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.

Answer 1

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?

Answer 2

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.