Prove $f(z)=c$ when $\operatorname{Re}(f)$ has an upper bound ($f$ is analytic)

Question

As the title says

Prove $f(z)=c$ when $\operatorname{Re}(f)$ has an upper bound, given that $f$ is analytic

I am just getting introduced on complex analysis, and I found proving that $f(z)=c$ when $f'(z)=0$ (some previous exercise) relatively easy, but this one looks very strange since $\operatorname{Re}(f)$ being upper bounded sounds very generic.

Any help will be appreciated.

score 4 · Accepted Answer · answered Sep 14 '18 at 09:28

4

Let $g(z)=e^{f(z)}$. Then $\bigl\lvert g(z)\bigr\lvert=e^{\operatorname{Re}f(z)}$ and therefore $g$ is bounded. Can you take it from here?

answered Sep 14 '18 at 09:28

José Carlos Santos

427,504

I am afraid not. Perhaps my way of thinking about it is wrong, but i am trying to think of a way to make the partial derivatives turn out zero, so i try to think of a way g being bounded does that. – Manouil Sep 14 '18 at 09:48
@Manouil Just apply Liouville's theorem. – José Carlos Santos Sep 14 '18 at 09:52
Ahh,i was not aware of that theorem, now it all makes sense! Thanks a lot sir! – Manouil Sep 14 '18 at 09:55
@Manouil I'm glad I could help. – José Carlos Santos Sep 14 '18 at 09:58

score 0 · Answer 2 · answered Sep 14 '18 at 09:50

0

You will need to argue that $|g(z)|=e^{\Re f(z)}$ has an upper bound, say $k$.(can you see why?) Thereafter, you can apply Liouville's Theorem which asserts that every bounded entire function must be constant.

answered Sep 14 '18 at 09:50

Cleytus

457
3
17

Prove $f(z)=c$ when $\operatorname{Re}(f)$ has an upper bound ($f$ is analytic)

2 Answers2