Math 6350 Homework #7 Solution 1. Ahlfors p. 123 #4: If f ( 1 /(1 − |z |), find the best f (z ) is analytic for | z | < 1 < 1 and | f ( f (z )| ≤ 1/ ( n) estimate of |f (0)| that Cauchy’ Cauchy’ss inequalit inequality y will yield. yield. (That is, find the optimal r in (25).) Solution: On
the circle C of of radius r radius r around zero, we have f (ζ )| ≤ |f (
1 1−r
by assumption. Thus we use M use M = 1/(1 − r) to get n ! |f (n) (0)| ≤ n!
1 . rn (1 − r)
We need to minimize the right hand side as a function of r, which is equivalent to maximizing r n (1 − r) (0, 1). Since Since the deriv derivative ative is nrn−1 − (n ( n + 1)r 1)rn , the r ) for r ∈ (0, n maximum occurs at r = n+1 , and we get !(n + 1)n+1 n!(n |f (0)| ≤ . nn (n)
2. Suppose you want want to find an estimate estimate for the error in approximat approximating ing f ( f (x) = ex sin x at x = 1/2 by P 3 (x), the Maclaurin polynomial f (0) 2 f (3) (0) 3 + f (0)x (0)x + P 3(x) = f (0) f (0) + f x + x.
2! 3! Estimate the error using Calculus II methods, then estimate the error using the remainder formula (29). Which method does better? Calculus II we would compute the maximum M maximum M 4 of f interval f (4) (x) on the interval 1 [0, [0, 2 ] and obtain the bound
Solution: In
|f ( f ( 12 ) − P 3 ( 12 )| ≤
M 4 . 4! · 24
We easily compute that f that f (4) (x) = − 4ex sin x and the maximum of this is 16, M 4 = |f (4) ( 12 )| = 4e1/2 sin 12 = 3.16, which leads to the bound | f ( 12 ) − P 3 ( 12 )| ≤ 0. 0 .008. To use the remainder formula 1 R4 ( 12 ) = ( 12 − 0)4 f n ( 12 ) = 4 2 · 2πi
f ( f (ξ ) dξ 1 n, ( )ξ ξ − − C 2
we need to pick a circle C of radius r larger than 12 . Using the formula C of radius r iθ
= e f (re )| = e |f (
r cos θ
cosh(r cosh(r sin θ) − cos (r cos θ) 2
≤ e r cos θ cosh r sin θ = 12 (er cos θ+r sin θ + er cos θ−r sin θ ) ≤ e
√
r 2
, 1
we obtain
√
er 2 1 |R4 ( 2 )| ≤ 32π
The maximum of (r − 12 )r3 e−r we get
√ 2
√
1 er 2 |dξ | 1 = . 1 4 16 (r − 12 )r3 C |ξ | − 2 |ξ |
occurs around r = 3 where it’s about equal to one, so
|R4 ( 12 )| ≤
1 ≈ 0.06. 16
Certainly the Calculus II method does better. 3. Ahlfors p. 130 #2: Show that a function which is analytic in the whole plane and has a nonessential singularity at ∞ reduces to a polynomial. Let f be such a function, and define g(z ) = f (1/z ). Then by definition of essential singularity at infinity, we know that g has a nonessential singularity at z = 0. Hence either it is removable or it is a pole of g. Solution:
If the singularity is removable then g is bounded near zero and so f is bounded at infinity, but the only analytic functions that are bounded on the entire complex plane are constants. On the other hand if the singularity is a pole, then we must have some singular part by the discussion on page 128–129, which we write as n
g(z ) = c z − + h(z ) k
k
k=1
where h is analytic at 0, n is some positive integer, and ck are some constants. Therefore we have n
f (z ) = g(1/z ) = c z + j(z ), k
k
k=1
where j (z ) = h(1/z ). Although a priori we are not sure if j is defined at z = 0, the expansion tells us that j(z ) is bounded in a neighborhood of zero since it’s just f (z ) minus a polynomial; hence j is actually analytic on the entire complex plane and has a finite limit h(0) as z → ∞ , which means it must actually be constant. Therefore f is a polynomial. 4. Ahlfors p. 130 #4: Show that any function which is meromorphic in the extended plane is rational. Solution: We
first want to show there are only a finite number of poles. If there were an infinite number of poles, then they could not all be contained in a bounded set or else they would not be isolated. Hence there would have to be a sequence of them which converged to infinity. But then the function 1/f (1/z ) would have a sequence of zeroes which converged to the origin, and the origin would not be an isolated zero (which contradicts the fact that the original function is assumed to have an isolated singularity at infinity). 2
So there are a finite number of poles {b1 , . . . , bn } and each pole bk has some order rk , which implies that g(z ) = (z − b1 )r · · · (z − bn )r f (z ) 1
n
is analytic everywhere on the complex plane, and has a pole or zero of finite order at infinity since f does. By the previous problem, g(z ) must be a polynomial, and therefore f (z ) is a quotient of polynomials. 5. How many roots of f (z ) = z 4 + z + 1 are there in the unit disc? Draw a picture of Γ = f ◦ γ where γ is the unit circle to determine n(Γ, 0). If γ is the unit circle then n(Γ, 0) is the number of roots. The easiest thing to do is to ask a graphing utility to plot f ◦ γ . I get: Solution:
It is easy to look at this graph and count the winding around w = 0; the graph goes counterclockwise twice around w = 0 and the other loops don’t matter. Hence there must be exactly two roots of z 4 + z + 1 inside the unit disc. Numerical calculation shows that the roots are 0.7271 + 0.9341i,
−0.7271 + 0.4300i,
−0.7271 − 0.4300i,
0.7271 − 0.9341i.
Two of these have absolute value 1.184 and two of them have absolute value 0.8447, which agrees with what we found graphically. 6. Ahlfors p. 133 #1: Determine explicitly the largest disk about the origin whose image under the mapping w = z 2 + z is one to one. need to solve z 12 + z 1 = z 22 + z 2 and see what the first solution with = z 2 is. We have z 1 (z 1 − z 2 )(z 1 + z 2 ) + (z 1 − z 2 ) = 0, Solution: We
or (z 1 − z 2 )(z 1 + z 2 + 1) = 0. 3
If we want z 1 = z 2 then we must have z 1 + z 2 + 1 = 0. If | z 1 | < r and | z 2 | < r, then we have 1 = | z 1 + z 2 | < 2r and thus if r ≤ 12 we will get a contradiction; hence on the open disk of radius function is one-to-one.
1 2
the
On the other hand if r > 12 (and r < 32 ), we may choose z 1 = − 2r+1 and z 2 = 2r4−3 . 4 Observe that | z 1 | < r2 + 14 < 2r + r2 = r, and | z 2 | = 3−42r < 3−4 1 = 12 < r as long as r < 32 . So the largest radius is 12 . 7. Ahlfors p. 136 #1 and #3: Show by use of (36), or directly, that | f (z )| ≤ 1 for | z | ≤ 1 implies 1 |f (z )| . ≤ 1 − |f (z )|2 1 − |z |2 Prove that equality implies that f (z ) is a linear transformation. Solution: When M
= R = 1, equation (36) becomes
f (z ) − f (z ) z − z ≤ . 1 − f (z )f (z ) 1 − z z f (z ) − f (z ) 1 − f (z )f (z ) z − z ≤ 1 − z z . 0
0
0
0
Rewrite this in the form
0
0
0
0
Since this is true for all z = z 0 in the disc, it will still be true in the limit as z → z 0 . When this happens the left side becomes the absolute value of | f (z 0 )|, and we get 2
, 1 ) − | | f (z |f (z )| ≤ 1 − |z | 0
0
0
2
which is equivalent to what we want.
A direct proof involves applying linear transformations to an arbitrary disc-preserving analytic map so that it will become a map that also preserves the origin. (This is the same thing we do to derive formula (36).) So suppose |f (z )| ≤ 1 whenever |z | < 1, and that f (z 0 ) = w0 for some |z 0 | < 1 and |w0 | < 1. Let z − w0 z − z 0 and S (z ) = T (z ) = . 1 − w0 z 1 − z 0 z Then S and T are both invertible and preserve the unit disc, and S (w0 ) = 0 and T (z 0 ) = 0. Therefore the function
− g(z ) = S f T (z ) 1
preserves the unit disc and has g(0) = 0. We therefore have | g (0)| ≤ 1. 4
By the Chain Rule we have
|S (w0 )||f (z 0 )| 1 − , |g (0)| = |S (w0 )||f (z 0 )||(T ) (0)| = |T (z 0 )| which we can solve to get
|T (z 0 )||g (0)| |T (z 0 )| |f (z 0 )| = ≤ . |S (w0 )| |S (w0 )| We easily compute that S (w0 ) = 1−|1w | and T (z 0 ) = 1−|1z | , which gives us what we want. 0
2
0
2
Now to solve #3 (to see what happens when we have equality) it’s a bit easier to use the second derivation since that reduces it to the Schwarz Lemma and gives an equality criterion. So if | f (z 0 )| is equal to the right side, then we must have |g (0)| = 1 and this implies that g is just a rotation (multiplication by a complex number of unit modulus), which in particular is an LFT. Hence f = S −1 ◦ g ◦ T is a composition of LFTs and therefore also an LFT.
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