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$|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant

$f$ is entire such that $|\mathrm{Re}f(z)|<|\mathrm{Im}\,f(z)|$ for all $z \in \mathbb C$. To prove $f$ is a constant function.

I know I can prove $f$ do not have essential singularity at $\infty$ for this case which will eventually lead function to be polynomial. Then I can even go further with $f(z)= \alpha z^n $ for some large $z$. Then I don't know where to go. Am I going somewhere with that or just doing something not important. Any hint much appreciated.

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marked as duplicate by WimC, Alexander Gruber, Henry T. Horton, Nameless, QiL Dec 30 '12 at 22:18

This question was marked as an exact duplicate of an existing question.

So it sounds like the image of $f$ would be contained a half-plane. Have you considered whether the half-plane admits a bounded holomorphic map, say, to the disc? – user54535 Dec 30 '12 at 19:36
By Picard's theorem, an entire function that misses two values must be a constant function. Under the conditions you give for the image $f$ misses many values, so it must be a constant function. – user44441 Dec 30 '12 at 20:06
up vote 0 down vote accepted

The two straight $x=y$ an $x=-y$ determine 4 sets in the plan. Your function $f$ assume their values in the two sets(upper and lower strict) containing the points (0,1) and (0,-1). Now take a small ball centered in the point (-1,0) that do not intersect the straight $x=y$ and $x=-y$. How $f$ do not assume any value in this ball(say ratio $\epsilon$), we must $|f-(-1,0)|>\epsilon$. Therefore $\frac{1}{|f-(-1,0)|}<\epsilon$, $\frac{1}{f-(-1,0)}$ being holomorphic, must be constant( Liouville). And also $f$.

Another way not so elementary follow from comment above about Picard's theorem

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