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

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 $i$.

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## 3 Answers

$|\mathrm{Im}f(z)|\le |\mathrm{Re}f(z)|$ 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.

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How did |f(z)−i|≥ √ 2 /2 happen? – neno Nov 4 '12 at 14:15
I tried to do it but I ended with |f(z)-i|>=1. – neno Nov 4 '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 4 '12 at 14:25

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.

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follows directly from picard's little theorem

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