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

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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?

You've already found the exact duplicate. – Robert Israel Jan 01 '13 at 03:02 — Robert Israel, Jan 01 '13 at 03:02
I am sorry, I mean to switch the inequality – Deepak Jan 01 '13 at 03:04 — Deepak, Jan 01 '13 at 03:04

score 4 · Answer 1 · answered Jan 01 '13 at 03:05

4

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

answered Jan 01 '13 at 03:05

Sugata Adhya

3,979

Picard theorem? Any other way then that, I am not familier with that. – Deepak Jan 01 '13 at 03:06
2

Eric Naslund already found it. – Sugata Adhya Jan 01 '13 at 03:12

score 4 · Accepted Answer · answered Jan 01 '13 at 03:05

4

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

answered Jan 01 '13 at 03:05

Eric Naslund

72,099

That will do the job. Cool. – Deepak Jan 01 '13 at 03:07
1

And you goota take the reciprocal and argue using Louiville theorem. – Deepak Jan 01 '13 at 03:08
@Deepak: Exactly. – Eric Naslund Jan 01 '13 at 03:08

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

2 Answers2