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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 |z 0 | = 1. Let h(z) be a holomorphic function on { |z| < R } with h (z 0 ) 6 = 0. Let m be a positive integer and
f (z) =
h(z) (z − z 0 )m^
Show that if (^) ∞ ∑
n=
anzn
denotes the power series expansion of f on { |z| < 1 }, then
lim n→∞
an an+
= z 0.
Solution of Problem 1. First of all, we express f (z) in the form
∑^ m
k=
Ak (z − z 0 )k^
with∑ A 1 , · · · , Am ∈ C, where Am = h (z 0 ) 6 = 0 and g(z) is a power series ∞ n=0 bnz
n (^) 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−^1 z 0 )k (or
differentiating the geometric series (^) z−^1 z 0 in z (k − 1)-times) and noting that (n+k− 1 k− 1
(n+k− 1 n
, we have
an = bn +
∑^ m
k=
(−1)kAk
(n + k − 1)(n + k − 2) · · · (n + 2)(n + 1) (k − 1)! (z 0 )n+k^
In the computation of the limit
lim n→∞
an an+
since Am = h (z 0 ) 6 = 0 and (^1) r <
∣ (^) z^10
∣ and |bn| ≤ (^) rBn , the dominant term from
an is
(−1)mAm
(n + m − 1)(n + m − 2) · · · (n + 2)(n + 1) (m − 1)! (z 0 )n+m
and we get
lim n→∞
an an+
= lim n→∞
(−1)mAm (n+m−1)((mn−+1)!(m−z2) 0 )···n+(nm+2)(n+1)
(−1)mAm (n+m)((mn+−m1)!(−z1) 0 )···n(+1+n+3)(mn+2)
= 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) =
f
z¯
whenever |z| > 1, and argue as in the Schwarz reflection principle.
Solution of Problem 2. Define
g(z) =
f (z) for |z| ≤ 1
1 f (^) ( (^) z¯^1 )
for |z| ≥ 1.
This is well defined, because when |z| = 1, we have
1 z¯
= z
and 1 f
z¯
) =^
f (z)
= f (z)
(b) Fix 0 < α < 1. Show that the holomorphic function f defined by
f (z) =
∑^ ∞
n=
2 −nαz^2 n for |z| < 1
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 ∈ N let zp,k = ei^
2 πp 2 k^.
Then, since ei^2 πp^2 n−k = 1 for n ∈ N and n ≥ k,
it follows that for 0 < r < 1
f (rzk,p) =
∑^ k−^1
n=
r^2 nei^2 πp^2
n−k
∑^ ∞
n=k
r^2 n,
which becomes unbounded as r → 1 −. Since the set of points
{zk,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) =
∑^ ∞
n=
2 −nαei^2
nx
for x ∈ R is continuous but nowhere differentiable on R. The function
f (z) =
∑^ ∞
n=
2 −nαz^2
n for |z| < 1
is continuous on D = { |z| ≤ 1 }, because
∑^ ∞
n=
∣ 2 −nαz^2 n^
∑^ ∞
n=
2 −nα^ < ∞
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.
