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$f$ 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.

Gary
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Sugata Adhya
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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
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    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

4 Answers4

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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

Dennis Gulko
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  • why is $|e^{if}|=e^{-\text{Im(f)}}$? –  Nov 04 '19 at 14:13
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    Just to help somebody who can't figure out why $|e^{if}| = e^{- \Im (f)}$, assuming $f = u+iv$:

    $|e^{if}| = |e^{i (u+iv)}| = |e^{-v}| | \cos u + i \sin u| = |e^{-v}|$

     – Maxis Jaisi Jan 20 '21 at 10:30
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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.

Michael Hardy
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lee
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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.

ncmathsadist
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0

Another way is to use the conformal map $F:\mathbb{H}\rightarrow\mathbb{D}$

$$ F:z\mapsto \frac{i-z}{i+z}. $$ We can write $$ g := F\circ f:\mathbb{C}\rightarrow \mathbb{D}. $$ Then $g$ is entire and bounded by $1$, so it is constant. Hence, $f$ is constant as well.

David
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