example in class. Therefore the sequence of the partial QN products n=1 zn is an increasing, bounded sequence that converges in the real numbers. With log(zn ) > 0 for all n, the comparison with the sum of logarithms (Proposition 5.4) gives absolute convergence. Q 1 1 Exercise 7. Show that ∞ n=2 1 − n2 = 2 . Solution. Consider Pn
=
=
= = = So,
! Y ! ! Y ! ! n n 1 k−1 k+1 1 1 1− 1+ = 1− 2 = k k k k k k=2 k=2 k=2 ! ! ! !# " n n k−1 k+1 Y k−1 k+1 X exp log = exp log k k k k k=2 k=2 n n X X exp log(k − 1) − log(k) + exp log(k + 1) − log(k) k=2 k=2 exp log(1) − log(n) + exp − log(2) + log(n + 1) !) ( n+1 n+1 = . exp − log(2n) + log(n + 1) = exp log 2n 2n n Y
∞ Y n=2
1−
! 1 n+1 1 = . = lim Pn = lim n→∞ n→∞ 2n 2 n2 108
Q Q Exercise 8. For which values of z do the products (1 − zn ) and (1 + z2n ) converge? Is there an open set G such that the product converges uniformly on each compact subset of G? If so, give the largest such open. Q n Solution. Consider the infinite product ∞ n=1 (1 − z ). By definition and Corollary 5.6 this converges absoP∞ n lutely if and only if n=1 −z converges absolutely. From Calculus it is known that the convergence is also uniform for |z| ≤ r < 1 with r ∈ [0, 1). If z is an nth root of unity for a positive integer n, then one of the factors in the infinite product becomes 0 and the product converges to 0, but not absolutely according to Definition 5.5. If there is no n such that z is nth root of unity then no factor of the infinite product equals zero and there is a subsequence {(1 − znk }k Q n such that |1 − znk | > 1.5. We conclude that ∞ n=1 (1 + z ) does not converge. S th With the denseness of n∈N {z ∈ C : z is n root of unity} in the unit sphere it follows that the maximal set G such that the convergence is uniform on compact subsets is the unit disk. Q 2n 2 The behavior of ∞ n=1 (1 + z ) is almost the same as the previous one. Substitute w := z , then the convergence is uniform for |w| ≤ r < 1, so |z| ≤ sqrtr < 1. Also if |z| -and hence |w|- is greater than 1, then the infinite product diverges. Similar to the previous case one factor equals 0 if z is a 2nth root of −1. The maximal region G such that the convergence is uniform on compact sets is again the open unit disk. Exercise 9. Use Theorem 5.15 to show there is an analytic function f on D = {z : |z| < 1} which is not analytic on any open set G which properly contains D. Solution. Not available. Exercise 10. Suppose G is an open set and { fn } is a sequence in H(G) such that f (z) = in H(G). (a) Show that ∞ X 0 Y fn (z) fk (z) k=1
Q
fn (z) converges
n,k
0
converges in H(G) and equals f (z). (b) Assume that f is not the identically zero function and let K be a compact subset of G such that f (z) , 0 for all z in K. Show that ∞
f 0 (z) X fn0 (z) = f (z) f (z) n=1 n and the convergence is uniform over K. Solution. Not available. Exercise 11. A subset J of H(G), G a region, is an ideal iff: (i) f and g in J implies a f + bg is in J for all complex numbers a and b; (ii) f in J and g any function in H(G) implies f g is in J. J is called a proper ideal if J , (0) and J , H(G); J is a maximal ideal if J is a proper ideal and whenever L is an ideal with J ⊂ L then either L = J or L = H(G); J is a prime ideal if whenever f and g ∈ H(G) and f g ∈ J then either f ∈ J or g ∈ J. If f ∈ H(G) let Z( f ) be the set of zeros of f counted according to their multiplicity. So Z((z − a)3 ) = {a, a, a}. If S ⊂ H(G) then Z(S) = ∩{Z( f ) : f ∈ S}, where the zeros are again counted according to their multiplicity. So if S = {(z − a)3 (z − b), (z − a)2 } then Z(S) = {a, a}. (a) If f and g ∈ H(G) then f divides g (in symbols, f |g if there is an h in H(G) such that g = f h. Show that f |g iff Z ⊂ Z(g). (b) If S ⊂ H(G) and S contains a non-zero function then f is a greatest common divisor of S if: (i) f |g for each g in S and (ii) whenever h|g for each g in S, h| f . In symbols, f = g.c.d.S. Prove that f = g.c.d.S. iff Z( f ) = Z(S) and show that each non-empty subset of H(G) has a g.c.d. (c) If A ⊂ G let J(a) = { f ∈ H(G) : Z( f ) ⊃ A}. Show that J(A) is a closed ideal in H(G) and J(A) = (0) 109
iff A has a limit point in G. (d) Let a ∈ G and J = J({a}). Show that J is a maximal ideal. (e) Show that every maximal ideal in H(G) is a prime ideal. (f) Give an example of an ideal which is not a prime ideal. Solution. Not available. Exercise 12. Find an entire function f such that f (n + in) = 0 for every integer n (positive, negative or zero). Give the most elementary example possible (i.e., choose the pn to be as small as possible). Solution. Not available. Exercise 13. Find an entire function f such that f (m + in) = 0 for all possible integers m, n. Find the most elementary solution possible. Solution. Not available.
7.6 Factorization of the sine function Exercise 1. Show that cos πz =
Q∞ h n=1 1 −
4z2 (2n−1)2
i
.
Solution. We know by the double-angle identity of sine sin(2z) = 2 sin(z) cos(z) (this is proved easily by Q z2 using the definition) or sin(2πz) = 2 sin(πz) cos(πz). Since, we know sin(πz) = πz ∞ n=1 1 − n2 , we obtain ⇐⇒ ⇐⇒
sin(2πz) = 2 sin(πz) cos(πz) ! ! ∞ ∞ Y Y 4z2 z2 2πz 1 − 2 = 2πz 1 − 2 cos(πz) n n n=1 n=1 ! ! ! ∞ ∞ ∞ Y Y 4z2 Y 4z2 z2 2πz 1− = 2πz 1− 1 − 2 cos(πz) (2m)2 m=1 (2m − 1)2 n m=1 n=1
where the last statement follows by splitting the product into a product of the even and odd terms (rearrangement of the terms is allowed). Hence ! ∞ ! ! ∞ Y 4z2 4z2 Y z2 2πz = 2πz 1− 1− 1 − 2 cos(πz) (2n)2 n=1 (2n − 1)2 n n=1 n=1 ! ! ! ∞ ∞ ∞ Y Y 4z2 z2 z2 Y = 2πz cos(πz). 1− 1 − 2πz 1− 2 n n=1 (2n − 1)2 n2 n=1 n=1 ∞ Y
⇐⇒ Thus,
cos(πz) =
∞ Y n=1
1−
Exercise 2. Find a factorization for sinh z and cosh z.
110
! 4z2 . (2n − 1)2
Solution. We know ez − e−z e−i(iz) − e−(−i)(iz) = 2 2 −ei(iz) + e−i(iz) −ei(iz) + e−i(iz) =i = 2 2i ei(iz) − e−i(iz) = −i = −i sin(iz) 2i ! ! ∞ ∞ Y Y (iz)2 z2 = −i(iz) 1− 2 2 =z 1+ 2 2 nπ nπ n=1 n=1
sinh(z) =
since by p. 175 Equation 6.2 sin(πz) = πz
∞ Y n=1
1−
! ! ∞ Y z2 z2 ⇒ sin(z) = z . 1 − n2 n2 π2 n=1
Therefore, sinh(z) = z
∞ Y
1+
n=1
! z2 . n2 π2
We have ez + e−z e−i(iz) + e−(−i)(iz) = 2 2 ei(iz) + e−i(iz) = cos(iz) = 2 ! Y ! ∞ ∞ Y 4z2 4(iz)2 = = 1+ 1− (2n − 1)2 π2 (2n − 1)2 π2 n=1 n=1
cosh(z) =
since by Exercise 1 p. 176 cos(πz) =
∞ Y n=1
! ! ∞ Y 4z2 4z2 1− ⇒ cos(z) = . 1− (2n − 1)2 (2n − 1)2 π2 n=1
Therefore, cosh(z) =
∞ Y
1+
n=1
Exercise 3. Find a factorization of the function cos Solution. Not available. Exercise 4. Prove Wallis’s formula: Solution. Not available.
π 2
=
! 4z2 . (2n − 1)2 π2
πz 4
− sin
(2n)2 n=1 (2n−1)(2n+1) .
Q∞
111
πz 4 .
7.7 The gamma function Exercise 1. Show that 0 < γ < 1. (An approximation to γ is .57722. It is unknown whether γ is rational or irrational.) Solution. Not available. Exercise 2. Show that Γ(z)Γ(1 − z) = π csc πz for z not an integer. Deduce from this that Γ( 12 ) =
√
π.
Solution. Assume z is not an integer, then by Gauss’s Formula p. 178 we get n!nz m!m1−z lim n→∞ z(z + 1) · · · (z + n) m→∞ (1 − z)(2 − z) · · · (m + 1 − z) n!n!nz+1−z = lim n→∞ z(1 + z)(1 − z)(2 + z)(2 − z) · · · (n + z)(n − z)(n + 1 − z) (n!)2 n = lim n→∞ z(12 − z2 )(22 − z2 ) · · · (n2 − z2 )(n + 1 − z) (n!)2 n = lim 2 2 2 2 n→∞ z(n!n!) 1 − z 1 − 2z 2 1 − 2z 3 · · · 1 − nz 2 (n + 1 − z) 12
Γ(z)Γ(1 − z) =
=
lim
lim z 1−
n→∞
z2 12
1−
z2 22
1 1−
= z limn→∞ 1 − = =
1− 12 z2
z2 32
··· 1 −
z2 n2
1 2 1 − 3z 2 · · · 1 − 22 z2
1 π = Q Q∞ ∞ z2 z n=1 1 − n2 πz n=1 1 − π = π csc(πz) sin(πz)
z2 n2
n+1−z n
z2 lim n2 n→∞
|
n+1−z {zn } =1
where the step before the last step follows by p. 175 Equation 6.2. So Γ(z)Γ(1 − z) = π csc(πz) for z not an integer. Now, let z = 21 , then ! ! π 1 1 Γ 1− = π csc Γ 2 2 2 which implies
Exercise 3. Show:
! √ √ 1 Γ = π· 1 = π. 2 √ πΓ(2z) = 22z−1 Γ(z)Γ(z + 21 ). (Hint: Consider the function Γ(z)Γ(z + 21 )Γ(2z)−1 .)
Solution. Following the hint define a function f on its domain G = {z ∈ C : z < (Z − N)} by ! 2z−1 z z 1 . + Γ f (z) = √ Γ 2 2 π 2 112
Let x be a positive real number. It suffices to show that f (x) satisfies the Bohr-Mollerup Theorem. By the logarithmic convexity of Γ is follows that ! x √ x 1 + + log Γ log f (x) = (x − 1) log 2 − log π + log Γ 2 2 2 is a convex function. Also f satisfies the functional equation of the Gamma-function: ! x 1 x 2x Γ +1 + f (x + 1) = √ Γ 2 π 2 2 ! x 2 x 1 x x = √ Γ + Γ π 2 2 2 2 ! x2 x−1 x 1 x Γ + = √ Γ 2 2 2 π = x f (x). Γ( 1 ) Lastly, also f (1) = √2π = 1; thus by Theorem 7.13, f agrees with Γ on the positive real numbers. With the Identity Theorem then f agrees with Γ on the whole domain G.
Exercise 4. Show that log Γ(z) is defined for z in C − (−∞, 0] and that log Γ(z) = − log z − γz −
∞ X n=1
z z log 1 + − . n n
Solution. Not available. Exercise 5. Let f be analytic on the right half plane Re z > 0 and satisfy: f (1) = 1, f (z + 1) = z f (z), and limn→∞ nf (z+n) z f (n) = 1 for all z. Show that f = Γ. Solution. Not available. Exercise 6. Show that Γ(z) =
Z ∞ ∞ X (−1)n + e−t tz−1 dt n!(z + n) 1 n=0
for z , 0, −1, −2, . . . (not for Re z > 0 alone). Solution. Write Γ(z) = Φ(z) + Ψ(z) where Φ(z) =
Z
1
e−t tz−1 dt
0
and Ψ(z) =
Z
∞
e−t tz−1 dt.
(7.11)
1
We can write Φ(z) =
Z
0
=
1
e−t tz−1 dt =
Z
0
∞ 1X n=0
∞
X (−1)n (−t)n z−1 t dt = n! n! n=0
" #1 X ∞ ∞ X (−1)n (−1)n tn+z = n! n + z 0 n=0 n!(z + n) n=0 113
Z
1
tn tz−1 dt 0
Thus, Γ(z) =
Z ∞ ∞ X (−1)n + e−t tz−1 dt n!(z + n) 1 n=0
(7.12)
Claim 1: Γ(z) given by (7.12) is the analytic continuation of (7.11), that is Γ(z) given by (7.12) is defined for all z ∈ C − {0, −1, −2, . . .}. Proof of Claim 1: We know from the book that Ψ(z) is analytic for Re (z) > 0. Claim 2: Ψ(z) is analytic for Re (z) ≤ 0. Thus Ψ(z) is analytic on C. Proof of Claim 2: Assume Re (z) ≤ 0. Then |tz−1 | = tRe(z)−1
≤ t−1 . t∈[1,∞),Re(z)≤0
1 1 But since e− 2 t tRe (z)−1 → 0 as t → ∞, there exists a constant C > 0 such that tRe(z)−1 ≤ Ce 2 when t ≥ 1. Hence, we have 1 1 |e−t tz−1 | ≤ |e−t |· |tz−1 | = e−t tRe(z)−1 ≤ e−t Ce 2 t = Ce− 2 t 1
and therefore Ce− 2 t is integrable on (1, ∞). By Fubini’s Theorem for any {γ} ⊂ G = {z : Re(z) ≤ 0}, Z Z ∞ Z ∞Z e−t tz−1 dt dz = e−t tz−1 dz dt = 0 γ
1
1
which implies
Z
∞
1
In summary, Ψ(z) =
γ
e−t tz−1 dt ∈ H(G).
Z
∞ 1
Thus, Claim 2 is proved. It remains to show that Φ(z) =
e−t tz−1 dt ∈ H(C).
∞ X (−1)n n!(z + n) n=0
is analytic on C − {0, −1, −2, . . .}. Note that Φ(z) is uniformly and absolutely convergent as a series in any closed domain which contains none of the points 0, −1, −2, . . . and thus provides the analytic continuation of Φ(z). Since Z ∞ ∞ X (−1)n + e−t tz−1 dt n!(z + n) 1 n=0 is analytic and we know (Theorem 7.15 p. 180) Γ(z) = for Re(z) > 0, we get Γ(z) =
Z
∞
e−t tz−1 dt
0
Z ∞ ∞ X (−1)n + e−t tz−1 dt n!(z + n) 1 n=0
is the analytic continuation of (7.11) for z ∈ C − {0, −1, −2, . . .}. 114
Exercise 7. Show that
Z
Solution. We have shown that Γ
∞
2
sin(t ) dt =
0
Z
∞ 0
1 cos(t ) = 2 2
r
1 π. 2
√ = π (see Exercise 2 p. 185). Thus
1 2
! Z ∞ Z ∞ √ 1 2 2 −t − 12 π=Γ = e t dt = 2a e−a t dt 2 0 0 (because of Theorem 7.15 p. 180). The last step involves integrating a complex integral. Hence, we have √ Z ∞ π 2 2 . (7.13) e−a t dt = 2a 0 Let a = Z
∞
1−i √ , 2 −
e
then a2 =
1−i √ 2
2
t2
dt
1−2i+i2 2
= −i and therefore Z
=
∞
Z0 ∞
0
= = (7.13)
−(−i)t2
e
dt =
Z
∞
2
eit dt
0
Z
Z ∞ cos(t2 ) dt + i sin(t2 ) dt 0 0 0 r r r √ √ π π 1 1 1 1 1 1 1+i = π(1 + i) = π+i π. = √ 2 2 2 2 √ 2 1−i 2 (1 − i)(1 + i) 2 2 2 2
cos(t ) + i sin(t2 ) dt =
Thus, Z
∞
2
cos(t ) dt + i
0
Z
∞
0
1 sin(t ) dt = 2
r
r
1 π 2
2
which implies Z
∞
0
and
Z
1 cos(t ) dt = 2
∞ 0
∞
2
1 sin(t ) = 2 2
r
1 1 π+i 2 2
r
1 π 2
1 π. 2
Exercise 8. Let u > 0 and v > 0 and express Γ(u)Γ(v) as a double integral over the first quadrant of the plane. By changing to polar coordinates show that Z π 2 Γ(u)Γ(v) = 2Γ(u + v) (cos θ)2u−1 (sin θ)2v−1 dθ. 0
The function B(u, v) =
Γ(u)Γ(v) Γ(u + v)
is called the beta function. By changes of variables show that Z 1 Z B(u, v) = tu−1 (1 − t)v−1 dt = 0
0
∞
tu−1 dt (1 + t)u+v
Can this be generalized to the case when u and v are complex numbers with positive real part? 115
Solution. Let u > 0 and v > 0, then by Theorem 7.15 p. 180 and three changes of variables (indicated by “CoV”) we obtain Z ∞ Z ∞ −s u−1 Γ(u)Γ(v) = e s ds e−t tv−1 dt 0 0 Z ∞ Z ∞ 2 2 = 2 e−x x2u−2 x dx· 2 e−y y2v−2 y dy CoV:s=x2 ,t=y2 0 0 Z ∞Z ∞ 2 2 = 4 e−x −y x2u−1 y2v−1 dx dy 0 0 Z π/2 Z ∞ 2 = 4 e−r r2u−1 (cos θ)2u−1 r2v−1 (sin θ)2v−1 r dr dθ CoV:x=r cos(θ),y=r sin θ 0 0 Z π/2 Z ∞ 2 = 4 e−r r2u+2v−1 (cos θ)2u−1 (sin θ)2v−1 r dr dθ 0 0 Z ∞ Z π/2 −r2 2u+2v−1 = 4 e r dr (cos θ)2u−1 (sin θ)2v−1 dθ 0 0 Z Z π/2 1 ∞ −t u+v−1 e t dt (cos θ)2u−1 (sin θ)2v−1 dθ = 4 2 0 CoV:t=r2 0 Z π/2 = 2Γ(u + v) (cos θ)2u−1 (sin θ)2v−1 dθ. 0
This implies B(u, v) =
Γ(u)Γ(v) =2 Γ(u + v)
Z
Z
1
Next, we show B(u, v) =
π/2
(cos θ)2u−1 (sin θ)2v−1 dθ. 0
tu−1 (1 − t)v−1 dt.
0
We have B(u, v)
= = CoV:t=cos2 θ
=
Z π/2 Γ(u)Γ(v) =2 (cos θ)2u−1 (sin θ)2v−1 dθ Γ(u + v) 0 Z 1 2v−2 √ 2u−2 √ t 1−t dt 2 0 Z 1 tu−1 (1 − t)v−1 dt. 0
Hence, B(u, v) =
Z
Finally, we show B(u, v) =
1
tu−1 (1 − t)v−1 dt.
0
Z
∞
0
tu−1 dt. (1 + t)u+v
116
We have B(u, v)
Z
= = t CoV:s= 1−t
= = =
Z Z
1
tu−1 (1 − t)v−1 dt
0
∞ 0 ∞
s u−1 s v−1 (1 + s)−2 ds 1− 1+s 1+s
1 1 (1 + s)−2 ds u−1 (1 + s)v−1 (1 + s) Z0 ∞ 1 su−1 (1 + s)−2 ds (1 + s)u+v−2 0 Z ∞ su−1 ds. u+v 0 (1 + s) su−1
Hence, B(u, v) =
Z
∞
0
tu−1 dt. (1 + t)u+v
Yes, the beta function can be generalized to the case when u and v are complex numbers with positive real part. We have seen that B(u, v) =
Z
0
∞
tu−1 dt = 2 (1 + t)u+v
Z
π/2
(cos θ)2u−1 (sin θ)2v−1 dθ 0
for u > 0 and v > 0. Thus it remains to show that Z ∞ tu−1 B(u, v) = dt ∈ H(G × G) u+v 0 (1 + t) where G × G ⊂ C × C with G = {z : Re(z) > 0}. Let Re(u) > 0 and Re(v) > 0. Z 1 Z ∞ Z ∞ tu−1 tu−1 tu−1 dt = dt dt . + u+v u+v (1 + t) (1 + t)u+v 0 (1 + t) {z } |0 {z } |1 II
I
I) We have
Z
Z
1
≤
1
≤
Z
1
≤
Z
1
|tu−1 (1 + t)−u−v | dt
0
|tu−1 |· |(1 + t)−u−v | dt
0
tRe(u)−1 (1 + t)−Re(u)−Re(v) dt
0
0
tRe(u)−1 < ∞
where the last step follows since Re(u) > 0 and the previous one since Re(u) > 0 and Re(v) > 0 and t ∈ (0, 1). This implies |tu−1 (1 + t)−u−v | is integrable on (0, 1). By Fubini’s Theorem for any {γ} ⊂ G × G Z Z γ
1 0
tu−1 dt dz = (1 + t)u+v
Z
1 0
Z
117
γ
tu−1 dz dt = 0 (1 + t)u+v
which implies Z
1 0
tu−1 dt ∈ H(G × G). (1 + t)u+v
II) We have Z
∞ 1
|t
u−1
(1 + t)
−u−v
| dt
≤ ≤ =
lim
n→∞
lim
n→∞
n
|tu−1 |· |(1 + t)−u−v | dt
Z
n
tRe(u)−1
n
lim
Z
n
lim
Z
lim
n→∞
≤
Z
|tu−1 (1 + t)−u−v | dt
1
tRe(u)−1 (1 + t)−Re(u)−Re(v) dt
lim
n→∞
=
n
Z1 n
n→∞
=
Z
n→∞
1
tRe(u)+Re(v)
1
1
1
1
t−1−Re(u) 1
t1+Re(v)
1 t
1 t
Re(u)+Re(v) +1
dt
1 Re(u)+Re(v) dt +1
dt < ∞
where the last step follows since Re(v) > 0 and the previous one since Re(u) + Re(v) > 0, t ≥ 1 and 1 therefore ≤ 1. This implies |tu−1 (1 + t)−u−v | is integrable on (1, ∞). By Fubini’s Theorem for Re (u)+Re(v) 1 ( t +1) any {γ} ⊂ G × G Z ∞Z Z Z ∞ tu−1 tu−1 dt dz = dz dt = 0 u+v u+v γ 1 (1 + t) 1 γ (1 + t) which implies
In summary
Z Z
∞
tu−1 dt ∈ H(G × G). (1 + t)u+v
∞
tu−1 dt ∈ H(G × G). (1 + t)u+v
1
0
Exercise 9. Let αn be the volume of the ball of radius one in Rn (n ≥ 1). Prove by induction and iterated integrals that αn = 2αn−1
Z
0
1
(1 − t2 )(n−1)/2 dt
Solution. Define the n-dimensional ball with radius r by Bn (r) = {x ∈ Rn : |x| ≤ r} and define the volume of Bn (r) by Vn (r). Clearly, V1 (r) = 2r, V2 (r) = πr2 , V3 (r) = V1 (1) = 2, V2 (1) and V3 = 43 π. Claim: For arbitrary n we have Vn (r) = rn Vn (1) 118
4 3 3 πr
and therefore
which implies Vn (r) = rn Vn (1) = rn αn
(7.14)
(def)
Assume the claim is true, then we write 2 Bn (1) = {x ∈ Rn : |x| ≤ 1} = {(x1 , x2 , . . . , xn−1 , t) ∈ Rn : x12 + x22 + . . . + xn−1 + t2 ≤ 1}.
For a fixed t ∈ [−1, 1] we have the balls √ 2 ≤ 1} Bn−1 1 − t2 = {(x1 , x2 , . . . , xn−1 ) ∈ Rn−1 : x12 + x22 + . . . + xn−1 and by (7.14) we obtain Vn−1
√
√ n−1 n−1 1 − t2 = 1 − t2 Vn−1 (1) = 1 − t2 2 αn−1 .
Clearly, the ball Bn (1) is the union of all disjoint balls Bn−1 αn = Vn (1) =
Z
1
−1
√ 1 − t2 as t varies over [−1, 1]. Hence
Z n−1 2 2 αn−1 dt = 2αn−1 1−t
0
1
n−1 1 − t2 2 dt
which proves the formula above. Finally, we prove the claim Vn (r) = rn Vn (1). We have by using a symmetry argument Z Z √2 2Z √2 2 2 Z √ 2 Pn−1 2 r −x1
r
Vn (r)
=
=
CoV:xi =ryi
= =
r −
2n
0
0
rn
Z
1 −1
1−y1 −y2
1−y1
r
=
r −x1 −x2
i=1
xi
1 dxn dxn−1 · · · dx2 dx1 ··· √ P √ √ 2 −r − r2 −x12 − r2 −x12 −x22 − r2 − n−1 i=1 xi Z r Z √r2 −x2 Z √r2 −x2 −x2 Z √r2 −Pn−1 2 i=1 xi 1 1 2 n 2 ··· 1 dxn dxn−1 · · · dx2 dx1 0 0 0 0 √ √ √ Pn−1 2 Z Z Z 2 Z 2 2 0
Z √1−y2 Z √1−y2 −y2
rn Vn (1).
1
−
√
1−y21
1
−
√
2
1−y21 −y22
1−
···
i=1
yi
0
Z √1−Pn−1 2 i=1 y i
···
−
√
1−
Pn−1
2 i=1 yi
dyn dyn−1 · · · dy2 dy1 1· |{z} rn Jacobian
1 dyn dyn−1 · · · dy2 dy1
This implies Vn (r) = rn Vn (1). (Note that this is kind of obvious, since we scale the unit ball in Rn by r in each dimension to obtain the ball with radius r in Rn ). We could have proved the formula by mathematical induction. Using spherical coordinates simplifies the computation if preferred. Exercise 10. Show that αn =
πn/2 (n/2)Γ(n/2)
where αn is defined in problem 9. Show that if n = 2k, k ≥ 1, then αn = πk /k! Solution. Not available.
119
Exercise 11. The Gaussian psi function is defined by Ψ(z) =
Γ0 (z) Γ(z)
(a) Show that Ψ is meromorphic in C with simple poles at z = 0, −1, . . . and Res(Ψ; −n) = −1 for n ≥ 0. (b) Show that Ψ(1) = −γ. (c) Show that Ψ(z + 1) − Ψ(z) = 1z . (d) Show that Ψ(z) − Ψ(1 − z) = −π cot πz. (e) State and prove a characterization of Ψ analogous to the Bohr-Mollerup Theorem. Solution. Not available.
7.8 The Riemann zeta function 1
Exercise 1. Let ξ(z) = z(z − l)π− 2 z ζ(z)Γ( 21 z) and show that ξ is an entire function which satisfies the functional equation ξ(z) = ξ(1 − z). Solution. The function ξ is analytic on C − ({−2k : k ∈ N0 } ∪ {1}) as a product of analytic functions. The 1 factors z, z − 1 and π− 2 z are entire. The zeta-function has a simple pole at z = 1, hence limz→1 (z − 1)ζ(z) exists in C and the function ξ is well-defined at z = 1. The function Γ 21 z has simple poles at the non-positive even integers. For z = 0, limz→0 zΓ 21 z ∈ C. For negative even integers z = 2k, −k ∈ N the poles of Gamma coincide with the simple zeros of the zetafunction, hence limz→2k ζ(z)Γ 21 z exists in C also. Therefore the function ξ is analytic at each point. The functional equation of ξ is related to the functional equation of ζ (8.3, p. 192), evaluated at 1 − z, i.e. ! 1 −z (7.15) ζ(1 − z) = 2(2π) Γ(z)ζ(z) sin π(1 − z) . 2 Furthermore the proof uses variants of problems 1, 2 of the homework set 5, in particular ! ! ! 1 z 1 z 1 Γ − + Γ = π csc π(1 − z) 2 2 2 2 2 ! √ z 1 z z−1 πΓ(z) = 2 Γ . + Γ 2 2 2
120
(7.16) (7.17)
Then the following chain of equalities holds. ! 1 z − ζ(1 − z)Γ ξ(1 − z) = (1 − z)(−z)π 2 2 ! ! 1 z 1−z −z 1 (7.15) − 12 + 2z = z(z − 1)π Γ − 2 π ζ(z)Γ(z) sin π(1 − z) 2 2 2 ! z 1 Γ(z)Γ 2 − 2 1 1 = ξ(z)π− 2 21−z sin π(1 − z) z 2 Γ 2 ! π csc 21 π(1 − z) 1 1 Γ(z) (7.16) − 2 1−z sin π(1 − z) = ξ(z)π 2 2 Γ z Γ 1+z − 12 + 2z
2
2
Γ(z) = ξ(z)π 21−z z Γ 2 Γ 12 + 2z
2
1 2
1 2z−1 1 z 1 (7.17) 1 Γ = ξ(z)π 2 21−z + 2 2 Γ 21 + 2z π 2
!
= ξ(z).
Exercise 2. Use Theorem 8.17 to prove that number of primes.
P
p−1 n = ∞. Notice that this implies that there are an infinite
Q 1 Solution. Euler’s Theorem asserts that for real values x > 1, ζ(x) = ∞ , where pn are the prime n=1 1−p−x n numbers. The left hand side is unbounded as x & 1 and so is the right hand side. P 1 Now the argument goes as follows. Seeking contradiction suppose that ∞ n=1 pn is finite. All summands P 1 are positive, therefore the sum converges absolutely. By Proposition 5.4, p.165, also ∞ n=1 log 1 − pn converges absolutely. But this implies that ! ! ∞ ∞ Y X 1 1 = log = log ζ(1) < ∞ − log 1 − pn 1 − pn n=1 n=1 which gives the desired contradiction because the log(ζ(z)) has a simple pole at z = 1. P d(n) Exercise 3. Prove that ζ 2 (z) = ∞ n=1 nz for Re z > 1, where d(n) is the number of divisors of n.
Solution. Not available.
Exercise 4. Prove that ζ(z)ζ(z − 1) = Solution. Not available. Exercise 5. Prove that ζ(z−1) ζ(z) = which are relatively prime to n.
P∞
σ(n) n=1 nz ,
ϕ(n) n=1 nz
P∞
for Re z > 1, where σ(n) is the sum of the divisors of n.
for Re z > 1, where ϕ(n) is the number of integers less than n and
Solution. Not available. P µ(n) km k1 k2 1 = ∞ Exercise 6. Prove that ζ(z) n=1 nz for Re z > 1, where µ(n) is defined as follows. Let n = p1 p2 · · · pm be the factorization of n into a product of primes p1 , . . . , pm and suppose that these primes are distinct. Let µ(1) = 1; if k1 = . . . = km = 1 then let µ(n) = (−1)m ; otherwise let µ(n) = 0. Solution. Not available. 121
0
(z) =− Exercise 7. Prove that ζζ(z) m ≥ 1; and Λ(n) = 0 otherwise.
P∞
n=1
Λ(n) nz
for Re z > 1, where Λ(n) = log p if n = pm for some prime p and
Solution. Not available. Exercise 8. (a) Let η(z) = ζ 0 (z)/ζ(z) for Re z > 1 and show that limz→z0 (z − z0 )η(z) is always an integer for Re z0 ≥ 1. Characterize the point z0 (in its relation to ζ) in terms of the sign of this integer. (b) Show that for > 0 ∞ X Re η(1 + + it) = − Λ(n)n−1(1+) cos(t log n) n=1
where Λ(n) is defined in Exercise 7. (c) Show that for all > 0, 3Re η(1 + ) + 4Re η(1 + + it) + Re η(1 + + 2it) ≤ 0. (d) Show that ζ(z) , 0 if Re z = 1 (or 0). Solution. Not available.
122
Chapter 8
Runge’s Theorem 8.1 Runge’s Theorem Exercise 1. Prove Corollary 1.14 if it is only assumed that E − meets each component of C∞ − G. Solution. Not available. Exercise 3. Let G be the open unit disk B(0; 1) and let K = {z : 14 ≤ |z| ≤ 34 }. Show that there is a function f analytic on some open subset G1 containing K which cannot be approximated on K by functions in H(G). Remarks. The next two problems are concerned with the following question. Given a compact set K contained in an open set G1 ⊂ G, can functions in H(G1 ) be approximated on K by functions in H(G)? Exercise 2 says that for an arbitrary choice of K, G, and G1 this is not true. Exercise 4 below gives criteria for a fixed K and G such that this can be done for any G1 . Exercise 3 is a lemma which is useful in proving Exercise 4. Solution. Not available. Exercise 3. Let K be a compact subset of the open set G and suppose that any bounded component D of G − K has D− ∩ ∂G , . Then every component of C∞ − K contains a component of C∞ − G. Solution. Not available. Exercise 4. Let K be a compact subset of the open set G; then the following are equivalent: (a) If f is analytic in a neighborhood of K and > 0 then there is a g in H(G) with | f (z) − g(z)| < for all z in K; (b) If D is a bounded component of G − K then D− ∩ ∂G , ; (c) If z is any point in G − K then there is a function f in H(G) with | f (z)| > sup{| f (w)| : w in K}. Solution. Not available. ˆ Exercise 5. Can you interpret part (c) of Exercise 4 in terms of K? Solution. Not available.
123
Exercise 6. Let K be a compact subset of the region G and define Kˆ G = {z ∈ G : | f (z)| ≤ k f kK for all f in H(G)}. ˆ (a) Show that if C∞ − G is connected then Kˆ G = K. ˆ (b) Show that d(K, C − G) = d(KG , C − G). (c) Show that Kˆ G ⊂ the convex hull of K ≡ the intersection of all convex subsets of C which contain K. (d) If Kˆ G ⊂ G1 ⊂ G and G1 is open then for every g in H(G1 ) and > 0 there is a function f in H(G) such that | f (z) − g(z)| < for all z in Kˆ G . (Hint: see Exercise 4.) (e) Kˆ G = the union of K and all bounded components of G − K whose closure does not intersect ∂G. Solution. Not available.
8.2 Simple connectedness Exercise 1. The set G = {reit : −∞ < t < 0 and 1 + et < r < 1 + 2et } is called a cornucopia. Show that G is simply connected. Let K = G− ; is intKˆ connected? Solution. he result follows once it is established that every closed piecewise smooth curve in G is 0homotopic. This requires the following subclaim: let z ∈ G then the argument of z is uniquely defined. Suppose otherwise then there are real r1 , r2 , t1 , t2 such that z = r1 eit1 = r2 eit2 . Suppose WLOG that t1 < t2 < 0 then there is a positive integer k such that t1 = 2kπ = t2 and therefore r2 > 1+et2 = 1+et1 ekπ > 1+2et1 > r1 contradicting the fact that |z| = r1 = r2 . Define 3 R : (−∞, 0) → G; R(t) = 1 + et . 2 Let γ : [0, 1] → G be an arbitrary closed piecewise smooth curve in G with radius function ρ(s) and argument function τ(s) such that γ(s) = ρ(s)eiτ(s) , and define Γ : [0, 1] × [0, 1] → G, Γ(s, 0) = γ(s), Γ(s, 1) = R(τ(s)), Γ(0, u) = Γ(1, u)
∀ u ∈ [0, 1].
Also define a function Θ : [0, 1] × [0, 1] → G,
Θ(s, 0) = R(τ(s)), !! 1 Θ(s, 1) = R τ , 2 Θ(0, u) = Θ(1, u)
∀ u ∈ [0, 1].
The function Θ ◦ Γ satisfies Definition IV 6.1 (p.88), the curve γ is homotopic to the point R τ 12 and by Definition IV 6.14 (p.93) the set G is simply connected. Exercise 2. If K is polynomially convex, show that the components of the interior of K are simply connected. Solution. Not available. 124
8.3 Mittag-Leffler’s Theorem Exercise 1. Let G be a region and let {an } and {bm } be two sequences of distinct points in G without limit points in G such that an an , bm for all n, m. Let S n (z) be a singular part at an and let pm be a positive integer. Show that there is a meromorphic function f on G whose only poles and zeros are {an } and {bm } respectively, the singular part at z = an is S n (z), and z = bm is a zero of multiplicity pm . Solution. Let G be a region and let {bm } be a sequence of distinct points in G with no limit point in G; and let {pm } be a sequence of integers. By Theorem 5.15 p.170 there is an analytic function g defined on G whose only zeros are at the points bm ; furthermore, bm is a zero of g of multiplicity pm . Since g ∈ H(g) and {an } ∈ G, g has a Taylor series in a neighborhood B(an ; Rn ) of each an , that is gn (z) =
∞ X k=0
where αk = sn (z) or
1 (k) k! g (an ).
rn (z)
∞ X k=0
αk (z − an )k ∈ B(an ; Rn )
Goal: Try to use this series to create a singular part rn (z) at an such that rn (z)gn (z) = αk (z − an )k =
mn X j=1
mn ∞ X X A jn A jn ⇐⇒ = αk rn (z)(z − an )k . j (z − an ) j (z − a ) n j=1 k=0
Claim: rn (z) =
mn X
B jk (z − an )− j−k
j=1
(8.1)
works. Proof of the claim: ∞ X k=0
αk rn (z)(z − an )k
=
∞ X
=
=
αk
k=0
mn X
B jk (z − an )− j−k (z − an )k
∞ X
αk
mn X
B jk (z − an )− j
k=0 mn X j=1
where the last step follows by choosing
∞ X
j=1
j=1
(z − an )− j
∞ X k=0
αk B jk =
mn X j=1
A jn (z − an ) j
αk B jk = A jn .
k=0
Since G is a region, G is open. Let {an } be a sequence of distinct points without a limit point in G and such that an , bm for all n, m. Let {rn (z)} be the sequence of rational functions given by rn (z) =
mn X j=1
B jk (z − an ) j+k
(see (8.1)). By Mittag-Leffler’s Theorem, there is a meromorphic function h on G whose poles are exactly the points {an } and such that the singular part of h at an is rn (z). Set f = g· h. Then by construction f is the meromorphic function on G whose only poles and zeros are {an } and {bm } respectively, the singular part at z = an is S n (z), and z = bm is a zero of multiplicity pm . (Note that the zeros do not cancel the poles since by assumption an , bm ∀ n, m). 125
Exercise 2. Let {an } be a sequence of points in the plane such that |an | → ∞, and let {bn } be an arbitrary sequence of complex numbers. (a) Show that if integers {kn } can be chosen such that !k ∞ X r n bn an an k=n
(3.4)
converges absolutely for all r > 0 then !k ∞ X r n bn an z − an k=n
(3.5)
converges in M(C) to a function f with poles at each point z = an . (b) Show that if lim sup |bn | < ∞ then (3.4) converges absolutely if kn = n for all n. (c) Show that if there is an integer k such that the series ∞ X bn ak+1 n=1 n
(3.6)
converges absolutely, then (3.4) converges absolutely if kn = k for all n. P (d) Suppose there is an r > 0 such that |an − am | ≥ r for all n , m. Show that |an |−3 < ∞. In particular, if the sequence {bn } is bounded then the series (3.6) with k = 2 converges absolutely. (This is somewhat involved and the reader may prefer to prove part (f) directly since this is the only application.) (e) Show that if the series (3.5) converges in M(C) to a meromorphic function f then ! !k −1 ∞ X bn z n z bn f (z) = + + ... + 1+ z − an an an an n=1
(f) Let ω and ω0 be two complex numbers such that Im(ω0 /ω) , 0. Using the previous parts of this exercise show that the series ! 1 1 z 1 X 0 , + + ζ(z) = + z z − w w w2 where the sum is over all w = 2nω + 2n0 ω for n, n = 0, ±1, ±2, . . . but not w = 0, is convergent in M(C) to a meromorphic function ζ with simple poles at the points 2nω + 2n0 ω0 . This function is called the Weierstrass zeta function. (g) Let P(z) = −ζ 0 (z); P is called the Weierstrass pe function. Show that ! 1 X 0 1 1 P(z) = 2 + − z (z − w)2 w2 where the sum is over the same w as in part (f). Also show that P(z) = P(z + 2nω + 2n0 ω0 ) for all integers n and n0 . That is, P is doubly periodic with periods 2ω and 2ω0 . Solution. Not available.
126
Exercise 3. This exercise shows how to deduce Weierstrass’s Theorem for the plane (Theorem VII. 5.12) from Mittag-Leffler’s Theorem. (a) Deduce from Exercises 2(a) and 2(b) that for any sequence {an } in C with lim an = ∞ and an , 0 there is a sequence of integers {kn } such that ! !k −1 ∞ X 1 1 z 1 z n 1 + + + ... + h(z) = z − an an an an an an n=1
is a meromorphic function on C with simple poles at a1 , a2 , . . .. The remainder of the proof consists of showing that there is a function f such that h = f 0 / f . This function f will then have the appropriate zeros. (b) Let z be an arbitrary but fixed point in C − {a1 , a2 , . . .}. Show that if γ1 and γ2 are any rectifiable curves in C − {a1 , a2 , . . .} from 0 to z and h is the function obtained in part (a), then there is an integer m such that Z Z h− h = 2πim. γ1
γ2
(c) Again let h be the meromorphic function from part (a). Prove that for z , a1 , a2 , . . . and γ any rectifiable curve in C − {a1 , a2 , . . .}, Z ! h
f (z) = exp
γ
defines an analytic function on C − {a1 , a2 , . . .} with f 0 / f = h. (That is, the value of f (z) is independent of the curve γ and the resulting function f is analytic. (d) Suppose that z ∈ {a1 , a2 , . . .}; show that z is a removable singularity of the function f defined in part (c). Furthermore, show that f (z) = 0 and that the multiplicity of this zero equals the number of times that z appears in the sequence {a1 , a2 , . . .}. (e) Show that ! !2 !k ∞ Y z 1 z 1 z n z (3.7) f (z) = exp + + ... + 1− an an 2 an 2 an n=1
Remark. We could have skipped parts (b), (c), and (d) and gone directly from (a) to (e). However this would have meant that we must show that (3.7) converges in H(C) and it could hardly be classified as a new proof. The steps outlined in parts (a) through (d) give a proof of Weierstrass’s Theorem without introducing infinite products. Solution. Not available. Exercise 4. This exercise assumes a knowledge of the terminology and results of Exercise VII. 5.11. (a) Define two functions f and g in H(G) to be relatively prime (in symbols, ( f, g) = 1) if the only common divisors of f and g are non-vanishing functions in H(G). Show that ( f, g) = 1 iff Z( f ) ∩ Z(g) = . (b) If ( f, g) = 1, show that there are functions f1 , g1 in H(G) such that f f1 + gg1 = 1 (Hint: Show that there is a meromorphic function ϕ on G such that f1 = ϕg ∈ H(G) and g|(1 − f f1 ).) (c) Let f1 , . . . , fn ∈ H(G) and g = g.c.d { f1 , . . . , fn }. Show that there are functions ϕ1 , . . . , ϕn in H(G) such that g = ϕ1 f1 + . . . + ϕn fn . (Hint: Use (b) and induction.) T (d) If {Jα } is a collection of ideals in H(G), show that J = α Jα is also an ideal. If S ⊂ H(G) then let J = ∩{S : S is an ideal of H(G) and S ⊂ J}. Prove that J is the smallest ideal in H(G) that contains S and J = {ϕ1 f1 + . . . + ϕn fn : ϕk ∈ H(G), fk ∈ S for 1 ≤ k ≤ n}. J is called the ideal generated by S and is denoted by J = (S). If S is finite then (S) is called a finitely generated ideal. If S = { f } for a single function f then ( f ) is called a principal ideal. (e) Show that every finitely generated ideal in H(G) is a principal ideal. 127
(f) An ideal J is called a fixed ideal if Z(J) , ; otherwise it is called a free ideal. Prove that if J = (S) then Z(J) = Z(S) and that a proper principal ideal is fixed. (g) Let fn (z) = sin(2−n z) for all n ≥ 0 and let J = ({ f1 , f2 , . . .}). Show that J is a fixed ideal in H(C) which is not a principal ideal. (h) Let J be a fixed ideal and prove that there is an f in H(G) with Z( f ) = Z(J) and J ⊂ ( f ). Also show that J = ( f ) if J is finitely generated. (i) Let M be a maximal ideal that is fixed. Show that there is a point a in G such that M = ((z − a)). (j) Let {an } be a sequence of distinct points in G with no limit point in G. Let J = { f ∈ H(G) : f (an ) = 0 for all but a finite number of the an }. Show that J is a proper free ideal in H(G). (k) If J is a free ideal show that for any finite subset S of J, Z(S) , . Use this to show that J can contain no polynomials. l) Let J be a proper free ideal; then J is a maximal ideal iff whenever g ∈ H(G) and Z(g) ∩ Z( f ) , for all f in J then g ∈ J. Solution. Not available. Exercise 5. Let G be a region and let {an } be a sequence of distinct points in G with no limit point in G. For each integer n ≥ 1 choose integers kn ≥ 0 and constants An(k) , 0 ≤ k ≤ kn . Show that there is an analytic function f on G such that f (k) (an ) = k!An(k) . (Hint: Let g be an analytic function on G with a zero at an of multiplicity kn + 1. Let h be a meromorphic function on G with poles at each an of order kn + 1 and with singular part S n (z). Choose the S n so that f = gh has the desired property.) Solution. Not available. Exercise 6. Find a meromorphic function with poles of order 2 at 1, √ each pole is 0 and lim(z − n)2 f (z) = 1 for all n.
√ √ 2, 3, . . . such that the residue at
Solution. We claim that the function √ ∞ X 3 nz2 − 2z3 √ n3/2 (z − n)2 k=1 √ has the desired properties. f has poles of order 2 at z = n, n ∈ N (and no other poles) and it is analytic for every z ∈ C that is√not one of the poles. Moreover f converges in M(C). To see this let r > 0 and choose N ∈ N so large that n > 2r for all n ≥ N. Then for |z| ≤ r and n ∈ N √ √ √ 3 nz2 − 2z3 r2 (3 n + n) 16r2 | 3/2 ≤ 2 =: Mn . √ 2 | ≤ 3/2 √ n n (z − n) n ( n − r)2 f (z) =
This is the majorant independent of z for the Weierstrass’ Criterion. √ Next fix m ∈ N and consider the limit in equation limz→ √n (z − n)2 f (z) = 1, √ ∞ √ 2 √ 2X 3 nz2 − 2z3 m) f (z) = lim m) lim √ √ (z − √ (z − z→ m z→ m n3/2 (z − n)2 n=1 √ √ m−1 X (z − m)(3 nz2 − 2z3 ) 3m3/2 − 2m3/2 + √ m3/2 z→ m n3/2 (z − n)2 n=1 √ √ ∞ X (z − n)(3 nz2 − 2z3 ) + lim √ √ z→ m n3/2 (z − n)2 m+1
= lim √
= 0 + 1 + 0 = 1. 128
Lastly we have √ to check that the residuals are 0 at the poles. By Proposition V 2.4, p.113, for a function g(z) := (z − m)2 f (z), it has to be verified that g0 (z)|z= √m = 0. First find the derivative of f as √ ∞ X 6 nz2 − 6nz − 2z3 0 f (z) = √ n3/2 (z − m)3 n=1 √ then compute the residuum at z = m √ √ √ Res( f, m) = g0 (z)|z= √m = 2(z − m) f (z) + (z − m)2 f 0 (z) z= √m ∞ √ √ X 2(z − m)(3 nz2 − 2z3 ) = √ n3/2 (z − n)2 n=1 √ √ ∞ X (z − m)2 (6 nz2 − 6nz − 2z3 ) + √ n3/2 (z − m)3 √ n=1 z= m ∞ p X √ √ √ = 2(z − (n))(3 nz2 − 2z3 ) + (z − m)(6 nz2 ) − 6nz − 2z3 ) n=1
# √ z− m · 3/2 √ n (z − n)3 z= √m
√ √ m, the last factor equals zero, in the case n = m, (z − m)2 cancels directly and we are √ √ 12 m − 2z3 − 6mz −6z(z − m) = 0. = √ n3/2 n3/2 (z − m) z=m z=m √ The residuum vanishes for all m, m ∈ N and therefore the function considered in this exercise has all the required properties. If n , m and z = left with
129
Chapter 9
Analytic Continuation and Riemann Surfaces 9.1 Schwarz Reflection Principle Exercise 1. Let γ be a simple closed rectifiable curve with the property that there is a point a such that for all z on γ the line segment [a, z] intersects {γ} only at z; i.e. [a, z] ∩ {γ} = {z}. Define a point w to be inside γ if [a, w] ∩ {γ} = and let G be the collection of all points that are inside γ. (a) Show that G is a region and G− = G ∪ {γ}. R (b) Let f : G− → C be a continuous function such that f is analytic on G. Show that γ f = 0. (c) Show that n(γ; z) = ±1 if z is inside γ and n(γ; z) = 0 if z < G− . Remarks. It is not necessary to assume that γ has such a point a as above; each part of this exercise remains true if γ is only assumed to be a simple closed rectifiable curve. Of course, we must define what is meant by the inside of γ. This is difficult to obtain. The fact that a simple closed curve divides the plane into two pieces (an inside and an outside) is the content of the Jordan Curve Theorem. This is a very deep result of topology. Solution. Not available. Exercise 2. Let G be a region in the plane that does not contain zero and let G∗ be the set of all points z such that there is a point w in G where z and w are symmetric with respect to the circle |ξ| = 1. (See III. 3.17.) (a) Show that G∗ = {z : (1/¯z) ∈ G}. (b) If f : G → C is analytic, define f ∗ : G∗ → C by f ∗ (z) = f (1/¯z). Show that f ∗ is analytic. (c) Suppose that G = G∗ and f is an analytic function defined on G such that f (z) is real for z in G with |z| = 1. Show that f = f ∗ . (d) Formulate and prove a version of the Schwarz Reflection Principle where the circle |ξ| = 1 replaces R. Do the same thing for an arbitrary circle. Solution. Not available. Exercise 3. Let G, G+ , G− , G0 be as in the statement of the Schwarz Reflection Principle and let f : G+ ∪ G0 → C∞ be a continuous function such that f is meromorphic on G+ . Also suppose that for x in G0 f (x) ∈ R. Show that there is a meromorphic function g : G → C∞ such that g(z) = f (z) for z in G+ ∪ G0 . Is it possible to allow f to assume the value ∞ on G0 ? 130
Solution. Not available.
9.2 Analytic Continuation Along a Path Exercise 1. The collection {D0 , D1 , . . . , Dn } of open disks is called a chain of disks if D j−1 ∩ D j , for 1 ≤ j ≤ n. If {( f j , D j ) : 0 ≤ j ≤ n} is a collection of function elements such that {D0 , D1 , . . . , Dn } is a chain of disks and f j−1 (z) = f j (z) for z in D j−1 ∩ D j , 1 ≤ j ≤ n; then {( f j , D j ) : 0 ≤ j ≤ n} is called an analytic continuation along a chain of disks. We say that ( fn , Dn ) is obtained by an analytic continuation of ( f0 , D0 ) along a chain of disks. (a) Let {( f j , D j ) : 0 ≤ j ≤ n} be an analytic continuation along a chain of disks and let a and b be the centers of the disks D0 and Dn respectively. Show that there is a path γ from a to b and an analytic continuation S {(gt , Bt )} along γ such that {γ} ⊂ nj=0 D j , [ f0 ]a = [g0 ]a and [ fn ]b = [g1 ]b . (b) Conversely, let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along a path γ : [0, 1] → C and let a = γ(0), b = γ(1). Show that there is an analytic continuation along a chain of disks {(g j , B j ) : 0 ≤ j ≤ n} S such that {γ} ⊂ nj=0 B j , [ f0 ]a = [g0 ]a and [ f1 ]b = [gn ]b . Solution. Not available. √ Exercise 2. Let D0 = B(1; 1) and let f0 be the restriction of the principal branch of z to D0 . Let γ(t) = exp(2πit) and σ(t) = exp(4πit) for 0 ≤ t ≤ 1. (a) Find an analytic continuation {( ft , Dt ) : 0 ≤ t ≤ 1} of ( f0 , D0 ) along γ and show that [ f1 ]1 = [− f0 ]1 . (b) Find an analytic continuation {(gt , Bt ) : 0 ≤ t ≤ 1} of ( f0 , D0 ) along σ and show that [g1 ]1 = [g0 ]1 . Solution. Not available. Exercise 3. Let f be an entire function, D0 = B(0; 1), and let γ be a path from 0 to b. Show that if {( ft , Dt ) : 0 ≤ t ≤ 1} is a continuation of ( f, D0 ) along γ then f1 (z) = f (z) for all z in D1 (This exercise is rather easy; it is actually an exercise in the use of the terminology.) Solution. Not available. Exercise 4. Let γ : [0, 1] → C be a path and let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along γ. Show that {( ft0 , Dt ) : 0 ≤ t ≤ 1} is also a continuation along γ. Solution. Not available. Exercise 5. Suppose γ : [0, 1] → C is a closed path with γ(0) = γ(1) = a and let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along γ such that [ f1 ]a = [ f00 ]a and f0 , 0. What can be said about ( f0 , D0 )? Solution. Not available. Exercise 6. Let D0 = B(1; 1) and let f0 be the restriction to D0 of the principal branch of the logarithm. For an integer n let γ(t) = exp(2πint), 0 ≤ t ≤ 1. Find a continuation {( ft , Dt ) : 0 ≤ t ≤ 1} along γ of ( f0 , D0 ) and show that [ f1 ]1 = [ f0 + 2πin]. Solution. Not available. Exercise 7. Let γ : [0, 1] → C be a path and let {( ft , Dt ) : 0 ≤ t ≤ 1} be an analytic continuation along γ. Suppose G is a region such that ft (Dt ) ⊂ G for all t, and suppose there is an analytic function h : G → C such that h( f0 (z)) = z for all z in D0 . Show that h( ft (z)) = z for all z in Dt and for all t. Hint: Show that T = {t : h( ft (z)) = z for all z in Dt } is both open and closed in [0, 1]. 131
Solution. Not available. Exercise 8. Let γ : [0, 1] → C be a path with γ(0) = 1 and γ(t) , 0 for any t. Suppose that {( ft , Dt ) : 0 ≤ t ≤ 1} is an analytic continuation of f0 (z) = log z. Show that each ft is a branch of the logarithm. Solution. Not available.
9.3 Monodromy Theorem Exercise 1. Prove that the set T defined in the proof of Lemma 3.2 is closed. Solution. Not available. Exercise 2. Let ( f, D) be a function element and let a ∈ D. If γ : [0, 1] → C is a path with γ(0) = a and γ(1) = b and {( ft , Dt ) : 0 ≤ t ≤ 1} is an analytic continuation of ( f, D) along γ, let R(t) be the radius of convergence of the power series expansion of ft at z = γ(t). (a) Show that R(t) is independent of the choice of continuation. That is, if a second continuation {(gt , Bt )} along γ is given with [g0 ]a = [ f ]a and r(t) is the radius of convergence of the power series expansion of gt about z = γ(t) then r(t) = R(t) for all t. (b) Suppose that D = B(1; 1), f is the restriction of the principal branch of the logarithm to D, and γ(t) = 1 + at for 0 ≤ t ≤ 1 and a > 0. Find R(t). (c) Let ( f, D) be as in part (b), let 0 < a < 1 and let γ(t) = (1 − at) exp(2πit) for 0 ≤ t ≤ 1. Find R(t). (d) For each of the functions R(t) obtained in parts (b) and (c), find min{R(t) : 0 ≤ t ≤ 1} as a function of a and examine the behavior of this function as a → ∞ or a → 0. Solution. Not available. Exercise 3. Let Γ : [0, 1] × [0, 1] → C be a continuous function such that Γ(0, u) = a, Γ(1, u) = b for all u. Let γu (t) = Γ(t, u) and suppose that {( ft,u , Dt,u ) : 0 ≤ t ≤ 1} is an analytic continuation along γu such that [ f0,u ]a = [ f0,v ]a for all u and v in [0, 1]. Let R(t, u) be the radius of convergence of the power series expansion of ft,u about z = Γ(t, u). Show that either R(t, u) ≡ ∞ or R : [0, 1] × [0, 1] → (0, ∞) is a continuous function. Solution. Not available. Exercise 4. Use Exercise 3 to give a second proof of the Monodromy Theorem. Solution. Not available.
9.4 Topological Spaces and Neighborhood Systems Exercise 1. Prove the propositions which were stated in this section without proof. Solution. Not available. Exercise 2. Let (X, T ) and (Ω, S) be topological spaces and let Y ⊂ X. Show that if f : X → Ω is a continuous function then the restriction of f to Y is a continuous function of (Y, TY ) into (Ω, S). Solution. Not available.
132
Exercise 3. Let X and Ω be sets and let {N x : x ∈ X} and {Mω : ω ∈ Ω} be neighborhood systems and let T and S be the induced topologies on X and Ω respectively. (a) Show that a function f : X → Ω is continuous iff when x ∈ X and ω = f (x), for each ∆ in Mω there is a U in N x such that f (U) ⊂ ∆. (b) Let X = Ω = C and let Nz = Mz = {B(z; ) : > 0} for each z in C. Interpret part (a) of this exercise for this particular situation. Solution. Not available. Exercise 4. Adopt the notation of Exercise 3. Show that a function f : X → Ω is open iff for each x in X and U in N x there is a set ∆ in Mω (where ω = f (x)) such that ∆ ⊂ f (U). Solution. Not available. Exercise 5. Adopt the notation of Exercise 3. Let Y ⊂ X and define UY = {Y ∩ U : U ∈ NY } for each y in Y. Show that {UY : y ∈ Y} is a neighborhood system for Y and the topology it induces on Y is TY . Solution. Not available. Exercise 6. Adopt the notation of Exercise 3. For each point (x, ω) in X × Ω let U(x,ω) = {U × ∆ : U ∈ N x , ∆ ∈ Mω } (a) Show that {U(x,ω) : (x, ω) ∈ X × Ω} is a neighborhood system on X × Ω and let P be the induced topology on X × Ω. (b) If U ∈ T and ∆ ∈ S, call the set U × ∆ an open rectangle. Prove that a set is in P iff it is the union of open rectangles. (c) Define p1 : X × Ω → X and p2 : X × Ω → X by p1 (x, ω) = x and p2 (x, ω) = ω. Show that p1 and p2 are open continuous maps. Furthermore if (Z, R) is a topological space show that a function f : (Z, R) → (X × Ω, P) is continuous iff p1 ◦ f : Z → X and p2 ◦ f : Z → Ω are continuous. Solution. Not available.
9.5 The Sheaf of Germs of Analytic Functions on an Open Set Exercise 1. Define F : S(C) → C by F(z, [ f ]z ) = f (z) and show that F is continuous. Solution. Not available. Exercise 2. Let F be the complete analytic function obtained from the principal branch of the logarithm and let G = C − {0}. If D is an open subset of G and f : D → C is a branch of the logarithm show that [ f ]a ∈ F for all a in D. Conversely, if ( f, D) is a function element such that [ f ]a ∈ F for some a in D, show that f : D → C is a branch of the logarithm. (Hint: Use Exercise 2.8.) Solution. Not available. Exercise 3. Let G = C − {0}, let F be the complete analytic function obtained from the principal branch of the logarithm, and let (R, ρ) be the Riemann surface of F (so that G is the base of R). Show that R is homeomorphic to the graph Γ = {(z, ez ) : z ∈ G} considered as a subset of C × C. (Use the map h : R → Γ defined by h(z, [ f ]z ) = ( f (z), z) and use Exercise 2.) State and prove an analogous result for branches of z1/n . 133
Solution. Not available. Exercise 4. Consider the sheaf S(C), let B = {z : |z − 1| < 1}, let l be the principal branch of the logarithm defined on B, and let l1 (z) = l(z) + 2πi for all z in B. (a) Let D = {z : |z| < 1} and show that ( 21 , [l] 21 ) and ( 12 , [l1 ] 12 ) belong to the same component of ρ−1 (D). (b) Find two disjoint open subsets of S(C) each of which contains one of the points ( 21 , [l] 21 ) and ( 21 , [l1 ] 21 ). Solution. Not available.
9.6 Analytic Manifolds Exercise 1. Show that an analytic manifold is locally compact. That is, prove that if a ∈ X and U is an open neighborhood of a then there is an open neighborhood V of a such that V − ⊂ U and V − is compact. Solution. Not available. Exercise 2. Which of the following are analytic manifolds? What is its analytic structure if it is a manifold? (a) A cone in R3 . (b) {(x1 , x2 , x3 ) ∈ R3 : x12 + x22 + x32 = 1 or x12 + x22 > 1 and x3 = 0}. Solution. Not available. Exercise 3. The following is a generalization of Proposition 6.3(b). Let (X, Φ) be an analytic manifold, let Ω be a topological space, and suppose there is a continuous function h of X onto Ω that is locally one-one (that is, if x ∈ X there is an open set U such that x ∈ U and h is one-one on U). If (U, ϕ) ∈ Φ and h is one-one on U let ∆ = h(U) and let ψ : ∆ → C be defined by ψ(ω) = ϕ ◦ (h/U)−1 (ω). Let Ψ be the collection of all such pairs (∆, ψ). Prove that (Ω, Ψ) is an analytic manifold and h is an analytic function from X to Ω. Solution. Not available. Exercise 4. Let T = {z : |z| = 1} × {z : |z| = 1}; then T is a torus. (This torus is homeomorphic to the usual hollow doughnut in R3 .) If ω and ω0 are complex numbers such that Im (ω/ω0 ) , 0 then ω and ω0 , considered as elements of the vector space C over R, are linearly independent. So each z in C can be 0 uniquely represented as z = tω + t0 ω0 ; t, t0 in R. Define h : C → T by h(tω + tω0 ) = (e2πit , e2πit ). Show that h induces an analytic structure on T . (Use Exercise 3.) (b) If ω, ω0 and ζ, ζ 0 are two pairs of complex numbers 0 0 such that Im (ω/ω0 ) , 0 and Im (ζ/ζ 0 ) , 0, define σ(sζ + s0 ζ 0 ) = (e2πis , e2πis ) and τ(tω+t0 ω0 ) = (e2πit , e2πit ). Let G = {tω + t0 ω0 : 0 < t < 1, 0 < t0 < 1} and Ω = {sζ + s0 ζ 0 : 0 < s < 1, 0 < s0 < 1}; show that both σ and τ are one-one on G and Ω respectively. (Both G and Ω are the interiors of parallelograms.) If Φτ and Φσ are the analytic structures induced on T by τ and σ respectively, and if the identity map of (T, Φτ ) into (T, Φσ ) is analytic then show that the function f : G → Ω defined by f = σ−1 ◦ τ is analytic. (To say that the identity map of (T, Φτ ) into (T, Φσ ) is analytic is to say that Φτ and Φσ are equivalent structures.) (c) Let ω = 1, ω0 = i, ζ = 1, ζ = α where Im α , 0; define σ, τ, G, Ω and f as in part (b). Show that Φτ and Φσ are equivalent analytic structures if and only if α = i. (Hint: Use the Cauchy-Riemann equations.) (d) Can you generalize part (c)? Conjecture a generalization? Solution. Not available. Exercise 5. (a) Let f be a meromorphic function defined on C and suppose f has two independent periods ω and ω0 . That is, f (z) = f (z + nω + n0 ω0 ) for all z in C and all integers n and n0 , and Im (ω/ω0 ) , 0. Using the notation of Exercise 4(a) show that there is an analytic function F : T → C∞ such that f = F ◦ h. (For an example of a meromorphic function with two independent periods see Exercise VIII. 4.2(g).) (b) Prove that there is no non-constant entire function with two independent periods. 134
Solution. Not available. Exercise 6. Show that an analytic surface is arcwise connected. Solution. Not available. Exercise 7. Suppose that f : C∞ → C∞ is an analytic function. (a) Show that either f ≡ ∞ or f −1 (∞) is a finite set. (b) If f . ∞, let a1 , . . . , an be the points in C where f takes on the value ∞. Show that there are polynomials p0 , p1 , . . . , pn such that ! n X 1 pk f (z) = p0 (z) + z − ak k=1 for z in C. (c) If f is one-one, show that either f (z) = az + b (some a, b in C) or f (z) =
a z−c
+ b (some a, b, c in C).
Solution. Not available. Exercise 8. Furnish the details of the discussion of the surface for
√ z at the end of this section.
Solution. Not available. o n Exercise 9. Let G = z : − π2 < Re z < π2 and define f : G → C by f (z) = sin z. Give a discussion for f √ similar to the discussion of z at the end of this section. Solution. Not available.
9.7 Covering spaces Exercise 1. Suppose that (X, ρ) is a covering space of Ω and (Ω, π) is a covering space of Y; prove that (X, π ◦ ρ) is also a covering space of Y. Solution. Not available. Exercise 2. Let (X, ρ) and (Y, σ) be covering spaces of Ω and Λ respectively. Define ρ × σ : X × Y → Ω × Λ by (ρ × σ)(x, y) = (ρ(x), σ(y)) and show that (X × Y, ρ × σ) is a covering space of Ω × Λ. Solution. Not available. Exercise 3. Let (Ω, ψ) be an analytic manifold and let (X, ρ) be a covering space of Ω. Show that there is an analytic structure Φ on X such that ρ is an analytic function from (X, Φ) to (Ω, ψ). Solution. Not available. Exercise 4. Let (X, ρ) be a covering space of Ω and let ω ∈ Ω. Show that each component of ρ−1 (ω) consists of a single point and ρ−1 (ω)) has no limit points in X. Solution. Not available. Exercise 5. Let Ω be a pathwise connected space and let (X, ρ) be a covering space of Ω. If ω1 and ω2 are points in Ω, show that ρ−1 (ω1 ) and ρ−1 (ω2 ) have the same cardinality. (Hint: Let γ be a path in Ω from ω1 to ω2 ; if x1 ∈ ρ−1 (ω1 ) and γ˜ is the lifting of γ with initial point x1 = γ˜ (0), let f (x1 ) = γ˜ (1). Show that f is a one-one map of ρ−1 (ω1 ) onto ρ−1 (ω2 ).) 135
Solution. Not available. Exercise 6. In this exercise all spaces are regions in the plane. (a) Let (G, f ) be a covering of Ω and suppose that f is analytic; show that if Ω is simply connected then f is one-one. (Hint: If f (z1 ) = f (z2 ) let γ be a path in G from z1 to z2 and consider a certain analytic continuation along f ◦ γ; apply the Monodromy Theorem.) (b) Suppose that (G1 , f1 ) and (G2 , f2 ) are coverings of the region Ω such that both f1 and f2 are analytic. Show that if G1 is simply connected then there is an analytic function f : G1 → G2 such that (G1 , f ) is a covering of G2 and f1 = f2 ◦ f . That is the diagram is commutative. (c) Let (G1 , f1 ), (G2 , f2 ) and Ω be as in part (b) and, in addition, assume that both G1 and G2 are simply connected. Show that there is a one-one analytic function mapping G1 onto G2 . Solution. Not available. Exercise 7. Let G and Ω be regions in the plane and suppose that f : G → Ω is an analytic function such that (G, f ) is a covering space of Ω. Show that for every region Ω1 contained in Ω which is simply connected there is an analytic function g : Ω1 → G such that f (g1 (ω)) = ω for all ω in Ω1 . Solution. Not available. Exercise 8. What is a simply connected covering space of the figure eight? Solution. Not available. Exercise 9. Give two nonhomeomorphic covering spaces of the figure eight that are not simply connected. Solution. Not available. Exercise 10. Prove that the closed curve in Exercise IV.6.8 is not homotopic to zero in the doubly punctured plane. Solution. Not available.
136
Chapter 10
Harmonic Functions 10.1 Basic properties of harmonic functions Exercise 1. Show that if u is harmonic then so are u x =
∂u ∂x
and uy =
∂u ∂y .
Solution. Let u be harmonic, that is u xx + uyy = 0.
(10.1)
Since u : G → R is harmonic, u is infinitely times differentiable by Proposition 1.3 p. 252. Thus, the third partial derivatives of u exist and are continuous (therefore u xyy = uyyx and uyxx = u xxy ). Now, we show that u x is harmonic, that is (u x ) xx + (u x )yy = 0. (u x ) xx + (u x )yy = u xxx + u xyy = u xxx + uyyx = (u xx + uyy ) x
= 0 x = 0. (10.1)
Thus, u x is harmonic. Finally, we show that uy is harmonic, that is (uy ) xx + (uy )yy = 0. (uy ) xx + (uy )yy = uyxx + uyyy = u xxy + uyyy = (u xx + uyy )y
= 0y = 0. (10.1)
Thus, uy is harmonic. Exercise 2. If u is harmonic, show that f = u x − iuy is analytic. Solution. Since, u is harmonic, we have u xx + uyy = 0.
(10.2)
In addition, u is infinitely times differentiable by Proposition 1.3 p. 252. Thus, all the partials are continuous (therefore u xy = uyx ). To show: f = u x − iuy is analytic which is equivalent to showing a) Re( f ) = u x and Im( f ) = −uy are harmonic and b) u x and −uy have to satisfy the Cauchy Riemann equations (by Theorem III 2.29). 137
a) By the previous exercise, we have seen that u x and uy are harmonic. Therefore, −uy is harmonic, too. Thus, part a) is proved. b) We have to show that the Cauchy-Riemann equations are satisfied, that is (u x ) x = −(uy )y
(u x )y = −(−uy ) x .
and
The first is true by (10.2) since (u x ) x = −(uy )y ⇐⇒ u xx + uyy = 0. The second is true since u xy = uyx . Therefore (u x )y = −(−uy ) x ⇐⇒ u xy = uyx ⇐⇒ u xy = uyx . In summary, f = u x − iuy is analytic provided u is harmonic. P Exercise 3. Let p(x, y) = nk,l=0 akl xk yl for all x, y in R. Show that p is harmonic iff: (a) k(k − 1)ak,l−2 + l(l − 1)ak−2,l = 0 for 2 ≤ k, l ≤ n; (b) an−1,l = an,l = 0 for 2 ≤ l ≤ n; (c) ak,n−1 = ak,n = 0 for 2 ≤ k ≤ n. Solution. Not available. Exercise 4. Prove that a harmonic function is an open map. (Hint: Use the fact that the connected subsets of R are intervals.) Solution. Not available. Exercise 5. If f is analytic on G and f (z) , 0 for any z show that u = log | f | is harmonic on G. Solution. Not available. ¯ R) ⊂ G. Show that Exercise 6. Let u be harmonic in G and suppose B(a; " 1 u(x, y) dx dy. u(a) = πR2 ¯ B(a;R) ¯ R) ⊂ D ⊂ G. There exist an f ∈ H(D) such that Re( f ) = u on D. Solution. Let D be a disk such that B(a; From Cauchy’s Integral Formula, we have f (a) =
1 2πi
Z
γ
f (w) 1 dw = w−a 2π
Z
2π
f (a + reiθ ) dθ
0
where γ(t) = a + reit , 0 ≤ t ≤ 2π. So 1 f (a) = 2π
Z
2π
1 f (a)r = 2π
Z
2π
f (a + reiθ ) dθ.
0
Multiply by r, we get f (a + reiθ )r dθ 0
138
and then integrate with respect to r yields Z 2π Z R 1 f (a + reiθ )r dθ dr f (a)r dr = 0 0 2π 0 Z R Z 2π r2 1 f (a) = f (a + reiθ )r dθ dr 2 2π 0 0 Z 2π Z R 1 f (a) = f (a + reiθ )r dr dθ πR2 0 0 " 1 f (x, y) dx dy. f (a) = πR2 ¯ B(a;R)
Z ⇐⇒ ⇐⇒ ⇐⇒
R
where the last step follows by changing from polar coordinates to rectangular coordinates. Now " " 1 1 u(a) = Re( f (a)) = Re( f (x, y)) dx dy = u(x, y) dx dy. πR2 πR2 ¯ ¯ B(a;R) B(a;R) Thus u(a) = Exercise 7. For |z| < 1 let
1 πR2
"
u(x, y) dx dy. ¯ B(a;R)
! 1 + z 2 . u(z) = Im 1−z
Show that u is harmonic and limr→1− u(reiθ ) = 0 for all θ. Does this violate Theorem 1.7? Why? 2 Solution. Let D = {z ∈ C : |z| < 1}. Let f (z) = 1+z 1−z . Clearly, f ∈ H(D) since z = 1 is not in D. By Theorem III 2.29 we obtain that v = Re( f ) and u = Im( f ) are harmonic. Therefore ! 1 + z 2 u(z) = Im 1−z is harmonic. Next, we will show
lim u(reiθ ) = 0
r→1−
139
for all θ. We have iθ
u(re ) = =
= = = = = =
! 1 + reiθ 2 Im 1 − reiθ ! ! 1 + reiθ 2 1 − reiθ 2 Im 1 − reiθ 1 − reiθ ! (1 + reiθ )(1 − reiθ ) 2 Im (1 − reiθ )(1 − reiθ ) " # (1 + reiθ − re−iθ − r2 )2 Im (1 − reiθ − re−iθ + r2 )2 # " (1 + 2ri sin θ − r2 )2 Im (1 − 2r cos θ + r2 )2 # " ((1 − r2 ) + 2ri sin θ)2 Im (1 − 2r cos θ + r2 )2 " # (1 − r2 )2 + 4ri sin θ(1 − r2 ) − 4r2 sin2 θ Im (1 − 2r cos θ + r2 )2 4r(1 − r2 ) sin θ . (1 − 2r cos θ + r2 )2
Note that the denominator is zero iff r = 1 and θ = 2πk, k ∈ Z ⇐⇒ z = 1. If we assume, that r = 1 and θ , 2πk, k ∈ Z, then lim− u(reiθ ) = u(eiθ ) =
r→1
4· 1· (1 − 12 ) sin θ 0 = 2 = 0. 2 2 (1 − 2· 1· cos θ + 1 ) 2 (1 − cos θ)2
If θ = 2πk, k ∈ Z, then lim− u(re2πk ) = lim− r→1
r→1
4r(1 − r2 ) sin(2πk) 0 = lim− = 0. 2 2 r→1 (1 − r)4 (1 − 2r cos(2πk) + r )
Thus, lim u(reiθ ) = 0.
r→1−
This does not violate Theorem 1.7. The reason is the following: If we pick u = u and v = 0, then the condition lim sup u(z) ≤ 0 z→a
does not hold ∀a ∈ ∂D. Claim: lim supz→1 u(z) > 0. Proof of the claim:
If we fix x = 1, then
! 2 2 1 + x + iy 2 = 4y(1 − x − y ) . u(z) = u(x, y) = Im 1 − x − iy (1 − 2x + x2 + y2 )2 u(1, y) =
4y(−y2 ) −4y3 4 = 4 =− y y4 y
But − 4y → ∞ as y → 0− . 140
∀y , 0.
Exercise 8. Let u : G → R be a function with continuous second partial derivatives and define U(r, θ) = u(r cos θ, r sin θ). (a) Show that # ! " 2 2 ∂U ∂2 U ∂ ∂U ∂2 U ∂ u ∂2 u 2∂ U = r + + r = r r + 2. + r2 2 2 2 2 ∂r ∂r ∂r ∂x ∂y ∂r ∂θ ∂θ So if 0 < G then u is harmonic iff
! ∂ ∂U ∂2 U r r + 2 = 0. ∂r ∂r ∂θ
(b) Let u have the property that it depends only on |z| and not arg z. That is, u(z) = ϕ(|z|). Show that u is harmonic iff u(z) = a log |z| + b for some constants a and b. Solution. Not available. Exercise 9. Let u : G → R be harmonic and let A = {z ∈ G : u x (z) = uy (z) = 0}; that is, A is the set of zeros of the gradient of u. Can A have a limit point in G? Solution. Let u be harmonic and not constant. By Exercise 1 p. 255, we know that u x and uy are harmonic. By exercise 2 p. 255, we know that f = u x − iuy is analytic. Define B = {z ∈ G : f (z) = 0} = {z ∈ G : u x (z) − iuy (z) = 0}. Clearly A = {z ∈ G : u x (z) = uy (z) = 0} = B since u x (z) − iuy (z) = 0 iff u x (z) = uy (z) = 0 (u x (z) and uy (z) are real). Since f is analytic, the zeros are isolated and therefore B cannot have a limit point in G (Corollary 3.10 p. 79). This implies that A cannot have a limit point in G. If u is constant, then A = B = G and thus A can have a limit point in G. (In this case f ≡ 0 and by Theorem 3.7 p. 78 A has a limit point in G). Exercise 10. State and prove a Schwarz Reflection Principle for harmonic functions. Solution. Not available. Exercise 11. Deduce the Maximum Principle for analytic functions from Theorem 1.6. Solution. Not available.
10.2 Harmonic functions on a disk Exercise 1. Let D = {z : |z| < 1} and suppose that f : D− → C is a continuous function such that both Re f and Im f are harmonic. Show that Z π 1 f (eit )Pr (θ − t) dt f (reiθ ) = 2π −π for all reiθ in D. Using Definition 2.1 show that f is analytic on D iff Z π f (eit )eint dt = 0 −π
for all n ≥ 1. 141
Solution. Let D = {z : |z| < 1} and suppose f = u + iv : D → C is a continuous function such that u and v are harmonic. Clearly u and v are continuous functions mapping D to R. By Corollary 2.9 we have Z π 1 iθ Pr (θ − t)u(eit ) dt for 0 ≤ r < 1 and all θ (10.3) u(re ) = 2π −π and v(reiθ ) =
Z
1 2π
π
−π
Pr (θ − t)v(eit ) dt
for 0 ≤ r < 1 and all θ.
(10.4)
Hence, Z π Z π 1 1 it f (re ) = u(re ) + iv(re ) = Pr (θ − t)u(e ) dt + i Pr (θ − t)v(eit ) dt 2π −π (10.3),(10.4) 2π −π Z π Z π h i 1 1 it iθ = Pr (θ − t) u(e ) + iv(r ) dt = Pr (θ − t) f (eit ) dt 2π −π 2π −π iθ
iθ
iθ
for 0 ≤ r < 1 and all θ. Thus, we have proved f (reiθ ) =
1 2π
Z
π
f (eit )Pr (θ − t) dt
−π
for all reiθ in D. Next, we will prove f is analytic ⇐⇒
Z
π
f (eiθ )eint dt = 0
−π
for all n ≥ 1. ⇒: Let f be analytic on D and assume n ≥ 1 (n integer). Define g(z) = f (z)zn−1 . Clearly g is analytic on D. Let γ = reit , −π ≤ t ≤ π, 0 < r < 1, then by Cauchy’s Theorem Z g(z) dz = 0 γ Z ⇐⇒ f (z)zn−1 dz = 0 γ Z π ⇐⇒ f (reit )ireit eit(n−1) rn−1 dt = 0 −π Z π ⇐⇒ i f (reit )eint rn dt = 0 −π Z π ⇐⇒ f (reit )eint rn dt = 0 −π
which implies lim−
r→1
Z
π
−π
f (reit )eint rn dt = 0
⇒
by Ex 3a p. 262
142
Z
π −π
f (eit )eint dt = 0 ∀n ≥ 1.
⇐: Assume
Rπ
−π
f (reit )eint rn dt = 0 for all n ≥ 1. We have for all z ∈ D Z π 1 iθ f (z) = f (re ) = f (eit )Pr (θ − t) dt Part 1 2π −π Z π ∞ X 1 f (eit ) r|n| ein(θ−t) dt = 2π −π n=−∞ Z π ∞ X 1 r|n| einθ f (eit )e−int dt = 2π n=−∞ −π Z Z −1 ∞ 1 X n inθ π 1 X |n| inθ π it −int r e f (e )e dt + r e f (eit )e−int dt = 2π n=−∞ 2π n=0 −π −π | {z } =0 by assumption
=
Z ∞ 1 X n π z f (eit )e−int dt. 2π n=0 −π
Now, for any closed rectifiable curve γ in D we get Z π Z Z Z ∞ ∞ Z 1 X 1 X n π f (eit )e−int dt z zn dz f (eit )e−int dt dz = f (z) dz = 2π n=0 γ −π −π γ γ 2π n=0 | {z } =0
n
since z is analytic for n ≥ 0. Thus f is analytic on D.
Exercise 2. In the statement of Theorem 2.4 suppose that f is piecewise continuous on ∂D. Is the conclusion of the theorem still valid? If not, what parts of the conclusion remain true? Solution. Not available. Exercise 3. Let D = {z : |z| < 1}, T = ∂D = {z : |z| = 1} (a) Show that if g : D− → C is a continuous function and gr : T → C is defined by gr (z) = g(rz) then gr (z) → g(z) uniformly for z in T as r → 1− . (b) If f : T → C is a continuous function define f˜ : D− → C by f˜(z) = f (z) for z in T and Z π ˜f (reiθ ) = 1 f (eit )Pr (θ − t) dt 2π −π (So Re f˜ and Im f˜ are harmonic in D). Define f˜r : T → C by f˜r (z) = f˜(rz). Show that for each r < 1 there is a sequence {pn , (z, z¯)} of polynomials in z and z¯ such that pn (z, z¯) → f˜r (z) uniformly for z in T . (Hint: Use Definition 2.1.) (c) Weierstrass approximation theorem for T. If f : T → C is a continuous function then there is a sequence {pn (z, z¯)} of polynomials in z and z¯ such that pn (z, z¯) → f (z) uniformly for z in T . (d) Suppose g : [0, 1] → C is a continuous function such that g(0) = g(1). Use part (c) to show that there is a sequence {pn } of polynomials such that pn (t) → g(t) uniformly for t in [0, 1]. (e) Weierstrass approximation theorem for [0,1]. If g : [0, 1] → C is a continuous function then there is a sequence {pn } of polynomials such that pn (t) → g(t) uniformly for t in [0, 1]. (Hint: Apply part (d) to the function g(t) + (1 − t)g(1) + tg(0).) (f) Show that if the function g in part (e) is real valued then the polynomials can be chosen with real coefficients. 143
Solution. Not available. Exercise 4. Let G be a simply connected region and let Γ be its closure in C∞ ; ∂∞G = Γ − G. Suppose there is a homeomorphism ϕ of Γ onto D− (D = {z : |z| < 1}) such that ϕ is analytic on G. (a) Show that ϕ(G) = D and ϕ(∂∞G) = ∂D. (b) Show that if f : ∂∞G → R is a continuous function then there is a continuous function u : Γ → R such that u(z) = f (z) for z in ∂∞G and u is harmonic in G. (c) Suppose that the function f in part (b) is not assumed to be continuous at ∞. Show that there is a continuous function u : G− → R such that u(z) = f (z) for z in ∂G and u is harmonic in G (see Exercise 2). Solution. Not available. Exercise 5. Let G be an open set, a ∈ G, and G0 = G − {a}. Suppose that u is a harmonic function on G0 such that limz→a u(z) exists and is equal to A. Show that if U : G → R is defined by U(z) = u(z) for z , a and U(a) = A then U is harmonic on G. Solution. Not available. Exercise 6. Let f : {z : Re z = 0} → R be a bounded continuous function and define u : {z : Re z > 0} → R by Z x f (it) 1 ∞ dt. u(x + iy) = 2 π −∞ x + (y − t)2 Show that u is a bounded harmonic function on the right half plane such that for c in R, f (ic) = limz→ic u(z). Solution. Not available. Exercise 7. Let D = {z : |z| < l} and suppose f : ∂D → R is continuous except for a jump discontinuity at z = 1. Define u : D → R¯ by (2.5). Show that u is harmonic. Let v be a harmonic conjugate of u. What can you say about the behavior of v(r) as r → 1− ? What about v(reiθ ) as r → 1− and θ → 0? Solution. Not available.
10.3 Subharmonic and superharmonic functions Exercise 1. Which of the following functions are subharmonic? superharmonic? harmonic? neither subharmonic nor superharmonic? (a) ϕ(x, y) = x2 + y2 ; (b) ϕ(x, y) = x2 − y2 ; (c) ϕ(x, y) = x2 + y; (d) ϕ(x, y) = x2 − y; (e) ϕ(x, y) = x + y2 ; (f) ϕ(x, y) = x − y2 . Rπ Rπ Rπ Rπ Solution. Note that −π sin θ dθ = 0, −π cos θ dθ = 0, −π sin2 θ dθ = π and −π cos2 θ dθ = π. For all a = (α, β) ∈ C and any r > 0 we have a) ϕ(a + reiθ ) = ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 + (β + r sin θ)2 = α2 + 2αr cos θ + r2 cos2 θ + β2 + 2βr sin θ + r2 sin2 θ = α2 + β2 + 2αr cos θ + 2βr sin θ + r2 . Thus 1 2π
Z
π
−π
ϕ(a + reiθ ) dθ
= α2 + β2 +
αr π
Z
Z βr π cos θ dθ + sin θ dθ +r2 π |−π {z } |−π {z } π
=0
= α2 + β2 + r2 ≥ α2 + β2 = ϕ(a). 144
=0
Hence, ϕ ∈ Subhar(G). b) ϕ(a + reiθ ) = ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 − (β + r sin θ)2 = α2 + 2αr cos θ + r2 cos2 θ − β2 − 2βr sin θ − r2 sin2 θ. Thus 1 2π
Z
π
ϕ(a + reiθ ) dθ
−π
= α2 − β2 +
αr π
Z
Z π Z r2 βr π cos θ dθ + cos2 θ dθ − sin θ dθ 2π −π π −π |−π {z } | {z } | {z } π
=π
=0
−
r2 2π
Z
π
=0
r2 r2 sin2 θ dθ = α2 − β2 + − = α2 − β2 = ϕ(a). 2 2 −π | {z } =π
Hence, ϕ ∈ Har(G), therefore also ϕ ∈ Subhar(G) and ϕ ∈ Superhar(G). c) ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 + (β + r sin θ) = α2 + 2αr cos θ + r2 cos2 θ + β + r sin θ.
Thus 1 2π
Z
π
ϕ(a + reiθ ) dθ
= α2 + β +
−π
αr π
Z
Z π Z π r2 r cos θ dθ + cos2 θ dθ + sin θ dθ 2π −π 2π −π |−π {z } | {z } | {z } π
=0
= α2 + β +
2
=π
=0
r ≥ α2 + β = ϕ(a). 2
Hence, ϕ ∈ Subhar(G). d) ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ)2 − (β + r sin θ) = α2 + 2αr cos θ + r2 cos2 θ − β − r sin θ.
Thus 1 2π
Z
π iθ
ϕ(a + re ) dθ
−π
αr = α −β+ π 2
Z
Z π Z π r2 r 2 cos θ dθ + cos θ dθ − sin θ dθ 2π −π 2π −π |−π {z } | {z } | {z } π
=0
= α2 − β +
2
=π
r ≥ α2 − β = ϕ(a). 2
Hence, ϕ ∈ Subhar(G). e) ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ) + (β + r sin θ)2 = α + r cos θ + β2 + 2βr sin θ + r2 sin2 θ. 145
=0
Thus 1 2π
Z
π
ϕ(a + reiθ ) dθ
= α + β2 +
−π
r 2π
Z
Z Z π βr π r2 cos θ dθ + sin θ dθ + sin2 θ dθ π −π 2π −π −π | {z } | {z } | {z } π
=0
=0
=π
r2 ≥ α + β2 = ϕ(a). = α + β2 + 2 Hence, ϕ ∈ Subhar(G). f) ϕ(a + reiθ )
= ϕ(α + r cos θ, β + r sin θ) = (α + r cos θ) − (β + r sin θ)2 = α + r cos θ − β2 − 2βr sin θ − r2 sin2 θ.
Thus 1 2π
Z
π
ϕ(a + reiθ ) dθ
−π
= α − β2 +
r 2π
Z
Z Z π βr π r2 cos θ dθ − sin θ dθ − sin2 θ dθ π 2π |−π {z } |−π {z } |−π {z } π
=0
=0
=π
r2 = α − β2 − ≤ α − β2 = ϕ(a). 2 Hence, ϕ ∈ Superhar(G).
Exercise 2. Let Subhar(G) and Superhar(G) denote, respectively, the sets of subharmonic and superharmonic functions on G. (a) Show that Subhar(G) and Superhar(G) are closed subsets of C(G; R). (b) Does a version of Harnack’s Theorem hold for subharmonic and superharmonic functions? Solution. Not available. Exercise 3. (This exercise is difficult.) If G is a region and if f : ∂∞G → R is a continuous function let u f be the Perron Function associated with f . This defines a map T : (∂∞G; R) → Har(G) by T ( f ) = u f . Prove: (a) T is linear (i.e., T (a1 f1 + a2 f2 ) = a1 T ( f1 ) + a2 T ( f2 )). (b) T is positive (i.e., if f (a) ≥ 0 for all a in ∂∞G then T ( f )(z) ≥ 0 for all z in G). (c) T is continuous. Moreover, if { fn } is a sequence in C(∂∞G; R) such that fn → f uniformly then T ( fn ) → T ( f ) uniformly on G. (d) If the Dirichlet Problem can be solved for G then T is one-one. Is the converse true? Solution. Not available. Exercise 4. In the hypothesis of Theorem 3.11, suppose only that f is a bounded function on ∂∞G; prove that the conclusion remains valid. (This is useful if G is an unbounded region and g is a bounded continuous function on ∂G. If we define f : ∂∞G → R by f (z) = g(z) for z in ∂G and f (∞) = 0 then the conclusion of Theorem 3.11 remains valid. Of course there is no reason to expect that the harmonic function will have predictable behavior near ∞ — we could have assigned any value to f (∞). However, the behavior near points of ∂G can be studied with hope of success.) Solution. Not available. Exercise 5. Show that the requirement that G1 is bounded in Corollary 3.5 is necessary. 146
Solution. Not available. Exercise 6. If f : G → Ω is analytic and ϕ : Ω → R is subharmonic, show that ϕ ◦ f is subharmonic if f is one-one. What happens if f 0 (z) , 0 for all z in G? Solution. Clearly f is continuous, since f : G → Ω is analytic and ϕ is continuous, since ϕ : Ω → R is subharmonic. Define u = ϕ ◦ f , then u : G → R is continuous. If we can show that for every bounded region G1 such that G¯ 1 ⊂ G and for every continuous function u1 : G¯ 1 → R that is harmonic in G1 and satisfies u(z) ≤ u1 (z) for z in ∂G1 , we have u(z) ≤ u1 for z in G1 , then u is subharmonic (by Corollary 3.5 p. 265). Hence, let G1 be a bounded region G1 such that G¯ 1 ⊂ G and let u1 : G¯ 1 → R that is harmonic in G1 and satisfies u(z) ≤ u1 (z) for z on ∂G1 . We have to show u(z) ≤ u1 (z) ∀z ∈ G1 . Since f is one-one, we know that f −1 exists and is analytic. So define ¯ →R ϕ1 = u1 ◦ f −1 : Ω ¯ 1 → G¯ 1 . Clearly ϕ1 is harmonic (harmonic composite analytic is harmonic). Since ϕ : Ω → R with f −1 : Ω is subharmonic, we can use Theorem 3.4 to obtain: For every region Ω1 contained in Ω and every harmonic function ϕ˜ on Ω1 , ϕ−ϕ˜ satisfies the Maximum Principle on Ω1 . In fact ϕ−ϕ1 satisfies the Maximum Principle on Ω1 . By assumption we have u(z) ≤ u1 (z) ∀z ∈ G1 which is the same as u(z) = ϕ( f (z)) ≤ u1 (z) = ϕ1 ( f (z)) ∀z ∈ ∂G1 since f maps G1 onto Ω1 , and thus ∂G1 onto ∂Ω1 (we choose this by assumption). Thus, ϕ( f (z)) − ϕ1 ( f (z)) ≤ 0 ∀z ∈ ∂G1 . Since ϕ − ϕ1 satisfies the Maximum Principle on Ω1 , this implies ϕ( f (z)) − ϕ1 ( f (z)) ≤ 0
∀z ∈ G1 ,
that is u(z) − u1 (z) ≤ 0 ∀z ∈ G1
and we are done. Therefore, u = ϕ ◦ f is subharmonic on G.
If f 0 (z) , 0 for all z in G, then the previous result holds locally for some neighborhood in G. ( f is locally one-one and onto and f −1 is analytic and hence u = ϕ ◦ f is locally subharmonic). Here is the claim: Let f : G → C be analytic, let z0 ∈ A, and let f 0 (z0 ) , 0. Then there is a neighborhood U of z0 and a neighborhood V of w0 = f (z0 ) such that f : U → V is one-one and onto and f −1 : V → U is analytic. Proof of the claim: f (z) − w0 has a simple zero at z0 since f 0 (z0 ) , 0. We can use Theorem 7.4 p. 98 to find > 0 and δ > 0 such that each w with |w − w0 | < δ has exactly one pre-image z with |z − z0 | < . Let V = {w| |w − w0 | < δ} and let U be the inverse image of V under the map f restricted to {z| |z − z0 | < }. By Theorem 7.4 p. 98, f maps U one-one onto V. Since f is continuous, U is a neighborhood of z0 . By the Open Mapping Theorem (Theorem 7.5 p. 99), f = ( f −1 )−1 is an open map, so f −1 is continuous from V to U. To finally show that it is analytic, we can use Proposition 3.7 p. 125. We have Z z f 0 (w) 1 −1 dz. f (w) = 2πi |z−z0 |= f (z) − w Note, if we assume that ϕ is twice differentiable, then the result holds globally. 147
10.4 The Dirichlet Problem Exercise 1. Let G = B(0; 1) and find a barrier for G at each point of the boundary. Solution. Not available. Exercise 2. Let G = C − (∞, 0] and construct a barrier for each point of ∂∞G. Solution. Not available. Exercise 3. Let G be a region and a a point in ∂∞G such that there is a harmonic function u : G → R with limz→a u(z) = 0 and lim inf z→w u(z) > 0 for all w in ∂∞G, w , a. Show that there is a barrier for G at a. Solution. Not available. Exercise 4. This exercise asks for an easier proof of a special case of Theorem 4.9. Let G be a bounded region and let a ∈ ∂G such that there is a point b with [a, b] ∩ G− = {a}. Show that G has a barrier at a. (Hint: Consider the transformation (z − a)(z − b)1 .) Solution. Not available.
10.5 Green’s Function Exercise 1. (a) Let G be a simply connected region, let a ∈ G, and let f : G → D = {z : |z| < 1} be a one-one analytic function such that f (G) = D and f (a) = 0. Show that the Green’s Function on G with singularity at a is ga (z) = − log | f (z)|. (b) Find the Green’s Functions for each of the following regions: (i) G = C − (∞, 0]; (ii) G = {z : Re z > 0}; (iii) G = {z : 0 < Im z < 2π}. Solution. Solution to part a): Let G be simply connected, let a ∈ G, and let f : G → D = {z : |z| < 1} be a one-one analytic function such that f (G) = D and f (a) = 0. To show: ga (z) = − log | f (z)| is a Green’s Function on G with singularity at a. Clearly, ga (z) has a singularity if | f (z)| = 0. This happens if f (z) = 0. By assumption f (z) = 0 iff z = a. (Note that there is no other point z ∈ G such that f (z) = 0 since f is assumed to be one-one). This implies ga (z) has a singularity at a. It remains to verify a), b) and c) of Definition 5.1. a) ga is harmonic in G − {a}. This follows directly by Exercise 5 p. 255 in G − {a}. b) G(z) = ga (z)+log |z−a| is harmonic in a disk about a. Let the disk be B(a; r). Clearly ga (z) = − log | f (z)| is harmonic on B(a; r)−{a}. By Exercise 5 p. 255, we can argue again that log |z−a| is harmonic on B(a; r)−{a} (choose f (z) = z − a and f (z) = 0 iff z = a). Hence G(z) = ga (z) + log |z − a| = − log | f (z)| + log |z − a| is harmonic on B(a; r) − {a} as a sum of two harmonic functions. But ! ! |z − a|· | f˜(z)| | f (z)| = − log = − log | f˜(z)| G(z) = − log | f (z)| + log |z − a| = − log |z − a| |z − a| where f (z) = (z − a) f˜(z) by assumption and f˜(z) is analytic on G with no zero. So G(z) has a removable singularity at a and therefore G(z) is harmonic on B(a; r). c) limz→w ga (z) = 0 for each w in ∂∞G. We have | f (z)| = 1 ∀z ∈ ∂∞G by assumption ( f (∂∞G) = ∂D). Hence lim ga (z) = lim − log | f (w)| = − log 1 = 0
z→w
w→a
148
for each a in ∂∞G. This implies ga (z) = − log | f (z)| is a Green’s Function on G with singularity at a. Solution to part b) ii): Clearly, G is a simply connected region (not the whole plane). If we can find a map, say h : G → D, then the Green’s Function, say ga , is given by ga (z) = − log |h(z)| by Part 1 a). Note that h has to satisfy the following assumptions to use Part 1 a): a) h : G → D one-one and analytic b) h(a) = 0 c) h(G) = D. The existence and uniqueness of this function h with the assumptions a), b) and c) is guaranteed by the Riemann Mapping Theorem p. 160 Theorem 4.2. It remains to find h: Claim: h : G → D given by h(z) = f (g(z)) where g(z) =
z − g(a) z−1 and f (z) = z+1 1 − g(a)z
works. Thus, the Green’s Function for the region G = {z : Re(z) > 0} is given by ga (z) = − log |h(z)| where h(z) is given by the Claim. Proof of the Claim: We will invoke the Orientation Principle (p. 53) to find g : G → D where g is one-one and analytic. Then g(z) =
z−1 z+1
(see p. 53 in the book for the derivation). From p. 162 we have seen that the Möbius Transformation (which is one-one and analytic) f (z) =
z − g(a)
1 − g(a)z
maps D onto D such that f (g(a)) = 0. Hence h(z) = f (g(z)) maps G onto D. In addition h is one-one, analytic and h(a) = 0 and therefore satisfies a), b) and c). Exercise 2. Let ga be the Green’s Function on a region G with singularity at z = a. Prove that if ψ is a positive superharmonic function on G − {a} with lim inf z→a [ψ(z) + log |z − a|] > −∞, then ga (z) ≤ ψ(z) for z , a. Solution. Not available. Exercise 3. This exercise gives a proof of the Riemann Mapping Theorem where it is assumed that if G is a simply connected region, G , C, then: (i) C∞ − G is connected, (ii) every harmonic function on G has a harmonic conjugate, (iii) if a < G then a branch of log(z − a) can be defined. (a) Let G be a bounded simply connected region and let a ∈ G; prove that there is a Green’s Function ga on G with singularity at a. Let u(z) = ga (z) + log |z − a| and let v be the harmonic conjugate of u. If ϕ = u + iv let f (z) = eiα (z − a)e−ϕ(z) for a real number α. (So f is analytic in G.) Prove that | f (z)| = exp(−ga (z)) and that limz→w | f (z)| = 1 for each w in ∂G (Compare this with Exercise 1). Prove that for 0 < r < 1, Cr = {z : | f (z)| = r} consists of a finite number of simple closed curves in G (see Exercise VI.1.3). Let Gr be a component of {z : | f (z)| < r} and apply Rouche’s Theorem to get that f (z) = 0 and f (z) − w0 = 0, 149
|w0 | < r, have the same number of solutions in Gr . Prove that f is one-one on Gr . From here conclude that f (G) = D = {z : |z| < 1} and f 0 (a) > 0, for a suitable choice of α. (b) Let G be a simply connected region with G , C, but assume that G is unbounded and 0, ∞ ∈ ∂∞G. Let l be a branch of log z on G, a ∈ G, and α = l(a). Show that l is one-one on G and l(z) , α + 2πi for any z in G. Prove that ϕ(z) = [l(z) − α − 2πi]−1 is a conformal map of G onto a bounded simply connected region in the plane. (Show that l omits all values in a neighborhood of α + 2πi.) (c) Combine parts (a) and (b) to prove the Riemann Mapping Theorem. Solution. Not available. Exercise 4. (a) Let G be a region such that ∂G = γ is a simple continuously differentiable closed curve. If f : ∂G → R is continuous and g(z, a) = ga (z) is the Green’s Function on G with singularity at a, show that Z ∂g (5.5) h(a) = f (z) (z, a) ds ∂n γ is a formula for the solution of the Dirichlet Problem with boundary values f ; where ∂g ∂n is the derivative of g in the direction of the outward normal to γ and ds indicates that the integral is with respect to arc length. (Note: these concepts are not discussed in this book but the formula is sufficiently interesting so as to merit presentation.) (Hint: Apply Green’s formula # " Z " ∂v ∂u [u∆v − v∆u] dx dy = u −v ds ∂n ∂n Ω ∂Ω with Ω = G − {z : |z − a| ≤ δ}, δ < d(a, {γ}), u = h, v = ga (z) = g(z, a).) (b) Show that if G = {z : |z| < 1} then (5.5) reduces to equation (2.5). Solution. Not available.
150
Chapter 11
Entire Functions 11.1 Jensen’s Formula Exercise 1. In the hypothesis of Jensen’s Formula, do not suppose that f (0) , 0. Show that if f has a zero at z = 0 of multiplicity m then (m) ! Z 2π n X f (0) r 1 + m log r = − log log + log | f (reiθ )| dθ. m! |a | 2π k 0 k=1 Solution. We have
r(z−ak ) r2 −¯ak z
maps B(0; r) onto itself and takes the boundary to the boundary. Let n rm Y r2 − a¯ k z . F(z) = f (z) m z k=1 r(z − ak )
Then F ∈ H(G) and F has no zeros in B(0; r) and |F(z)| = | f (z)| if |z| = r. Thus, by a known result, we have log |F(0)| =
1 2π
Z
2π
0
log |F(reiθ )| dθ =
1 2π
Z
2π 0
log | f (reiθ )| dθ.
Also g(z) ≡ =
P∞ f (i) (0)zi f (z) i=0 i! = zm zm P∞ f (i) (0)zi i=m
i!
zm ∞ X f (m) (0) 0 f (m+1) (0) 1 f (i) (0)zi−m = = z + z + ... i! m! (m + 1)! i=m ( f has a zero at z = 0 of multiplicity m) which implies g(0) = F(0) = g(0)rm
n Y k=1
151
−
f (m) (0) m! .
! r . ak
Thus,
Therefore |F(0)| = |g(0)|rm
n Y r |a k| k=1
implies (m) n X f (0) r r log |F(0)| = log |g(0)| + m log r + log = log . + m log r + log |a | m! |a k k| k=1 k=1 n X
Hence,
(m) Z 2π n X f (0) r 1 + m log r + log = log |F(0)| = log log | f (reiθ )| dθ m! |ak | 2π 0 k=1
which is equivalent to
(m) Z 2π n X f (0) 1 r log log | f (reiθ )| dθ + + m log r = − log m! |ak | 2π 0 k=1
which proves the assertion.
Exercise 2. Let f be an entire function, M(r) = sup{| f (reiθ )| : 0 ≤ θ ≤ 2π}, n(r) = the number of zeros of f in B(0; r) counted according to multiplicity. Suppose that f (0) = 1 and show that n(r) log 2 ≤ log M(2r). ¯ 2r). Suppose a1 , . . . , an are the zeros of f in Solution. Since f is an entire function, it is analytic on B(0; B(0; r) repeated according to multiplicity and b1 , . . . , bm are the zeros of f in B(0; 2r) − B(0; r). That is, a1 , . . . , an , b1 , . . . , bm are the zeros of f in B(0; 2r) repeated according to multiplicity. Since f (0) = 1 , 0, we have by Jensen’s formula ! X ! Z 2π m n X 2r 1 2r log − + log | f (2reiθ )| dθ log 0 = log 1 = log | f (0)| = − |a | |b | 2π k k 0 k=1 k=1 which implies n X
2r log |a k| k=1
!
m X
! Z 2π 2r 1 log − + log | f (2reiθ )| dθ |bk | 2π 0 k=1 | {z } ≥ 0 since r ≤ |bk | < 2r Z 2π 1 log | f (2reiθ )| dθ 2π 0 Z 2π 1 log |M(2r)| dθ = log |M(2r)|. 2π 0
=
≤ ≤ Def of M(r) Hence,
n X
! 2r log ≤ log |M(2r)|. |ak | k=1
But n X
2r log |ak | k=1
!
n(r) X
! ! n(r) Y 2r 2r = = log log |ak | |ak | k=1 k=1 n(r) n(r) Y Y r 2r n n(r) = log 2 + log = log 2 |ak | |ak | k=1
≥ n(r) log(2)
152
k=1
since 0 < |ak | < r implies
r |ak |
> 1 ∀k which means
Qn(r)
r k=1 |ak |
> 1 so the logarithm is greater than zero. Thus,
n(r) log(2) ≤ log |M(2r)| which proves the assertion. ¯ r) but only Exercise 3. In Jensen’s Formula do not suppose that f is analytic in a region containing B(0; that f is meromorphic with no pole at z = 0. Evaluate Z 2π 1 log | f (reiθ )| dθ. 2π 0 Solution. Not available. Exercise 4. (a) Using the notation of Exercise 2, prove that ! Z r n X n(t) r dt = log t |ak | 0 k=1 where a1 , . . . , an are the zeros of f in B(0; r). (b) Let f be meromorphic without a pole at z = 0 and let n(r) be the number of zeros of f in B(0; r) minus the number of poles (each counted according to multiplicity). Evaluate Z r n(t) dt. t 0 Solution. Not available. Exercise 5. Let D = B(0; 1) and suppose that f : D → C is an analytic function which is bounded. P (a) If {an } are the non-zero zeros of f in D counted according to multiplicity, prove that (1 − |an |) < ∞. (Hint: Use Proposition VII. 5.4). (b) If f has a zero at z = 0 of multiplicity m ≥ 0, prove that f (z) = zm B(z) exp(−g(z)) where B is a Blaschke Product (Exercise VII. 5.4) and g is an analytic function on D with Re g(z) ≥ − log M (M = sup{| f (z)| : |z| < 1}). Solution. Not available.
11.2 The genus and order of an entire function Exercise 1. Let f (z) =
P
cn zn be an entire function of finite genus µ; prove that lim cn (n!)1/(µ+1) = 0.
n→∞
(Hint: Use Cauchy’s Estimate.) Solution. Define M(r) = max{ f (reiθ ) : 0 ≤ θ ≤ 2π} = max|z|=r | f (z)|. For sufficiently large r = |z|, we have M(r) < eαr
µ+1
by Theorem 2.6 p. 283. By Cauchy’s estimate p. 73, we have |cn | ≤
M(r) rn
153
(11.1)
where cn =
f (n) (0) n! .
Hence, |cn | ≤
M(r) µ+1 < eαr . n r (11.1)
(11.2)
1
Define, r = n µ+1 λ, where λ is a positive constant, and choose λ > e for convenience. Then (11.2) can be written in the form αλµ+1 n µ+1 µ+1 e 1 eαnλ eαnλ n µ+1 . ⇐⇒ (n ) |cn | < |cn | < = 1 n λ λ (nn ) µ+1 λn
Now, if we choose α such that αλµ+1 = 1, the last inequality becomes e n 1 (nn ) µ+1 |cn | < < λ
for any > 0, provided n is taken large enough, n > N say. Since n! < nn , for n ≥ 2, we have 1
(n!) µ+1 |cn | < for n > N or lim cn (n!)1/(µ+1) = 0.
n→∞
Exercise 2. Let f1 and f2 be entire functions of finite orders λ1 and λ2 respectively. Show that f = f1 + f2 has finite order λ and λ ≤ max(λ1 , λ2 ). Show that λ = max(λ1 , λ2 ) if λ1 , λ2 and give an example which shows that λ < max(λ1 , λ2 ) with f , 0. Solution. Not available. Exercise 3. Suppose f is an entire function and A, B, a are positive constants such that there is a r0 with | f (z)| ≤ exp(A|z|a + B) for |z| > r0 . Show that f is of finite order ≤ a. Solution. Not available. Exercise 4. Prove that if f is an entire function of order λ then f 0 also has order λ. Solution. Let f be an entire function of order λ, then f has power series f (z) =
∞ X
cn zn .
n=0
By Exercise 5 e) p. 286 we have α = lim inf n→∞
Differentiating f yields f 0 (z) =
∞ X
− log |cn | 1 = . n log n λ
ncn zn−1 =
n=1
∞ X n=0
154
(n + 1)cn+1 zn . | {z } dn
Now, let λ0 be the order of f 0 , then 1 λ0
− log |dn | − log |(n + 1)cn+1 | = lim inf n→∞ n log n n log n − log(n + 1) − log |cn+1 | = lim inf n→∞ n log n " # − log(n + 1) log |cn+1 | (n + 1) log(n + 1) = lim inf − · n→∞ n log n n log n (n + 1) log(n + 1) − log(n + 1) − log |cn+1 | n + 1 log(n + 1) + · · = lim inf n→∞ n (n + 1) log(n + 1) |{z} n log(n) | n log {z } | {z } | {z } →1 1 = lim inf n→∞
→0
=
→λ
→1
1 . λ
Hence, λ0 = λ and therefore f 0 also has order λ. P Exercise 5. Let f (z) = cn zn be an entire function and define the number α by α = lim inf n→∞
− log |cn | n log n
(a) Show that if f has finite order then α > 0. (Hint: If the order of f is λ and β > λ show that |cn | ≤ r−n exp(rβ ) for sufficiently large r, and find the maximum value of this expression.) (b) Suppose that 0 < α < ∞ and show that for any > 0, < α, there is an integer p such that |cn |1/n < n−(α−) for n > p. Conclude that for |z| = r > 1 there is a constant A such that | f (z)| < Ar p +
∞ X r n nα− n=1
(c) Let p be as in part (b) and let N be the largest integer ≤ (2r)1/(α−) . Take r sufficiently large so that N > p and show that N ∞ X X r n r n < 1 and < B exp (2r)1/(α−) α− α− n n n=p+1 n=N+1
where B is a constant which does not depend on r. (d) Use parts (b) and (c) to show that if 0 < α < ∞ then f has finite order λ and λ ≤ α−1 . (e) Prove that f is of finite order iff α > 0, and if f has order λ then λ = α−1 . Solution. Not available. Exercise 6. Find the order of each of the following functions: (a) sin z; (b) cos z; (c) cosh P∞ −an n z where a > 0. (Hint: For part (d) use Exercise 5.) n=1 n
√
z; (d)
Solution. c) The order is λ = 21 . Note that sinh2 z = cosh2 z − 1 and that cos x is bounded for real values x, so | cos z|2 = cos2 x + sinh2 y ≤ cosh2 y ≤ cosh2 |z|. 155
Thus | cos z| ≤ cosh |z| = and therefore | cos
1 1 |z| (e + e−|z| ) ≤ e|z| ≤ e|z| , 2 2
√ z| ≤ exp(|z|1/2 ).
This implies that the order λ ≤ 21 . Next we want to show that λ = 12 . Seeking contradiction assume that there is > 0 and r > 0 such that √ 1 | cos z| ≤ exp(|z| 2 − ) (11.3) for all |z| > r. √ Let z = −n2 for an n ∈ N with n > max{ r, 4, 21/(2) } then in particular (i) |z| > r, (ii) − ln 2 + n > 2n , and (iii) 12 > n12 and | f (z)| = | cos
√ n 1 1 1 1 z| = |en + e−n | > en = e− log 2+n > e 2 > exp(n−2 n) = exp(n2( 2 −) ) = exp(|z| 2 − ). (ii) (iii) 2 2
This shows that no > 0 can be found that satisfies equation (11.3). This implies λ = 12 . Exercise 7. Let f1 and f2 be entire functions of finite order λ1 , λ2 ; show that f = f1 f2 has finite order λ ≤ max(λ1 , λ2 ). Solution. Define M(r) = max|z|=r | f (z)|, M1 (r) = max|z|=r | f1 (z)| and M2 (r) = max|z|=r | f2 (z)|. Since f1 and f2 are entire functions of finite order λ1 and λ2 , respectively, we have for > 0, M1 (r) < er
λ1 +/2
M2 (r) < er
λ2 +/2
(11.4)
and (11.5)
for sufficiently large r = |z| by Proposition 2.14. Hence M(r)
= max | f (z)| = max | f1 (z)· f2 (z)| |z|=r
|z|=r
≤ max | f1 (z)|· max | f2 (z)| |z|=r
=
|z|=r
M1 (r)M2 (r)
= er
<
er
λ1 +/2
er
λ2 +/2
by (11.4),(11.5)
λ1 +/2 +rλ2 +/2
≤
λ≤max(λ1 ,λ2 )
e2r
λ+/2
< er
λ+
provided r is sufficiently large. Hence, the order of f (z) = f1 (z)· f2 (z) has finite order λ ≤ max(λ1 , λ2 ). P Exercise 8. Let {an } be a sequence of non-zero complex numbers. Let ρ = inf{a : |an |−a < ∞}; the number called the exponent of convergence of {an }. (a) If f is an entire function of rank p then the exponent of convergence ρ of the non-zero zeros of f satisfies: p ≤ ρ ≤ p + 1. P P (b) If ρ0 = the exponent of convergence of {an } then for every > 0, |an |−(ρ+) < ∞ and |an |−(ρ−) = ∞. (c) Let f be an entire function of order λ and let {a1 , a2 , . . .} be the non-zero zeros of f counted according to multiplicity. If ρ is the exponent of convergence of {an } prove that ρ ≤ λ. (Hint: See the proof of (3.5) in the next section.) 156
Q (d) Let P(z) = ∞ n=1 E p (z/an ) be a canonical product of rank p, and let ρ be the exponent of convergence of {an }. Prove that the order of P is ρ. (Hint: If λ is the order of P, ρ ≤ λ; assume that |al | ≤ |a2 | ≤ . . . and fix z, |z| > 1. Choose N such that |an | ≤ 2|z| if n ≤ N and |an | > 2|z| if n ≤ N + 1. Treating the cases ρ < p + 1 and ρ = p + 1 separately, use (2.7) to show that for some > 0 ! ∞ X z ρ+ log E p < A|z| . an n=N+1
Prove that for |z| ≥ 21 , log |E p (z)| < B|z| p where B is a constant independent of z. Use this to prove that ! z ρ+ log E p < C|z| a n n=1
N X
for some constant C independent of z.) Solution. Not available.
Exercise 9. Find the order of the following entire functions: (a) ∞ Y f (z) = (1 − an z), 0 < |a| < 1 n=1
b) f (z) =
∞ Y n=1
1−
z n!
c) f (z) = [Γ(z)]−1 . Q z Solution. b) Considering g(z) := ∞ m=1 1 − n! one notes that g(z) is already in form of the canonical product with the entire function in the exponent being constantly zero and p = 0. Hence also the genus µ = 0. The simple zeros of g(z) are at z = n! and from Exercise 8a) it follows that λ ≤ 1. In fact we will show that λ = 0. It suffices to show that the exponent of convergence ρ = 0 and to employ Exercise 8d). Let m ≤ N, let n > 4m then n! ≥ n· (n − 1)· 2· (n − 2)· 3 . . . (n − (2k − 1))· 2m . | {z } | {z } | {z } ≥n
≥n
≥n
Each product of consecutive numbers is of the form (n − (k − 1))k, for 1 ≤ k ≤ 2m. By choice of n, nk ≤ 12 and n − k + 1 > 2m which justifies the individual estimates by n. In total this yields n! > n2m for n large, or equivalently ∞ ∞ X X 1 (n!)− m ≤ n−2 n=4m
n=4m
and the sum on the left converges for every positive integer m. We conclude that ρ = inf{m : the left sum converges} = 0. By Part d) of Exercise 8 we conclude that the order of g(z) is zero.
157
11.3 Hadamard Factorization Theorem Exercise 1. Let f be analytic in a region G and suppose that f is not identically zero. Let G0 = G − {z : f0 ∂h − i = on G f (z) = 0} and define h : G0 → R by h(z) = log | f (z)|. Show that ∂h 0. ∂x ∂y f Solution. Let f be analytic in a region G and suppose that f is not identically zero. Let G0 = G − {z : 0 f (z) = 0}, then h : G0 → R given by h(z) = log | f (z)| is well defined as well as ff is well defined on G0 . Let f = u(x, y) + iv(x, y) = u + iv. Since f is analytic, the Cauchy-Riemann (C-R) equations u x = vy and uy = −v x are satisfied. We have by p. 41 Equation 2.22 and 2.23 f 0 = u x + iv x
f 0 = −iuy + vy
and
and thus 2 f 0 = u x + iv x − iuy + vy implies f0 =
1 1 1 u x + iv x − iuy + vy = u x − iuy − iuy + u x = 2u x − 2iuy = u x − iuy . (C-R) 2 2 2
Therefore, f 0 u x − iuy = . f u + iv Next, we calculate get
∂h ∂x
= h x and
∂h ∂y
= hy where h(z) = log | f (z)| =
h x = hu u x + hv v x =
1 2
log(u2 + v2 ). Using the chain rule, we
u v ux + 2 vx u2 + v2 u + v2
and hy = hu uy + hv vy =
(11.6)
u2
v u uy + 2 vy 2 +v u + v2
and hence ∂h ∂h −i ∂x ∂y
= = = = (C-R)
= = = =
h x − ihy u v u v ux + 2 vx − i 2 uy − i 2 vy u2 + v2 u + v2 u + v2 u + v2 v u (u x − iuy ) + 2 (v x − ivy ) 2 2 u +v u + v2 v u (u x − iuy ) + 2 (−uy − iu x ) u2 + v2 u + v2 u iv (u x − iuy ) − 2 (u x − iuy ) u2 + v2 u + v2 u − iv (u x − iuy ) u2 + v2 u − iv (u x − iuy ) (u − iv)(u + iv) u x − iuy . u + iv 158
Compare this with (11.6) and we obtain the assertion ∂h ∂h f0 −i = . ∂x ∂y f Exercise 2. Refer to Exercise 2.7 and show that if λ1 , λ2 then λ = max(λ1 ; λ2 ). Solution. Not available. Exercise 3. (a) Let f and g be entire functions of finite order λ and suppose that f (an ) = g(an ) for a P sequence {an } such that |an |−(λ+1) = ∞. Show that f = g. P (b) Use Exercise 2.8 to show that if f , g and {an } are as in part (a) with |an |−(λ+) = ∞ for some > 0 then f = g. (c) Find all entire functions f of finite order such that f (log n) = n. (d) Give an example of an entire function with zeros {log 2, log 3, . . .} and no other zeros. Solution. a) Since f and g are entire functions of finite order λ, also F = f − g is an entire function of finite P order λ (see Exercise 2 p. 286). Suppose that f (an ) = g(an ) for a sequence {an } such that |an |−(λ+1) = ∞. Assume F . 0 ( f . g), we will derive a contradiction. Since F . 0, the sequence {an } are the zeros of F, because F(an ) = f (an ) − g(an ) = 0 since f (an ) = g(an ) by assumption. By Hadamard’s Factorization Theorem (p. 289), F has finite genus µ ≤ λ since F ∈ H(C) with finite order λ. Since by definition µ = max(p, q) where p is the rank of F, we certainly have p ≤ λ. By the definition of the rank (p. 281 Definition 2.1), we have ∞ X n=1
|an |−(p+1) < ∞,
where an ’s are the zeros of F. Therefore by the comparison test, we also have ∞ X n=1
|an |−(λ+1) ≤
p≤λ
∞ X n=1
|an |−(p+1) < ∞
P∞
P −(λ+1) so n=1 |an |−(λ+1) < ∞ contradicting the assumption ∞ = ∞. Hence F ≡ 0, and therefore we n=1 |an | obtain f = g. b) Since f and g are entire functions of finite order λ, we also have F = f − g is an entire function of finite P order λ (see Exercise 2 p. 286). Suppose that f (an ) = g(an ) for a sequence {an } such that |an |−(λ+) = ∞. Assume F . 0 ( f . g), we will derive a contradiction. Since F . 0, the sequence {an } are the zeros of F, because F(an ) = f (an ) − g(an ) = 0 since f (an ) = g(an ) by assumption. Let ρ be the exponent of convergence of {an }, then ρ≤λ by Exercise 8 c) p. 286-287. Hence, according to Exercise 8 b) p. 286-287, we have ∞ X n=1
|an |−(ρ+) < ∞. 159
Therefore by the comparison test, we also have ∞ X n=1
P∞
|an |−(λ+) ≤
Hence n=1 |an |−(λ+) < ∞ contradicting the fact assertion f = g.
ρ≤λ
P∞
∞ X n=1
n=1
|an |−(ρ+) < ∞.
|an |−(λ+) = ∞. Thus, F ≡ 0 and therefore we get the
160
Chapter 12
The Range of an Analytic Function 12.1 Bloch’s Theorem Exercise 1. Examine the proof of Bloch’s Theorem to prove that L ≥ 1/24. Solution. Mimic the proof of Block’s Theorem up to the application of Schwarz’s Lemma, i.e. set K(r) = 1 max{| f 0 (z)|||z| = r}, h(r) = (1 − r)K(r), r0 = sup{r : h(r) = 1}. Choose a with |a| = r0 , | f 0 (a)| = K(r0 ) = 1−r . 0 1 Also choose ρ0 = 2 (1 − r0 ) and infer that for all z ∈ B(a, ρ) | f 0 (z)| ≤
3|z − a| 1 , and | f 0 (z) − f 0 (a)| < . ρ0 2ρ20
For z ∈ B(0, ρ0 ) define g(z) = f (z + a) − f (a). By convexity of B(a, ρ0 ) the line segment γ = [a, a + z] is completely contained in B(a, ρ0 ). It follows that Z 1 |g(z)| = | γ f 0 (w) dw| ≤ |z| ≤ 1 =: M. ρ0 The function g is not necessarily injective on B(0, ρ0 ) but the problem only asks for an estimate on the Landau constant for which injectivity is not required. In this setting Lemma 1.2 gives ! ! ! ρ2 |g0 (0)|2 1 1 g(B(0, ρ0 )) ⊃ B 0, = B 0, = B 0, . 6M 4·6 24 This statement rewritten for f yields ! 1 . f (B(a, ρ0 )) ⊃ B f (a), 24 Exercise 2. Suppose that in the statement of Bloch’s Theorem it is only assumed that f is analytic on D. What conclusion can be drawn? (Hint: Consider the functions f s (z) = s−1 f (sz), 0 < s < 1.) Do the same for Proposition 1.10. Solution. Not available.
161
12.2 The Little Picard Theorem Exercise 1. Show that if f is a meromorphic function on C such that C∞ − f (C) has at least three points then f is a constant. (Hint: What if ∞ , f (C)?) Solution. First consider an entire function f that misses three points in C∞ , one on which is ∞. Then there are two distinct numbers a, b ∈ C that are not in the image of f . By Little Picard’s Theorem the function f must be a constant. Next consider a meromorphic function on C that is not entire. Hence ∞ ∈ f (C) and by assumption there are distinct a, b, c ∈ C − f (C). Define a function g(z) :=
1 , f (z) − a
z∈C
1 1 , c−a ∈ C − g(C). Again using Little Picard’s then g is entire because f (z) − a , 0 for all z ∈ C and b−a Theorem conclude that g is a constant function. It cannot be the zero-function, because f was assumed to be meromorphic, hence f (z) , ∞ for at least some z ∈ C. Thus g is a nonzero constant function and f (z) = g(z)−1 + a is also a constant function and the result is established.
Exercise 2. For each integer n ≥ 1 determine all meromorphic functions f and g on C with a pole at ∞ such that f n + gn = 1. Solution. Not available.
12.3 Schottky’s Theorem No exercises are assigned in this section.
12.4 The Great Picard Theorem Exercise 1. Let f be analytic in G = B(0; R) − {0} and discuss all possible values of the integral Z 1 f 0 (z) dz 2πi γ f (z) − a where γ is the circle |z| = r < R and a is any complex number. If it is assumed that this integral takes on certain values for certain numbers a, does this imply anything about the nature of the singularity at z = 0? Solution. Not available. Exercise 2. Show that if f is a one-one entire function then f (z) = az + b for some constants a and b, a , 0. Solution. Not available. Exercise 3. Prove that the closure of the set F in Theorem 4.1 equals F together with the constant functions ∞, 0, and 1. Solution. Not available.
162
