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$f$ be analytic such that $\Re(f)\ge 0$ then

  1. $\Im(f)=$ constant

  2. $\Im(f)\ge 0$

  3. $f=$ constant

  4. $\Re(f)=|z|$

what I have done is if $f=u+iv$ then consider, $g(z)=e^{u+iv}$ then $|g|=e^{u}$ as $g$ never vanishes so $g$ must be constant so 3 is correct in my guess. 2 is not necesarily true also 4, please help for rigoriousness.

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Two things that might help: Firstly, you need to know the set on which $f$ is analytic (and on which $Re(f)\ge 0$). e.g. is $f$ analytic in all of $\mathbb{C}$? Secondly, I think you will find that $|g|=e^{u}$, and not $|g|=e^{|u|}$. –  Old John Jun 15 '12 at 11:21
    
Dear sir, yes all $\mathbb{C}$ oh that was typo, I am correcting it. –  El Angel Exterminador Jun 15 '12 at 11:24
1  
$G(z) =\frac{1}{f(z)+1}$ is entire and bounded –  N. S. Jun 15 '12 at 11:38

1 Answer 1

up vote 4 down vote accepted

Then here is a sketch solution:

Put $g = e^{-f}$, and consider $|g|= e^{-u}\le 1$, so that $g$ is bounded in all of $\mathbb{C}$. Then use Liouville's Theorem.

Added: Note also that this only proves that $g$ is constant in $\mathbb{C}$, and you will need to do a bit more work (e.g. calculate $g'$) to show that $f$ is constant in $\mathbb{C}$.

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so $e^{-f}=c$ so $-f'e^{-f}=0$ so $f'=0$ so $f=constant$ am I ryt? –  El Angel Exterminador Jun 15 '12 at 12:14
    
Yes, that is right. –  Old John Jun 15 '12 at 12:54

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