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Does there exists an entire function such that for each $|z|>C$ for some constant $C$, $|f(z)|=|z|+1$?

Can I assume that since $f$ has a pole in infinity, $Imf$ or $Ref$ must have a pole at infinity too? If so, I can use this fact to define an exponent function of $Imf$ or $Ref$ (the one that goes to infinity) or their minus, and show that function is a bounded entire function, and therefore constant. Which gives a contradiction to the conditions about $f$.

  • 0
    Hint: Consider $1/f$... – 2017-02-14
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    I understand that $1/f$ is bounded for $|z|>1/C$, but how do I deal with the circle $|z|<1/C$ which might have poles? – 2017-02-14

1 Answers 1

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Define $g(z)=\frac{f(z)-f(0)}{z}$ is $z\not=0$ and $g(0)=f'(0)$. Easy to verify that $g$ is also entire.

Now we will see that $g$ is bounded and therefore by Liouville's theorem it is constant.

for $|z|>C$

$|g(z)|\leq 1+\frac{1+|f(0)|}{|z|}$ from your hypothesis. Therefore $g$ is bounded in {$|z|>C$}. {$|z|\leq C$} is compact, so $g$ is bounded here too.

Therefore $g(z)=C$, which implies $f(z)=Cz+f(0)$.