Does entire and v>0 implies f is constant?

Question

Let $f$ : $C$ to $C$ be a function. Assume that $f$ is entire and Im(f(z)) > 0 for all $z$ belongs to $C$. Does that imply $f$ is constant?

I think f will be constant,but can't deduce it Its easy when the imaginary part(v) is constant but does v>0 also implies that f is constant?

what if v is non-negative that is it can attain $0$ also, will the result remain same?

"$f=az+b$ is the only entire function from C to C" ??? This is completely false. What about $f=z^2$ for instance? — Wojowu, Jan 15 '19 at 12:06
Wojowu you are right,actually i mistaken conformal maps and entire functions,my bad — Nilesh Khatri, Jan 15 '19 at 13:49

score 0 · Accepted Answer · answered Jan 15 '19 at 13:35

0

If $f$ is a non-constant entire function then $f(\Bbb C)$ is dense in the plane.

Proof: Suppose that $f(\Bbb C)\cap D(p,r)=\emptyset$. Then $g=\dots$ is bounded, hence $g$ is constant, so $f$ is constant.

answered Jan 15 '19 at 13:35

David C. Ullrich

1 Answers1