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we need to pick out the cases where $f:\mathbb{C}\rightarrow\mathbb{C}$ is analytic but not neccessarily constant.

$1.$$ \Im(f'(z))>0$ for all $z$

$2.$ $f(n)=3\forall n\in\mathbb{Z}$

$3$ $f'(0)=0$ and $|f'(z)|\le 3\forall z$

My attempt:

$1$. take $f(z)=\sin z+ 2iz$

$2$. take $f(z)=3\sin (\frac{\pi z}{2})$

$3$. $f'$ must be constant as $f'$ is bounded analytic,hence constant so $f(z)=cz$, for some complex constant $c$.

Please tell me whether I am right.

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1  
I don't think $1$. works. At $z = 2 + 2i$ it looks like $\mathfrak{Im}(f'(z)) < 0$.. –  Cocopuffs Jan 14 '13 at 21:46
    
In 3 do you mean $f(0)=0$? –  user108903 Jan 14 '13 at 21:47
1  
For 1) you can take $f(z) = iz$. –  mrf Jan 14 '13 at 21:48

1 Answer 1

up vote 3 down vote accepted
  1. Simply let $f'(z)=i$ (constant). Then $f(z)=iz$ is a solution.
  2. is ok the way you do it
  3. You missed a little point: $f'(0)=0$ and $f'$ constant implies $f'(z)=0$ and $f$ constant. If the first condition were $f(0)=0$, then $f(z)=cz$ would be fine.
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1. In such a case, $f'$ is necessarily constant, so an answer similar to Hagen's is all that is possible. 2. Although something similar will work, what is written in the question is incorrect. E.g., try $n=0$. –  Jonas Meyer Jan 14 '13 at 21:56

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