# If $f$ is entire and does not take values in $|w|\leq 1$, then $f$ is constant [duplicate]

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Magnitude of differentiable complex function $f(z)$

I have the question

Let $f(z)$ be an entire function and assuming $f(z)$ does not take values in $|w|\leq 1$, show that $f(z)$ is identically constant.

I tried to prove with Liouville's theorem, but I couldn't find the correct implementation., because I did not understand the meaning of "$f(z)$ does not take values in $|w|\leq1$". Could you help me please? If you can, could you give the proof?

Thanks.

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## marked as duplicate by Marvis, Chris Eagle, t.b., Gerry Myerson, Asaf KaragilaJun 3 '12 at 15:59

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

If $f(z)$ does not take values in the unit disc, then it does not come arbitrarily close to $0$. What does that mean for $g(z)=\frac{1}{f(z)}$? – Gregor Bruns Jun 2 '12 at 14:23
yes your right . ı got solution from there. thanks – Brhn Jun 2 '12 at 14:33
Picards theorem : If $f$ is a non-constant entire function on $\mathbb{C}$ then then $f$ assumes all the values in $\mathbb{C}$ except possibly one value. Since your function misses out on all values inside the unit disc. It is automatically a constant function. – Aloysius Godinho Jun 17 at 8:47

## 1 Answer

If $|f(z)|\not\leq 1$, then $|f(z)|>1$ for all $z\in \mathbb{C}$. Take reciprocals of the inequality. Then you should see how to apply Liouville to $\frac{1}{f(z)}$.

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