0
$\begingroup$

Possible Duplicate:
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.

  • 0
    **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.2015-06-17

1 Answers 1

2

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 $\frac{1}{f(z)}$.