$f$ be a non constant entire, which of the following is possible?

Question

Re(f(z))=Im(f(z))
$|f|<1$
Im(f(z))< 0
$f\ne 0$

as $f$ non constant so all $1,2,3$ are false as they would imply $f$ as constant.

so true is $4$ say $f(z)=e^z\ne 0$. thank you for a confirmation.

But what is important is not the answer but the reasons. – Jan 05 '14 at 20:05 — , Jan 05 '14 at 20:05

score 3 · Accepted Answer · answered Jan 05 '14 at 20:07

3

Two words for all three: Liouville's theorem. A few more words: Though it applies directly to #2, you have to use some simple conformal maps to make it apply to #1 and #3.

answered Jan 05 '14 at 20:07

Ryan Reich

6,306

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For 1 I would use directly Cauchy's equations. It is less cumbersome. – Jan 05 '14 at 20:08
1

True enough! I had a hammer, I saw three nails. – Ryan Reich Jan 05 '14 at 20:10
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Even easier for $1$: The image lies on the line $y = x$, so the function must be constant by the open mapping theorem. – Ayman Hourieh Jan 05 '14 at 20:18
@Ayman Is the open mapping theorem easier? Once you leave the basics I don't think you can linearly rank these things; certainly, Wikipedia's proof of it quotes several major results, namely the isolation of zeros and Rouche's theorem. – Ryan Reich Jan 05 '14 at 20:22
I'm not saying that the proof of the open mapping theorem is easier. Applying it here is easier. – Ayman Hourieh Jan 05 '14 at 20:25
Oh, yes, it does definitely apply directly. – Ryan Reich Jan 05 '14 at 20:26

score 0 · Answer 2 · edited Dec 14 '17 at 10:33

0

Picard's Little Theorem can be directly applied to 1,2 and 3.!

edited Dec 14 '17 at 10:33

Raskolnikov

16,108

answered Mar 07 '15 at 15:48

Jesse P Francis

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$f$ be a non constant entire, which of the following is possible?

2 Answers2