ANALYSIS QUALIFYING EXAM PROBLEMS BRIAN LEARY ...

22 ANALYSIS QUALS
Thus, f(z1) = f(z2), so f is injective.
b) Let Ω = {z : − 3π
4
3π
< arg(z) < 4 }. This is clearly connected and
simply connected. Let f(z) = z 5/3 , with an appropriate branch cut so
that f is holomorphic on Ω.
Then f ′ (z) = 5
3 z2/3 . Let re iθ ∈ Ω. Then f ′ (re iθ ) = 5
3 r2/3 e 2iθ/3 .
But π
2
< 2
3
θ < π
2 , and so Re(f ′ (z)) > 0 for all z ∈ Ω.
However, f(e 7πi/10 ) = e 7πi/6 = e −5πi/6 = f(e −π/2 ). Hence f is not
injective on Ω.
Problem 9: Let f be a non-constant meromorphic function on the complex
plane C that obeys
f(z) = f(z + √ 2) = f(z + i √ 2).
(In particular, the poles of these three functions coincide.) Assume f has
at most one pole in the closed unit disc D.
a) Prove that f has exactly one pole in D.
b) Prove that this is not a simple pole.
Solution. a) Let z = e 5πi/4 . Then by the periodicity, f takes all its
possible values on the square with corners at z, z + √ 2, z + i √ 2, and
z + √ 2 + √ 2. This square is then contained in D, so it suffices to show
that there is a pole in this square, P . If not, then f is entire. But since
is compact, f is bounded, so by Liousville, f would be constant, a
contradiction. Thus, f has a pole in P .
b) Using the periodicity, we have that
and
z+ √ 2
z
z+i √ 2
z
Thus, we have that
f(w)dw =
f(w)dw =
∂P
z+ √ 2+i √ 2
z+i √ 2
z+ √ 2+i √ 2
z+ √ 2
f(w)dw = 0.
f(w)dw
f(w)dw.
But by the residue theorem, we have that
∂P f(w)dw = 2πi
Res(f, z0).
poles z0
Thus, Res(f) = 0. Thus, counting multiplicities, f has at least two
poles. But as f has only one pole, it cannot be simple.