# Proving $f$ is constant How I can answer this by considering the distance between $f(z)$ and $i$. [duplicate]

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

Let $f\colon\mathbb C \to \mathbb C$ be entire. Show that if $|\operatorname{Im}f(z)|\geqslant |\operatorname{Re}f(z)|$ for all $z \in \mathbb C$, then $f$ is constant on $\mathbb C$.

Can I answer this by considering the distance between $f(z)$ and $i$ like in this problem $|\operatorname{Im}f(z)|\leq |\operatorname{Re}f(z)|$ then $f$ is constant?

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## marked as duplicate by Robert Israel, Fabian, Ittay Weiss, Nameless, RahulJan 1 '13 at 9:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

You've already found the exact duplicate. –  Robert Israel Jan 1 '13 at 3:02
I am sorry, I mean to switch the inequality –  Deepak Jan 1 '13 at 3:04
So multiply by $i$. –  Robert Israel Jan 1 '13 at 3:06
Humm..Okay..... –  Deepak Jan 1 '13 at 3:08

## 2 Answers

Hint: Consider the function $$f(z)-1.$$ Notice that $|f(z)-1|\geq \frac{1}{\sqrt{2}}$.

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That will do the job. Cool. –  Deepak Jan 1 '13 at 3:07
And you goota take the reciprocal and argue using Louiville theorem. –  Deepak Jan 1 '13 at 3:08
@Deepak: Exactly. –  Eric Naslund Jan 1 '13 at 3:08

$(2,1),(3,2)\notin f(\mathbb C)$. Also every non-constant entire function assumes each complex number with one possible exception.

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Picard theorem? Any other way then that, I am not familier with that. –  Deepak Jan 1 '13 at 3:06
Eric Naslund already found it. –  Sugata Adhya Jan 1 '13 at 3:12