$f$ is an entire function with Im $f\geq 0$

$f$ is an entire function with $\operatorname{Im}f \geq 0$. Then which of the followings are true: 1) $f$ is constant. 2) $\operatorname{Re}f$ is constant. 3) $f = 0$. 4) $f'$ is a non-zero constant.

That (3) & (4) are wrong can be shown by using $f(z) = i$. But I'm clueless about the remaining two options.

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Would you be able to answer the question if we changed it to "$f$ is entire with its image lying in the unit disk?" –  froggie Nov 24 '12 at 15:00
Entire functions restricted "so much" usually want to be constants. Note also that 1)$\Rightarrow$2). –  Berci Nov 24 '12 at 15:01
I think I've got the point. Non-constant entire function comes arbitrarily close to each complex number. Then both 1 & 2 must be true. –  Sugata Adhya Nov 24 '12 at 15:03

Consider $g=\dfrac{1}{f+i}$(thanks to point out my mistake), then $g$ is entire and $|g|<1$, now you can conclude that $g$ is a constant with the Liouville's theorem(I forget it's name). Consequently $f$ is also a constant. Also by Picard's theorem, that the range of any nonconstant entire function must contain the complex except for at most one point.

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You rather mean $g=1/(f+i)$. –  Berci Nov 24 '12 at 15:08

Hints:
(1) If $f$ is entire then so is $e^{if}$
(2) $|e^{if}|=e^{-\operatorname{Im}(f)}\leq 1$
(3) Use Liouville's theorem

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Such a function $f$ would map the entire plane into the upper half-plane. The Picard theorem says this is impossible, since an entire nonconstant function must map the plane onto itself or onto the complex plane punctured by a "missing" point. The mapping $f$ must be constant.

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