Math 411: Honours Complex Variables - University of Alberta

is entire. Since p is a nonconstant polynomial, we have lim |p(z)|= ∞ and thus
|z|→∞
lim
|z|→∞
|f(z)|= 0. Let R > 0 be such that |f(z)|≤ 1 for |z|> R. Since f is continuous it
is bounded on BR[0], and by the choice of R, it is bounded on C\BR[0], too, and thus
bounded on all of C. By Liouville’s Theorem, f is thus constant, and so is therefore
p, which is a contradiction.
Problem 5.1. Let D ⊂ C be open.
(a) Suppose f : D → C has an antiderivative on D and 0 /∈ f(D). Show that there
exists a holomorphic function g: D → C such that f = exp◦g.
Hint: consider f′ (z)
f(z) .
(b) If D is star shaped, and f: D → C is holomorphic such that 0 /∈ f(D), show that
there exists a holomorphic function g: D → C such that f = exp◦g.
Problem 5.2. Let z0 ∈ C, let r > 0, and let f: Br[z0] → C be continuous such that
f|Br(z0) is holomorphic. Show that
f(z) = 1
�
f(ζ)
2πi ∂Br(z0) ζ −z dζ
for all z ∈ Br(z0). Hint: For θ ∈ (0,1), apply the Cauchy integral formula to
z ↦→ f(θ(z −z0)+z0) on ∂Br
θ (z0); then let θ → 1 − .
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