An entire function whose imaginary part is nonnegative must have a constant imaginary part

Question

Let $f(z)$ be entire such that $ \text{Im} (f(z)) \geq 0.$ Show that $\text{Im}(f(z))$ is constant.

So far what I have done is this:

Consider $|e^{if(z)}| = |e^{v(x,y)}| \geq 1.$

Then this implies $|e^{-v(x,y)}| \leq 1.$

I have shown that $e^{-v(x,y)}$ is bounded.

Since f is entire, does it follow that its real and imaginary parts are also entire?

Think along these lines. Find a function that is entire and bounded using this technique. — Sarvesh Ravichandran Iyer, Nov 03 '16 at 11:38
Hi I have considered taking the exponential. If f=u+iv is entire, does it follow that u and v are also entire? — stonehenge, Nov 03 '16 at 12:14
Yes. But the answer from Fred below is better than my idea. Actually, I saw a similar question: if an entire function has real part bounded, then it is a constant. In that case, you took exponential to prove this via Liouville. I looked at the question, and assumed a similar technique would work. — Sarvesh Ravichandran Iyer, Nov 03 '16 at 12:22
Precisely, my friend. That's the question. Only difference, I looked it up in Shakarchi and solved it myself with this trick. I didn't know the other tricks, boy they are interesting though. The techniques used in that question could be of help to you. — Sarvesh Ravichandran Iyer, Nov 03 '16 at 12:31

score 1 · Answer 1 · answered Nov 03 '16 at 11:38

1

One can say more: $f$ is constant. Let $g:=f+i$, then $Im(g(z)) \ge 1$, $g$ has no zeros , hence $h:=1/g$ is entire.

Now show that $|h| \le 1$ and do not forget Mr. Liouville.

answered Nov 03 '16 at 11:38

Fred

1 Answers1