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$f$ is an entire function, and $1 \leq |f(z)-i|$ for every $z \in \mathbb C$. Show that $f$ is constant.

I don't know how to approach this. I tried writing $f(z)= \Sigma _{n=0}^\infty a_nz^n$, but didn't really see how can I proceed.

Thanks in advance for your assistance!

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    $\begingroup$ Liouville's theorem states that any bounded entire function must be constant. Use it for $1/(f(z)-i)$, which is also entire function $\endgroup$ – Michael Galuza Jun 6 '15 at 12:35
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Observe, that $$ g(z)=\frac{1}{f(z)-i} $$ is also entire analytic, and $\lvert g(z)\rvert\le 1$.

Hence $g$ is constant, and so is $f$.

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g(z)=f(z)-2 i lives on or outside the unit circle. Thus i/g(z) is entire and bounded. Thus it is constant and hence f is constant.

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