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Let $f\colon\mathbb C \to \mathbb C$ be entire. Show that if $|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ for all $z \in \mathbb C$, then $f$ is constant on $\mathbb C$. How I can answer this by considering the distance between $f(z)$ and $i$.

neno
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    Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. – Did Nov 03 '12 at 11:08

3 Answers3

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$|\mathrm{Im}f(z)|\le |\mathrm{Re}f(z)|$ implies that $|f(z)-i|\ge \dfrac{\sqrt{2}}{2}$. It follows that $g(z)=\dfrac{1}{f(z)-i}$ is a bounded entire function, and due to Liouville's theorem, it must be a constant.

23rd
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  • How did |f(z)−i|≥ √ 2 /2 happen? – neno Nov 04 '12 at 14:15
  • I tried to do it but I ended with |f(z)-i|>=1. – neno Nov 04 '12 at 14:22
  • @neno: $\frac{\sqrt{2}}{2}$ is the distance between $i$ and the lines $y=\pm x$, i.e. the boundary of the region ${z\in\mathbb{C}:|\mathrm{Im}f(z)|\le |\mathrm{Re}f(z)|}$. – 23rd Nov 04 '12 at 14:25
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Alternatively, $\textrm{Re} \, f(z)^2 = \left(\textrm{Re} \, f(z) \right)^2 - \left(\textrm{Im} \, f(z) \right)^2 \geq 0$. This implies that $f(z)^2$ is constant.

WimC
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follows directly from picard's little theorem

Koushik
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