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Homework Assigned on February 17, 2009 due February 24, 2009 Problem 1 (from Stein & Shakarchi, p.67, #14). Let R > 1 and z 0 ∈ C with = 0. |z 0 | = 1. Let h(z ) be a holomorphic function on { |z | < R } with h (z 0 ) Let m be a positive integer and f (z ) =
Show that if
h(z ) . (z − z 0)m
∞
a z
n
n
n=0
denotes the power series expansion of f on { |z | < 1 }, then an = z 0 . n→∞ an+1
lim
Solution of Problem 1. First of all, we express f (z ) in the form m
k=1
Ak
(z − z 0 )k
+ g (z )
with A1 , · · · , Am ∈ C, where Am = h (z 0 ) = 0 and g (z ) is a power series ∞ n n=0 bn z with radius of convergence at least equal to R so that for any |z 0 | < r < R we can find a positive number B such that
|bn| ≤
B rn
for all nonnegative integer n. By using the binomial expansion of (z −1z differentiating the geometric series n+k−1 = n+nk−1 , we have k−1
(−1) A a = b + n
m
k
n
k=1
k
1
z −z0
(or
in z (k − 1)-times) and noting that
(n + k − 1)(n + k − 2) · · · (n + 2)(n + 1) . (k − 1)!(z 0 )n+k
In the computation of the limit an , n→∞ an+1
lim
k
0)
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< and |b | ≤
1 B since Am = h (z 0 ) = 0 and r , the dominant term from n z0 rn an is (n + m − 1)(n + m − 2) · · · (n + 2)(n + 1) (−1)m Am (m − 1)!(z 0 )n+m and we get (n+m−1)(n+m−2)···(n+2)(n+1) (m−1)!(z0 )n+m (n+m)(n+m−1)···(n+3)(n+2) (m−1)!(z0 )n+1+m
(−1)m Am an lim = lim n→∞ an+1 n→∞ (−1)m Am
= z 0 .
The dominant term from an means that an minus the dominant term and then divided by the dominant term would have limit zero when n → ∞. Problem 2 (from Stein & Shakarchi, p.67, #15). Suppose f is a nowhere vanishing continuous function on the closure D of the open unit disk D and f is holomorphic in D. Prove that if
|f (z )| = 1 whenver |z | = 1, then f is constant. Hint: Extend f to all of C by f (z ) =
1 f 1
z¯
whenever | z | > 1, and argue as in the Schwarz reflection principle.
Solution of Problem 2. Define
f (z ) for |z | ≤ 1 g (z ) = ( ) for |z | ≥ 1. 1
f
1
z ¯
This is well defined, because when |z | = 1, we have 1 = z z¯ and
1 f 1 z¯
=
1 f ( z )
= f (z )
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due to the assumption that |f (z )| = 1 for |z | = 1. The function g (z ) is continuous on all of C and is holomorphic outside the boundary of D, because f is holomorphic on D and continuous on D and g (z ) locally on C − D is a convergent series from its definition. By the continuity of g (z ) on D, for any bounded solid triangle G in C whose boundary is T and any 0 < ε < 1, if we denote by T ε the boundary of G − { 1 − ε < |z | < 1 + ε }, then lim
ε→0
By the holomorphicity of g on
g (z )dz =
T ε
g (z )dz.
T
C − D,
we have
g (z )dz = 0 for 0 < ε < 1 .
T ε
Hence
g (z )dz = 0.
T
By Morera’s theorem g (z ) is holomorphic on all of C. Since f (z ) is uniformly bounded on D, it follows that g (z ) is uniformly bounded on all of C and must be constant by Liouville’s theorem. Hence f is constant. Remark. An alternative way of proving the constancy of f (z ) is to apply the maximum modulus principle to f (z ) on D and then to f (1z) to show that |f (z )| ≡ 1 on D. The local constancy of the real part of a holomorphic function implies its local constancy. This alternative proof works even when D is replaced by any bounded connected open subset of D, because only the maximum modulus principle is used and not the Schwarz reflection principle. Problem 3 (from Stein & Shakarchi, p.67-68, Problem 1). For a function f defined on the open unit disk D, a point w of the boundary C of D is said to be regular for f if there is an open neighborhood U of w and a holomorphic function g on U , so that f = g on D ∩ U . A function f defined on D cannot be continued analytically past the unit disk if no point of C is regular for f . (a) Let ∞
2n
z f (z ) =
for |z | < 1 .
n=0
Notice that the radius of convergence of the above series is 1. Show that f cannot be continued analytically past the unit disk. Hint: Suppose θ = 22πp k , iθ iθ where p and k are positive integers. Let z = re ; then f re → ∞ as r → 1.
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(b) Fix 0 < α < 1. Show that the holomorphic function f defined by ∞
2 f (z ) =
−nα
n
z 2
for |z | < 1
n=0
extends continuously to the unit circle, but cannot be analytically continued past the unit circle. Hint: Use Euler’s formula eiθ = cos θ + i sin θ and the Weierstrass continuous nowhere differentiable function reproduced below from “Fourier Analysis” by Stein & Shakarchi, pp.113-118.
Solution of Problem 3. (a) For positive integers k, p ∈ Then, since n−k ei2πp2 = 1 for n ∈ N and n ≥ k,
N let z p,k
2πp
= e i 2k .
it follows that for 0 < r < 1 k−1
f ( rz k,p
∞
r )=
2n i2πp2n−k
r +
e
n=0
2n
,
n=k
which becomes unbounded as r → 1 −. Since the set of points
{z k,p } p,k∈N is dense in the unit circle { |z | = 1 }, it follows that f (z ) cannot be extended analytically past any point of the unit circle.
(b) According to Theorem 3.1 of the book “Fourier Analysis” by Stein & Shakarchi whose pages 113 – 118 are reproduced below, for 0 < α < 1 the function ∞
g ( x) = 2
−nα
n
ei2
x
n=0
for x ∈
R is
continuous but nowhere differentiable on
R.
∞
f (z ) = 2
−nα
n
z 2
for |z | < 1
n=0
is continuous on
D
= { |z | ≤ 1 }, because ∞
∞
2
−nα
n=0
2n
z
≤ 2
−nα
n=0
< ∞
The function
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for |z | ≤ 1 due to α > 0. Since f eiθ = g (θ ) for θ ∈ R and since g (θ) is nowhere differentiable as a function of θ ∈ R, it follows that no extension of f (z ) can be complex-differentiable at any point z on the unit circle. Thus f (z ) cannot be analytically continued past the unit circle.
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