Entire function constant on $|z|=1$

Question

Suppose $f(z)$ is entire function such that $|f(z)|=1$ on $C(0,1)$. Show that $f(z)=k z^n$ for some constant $k$,$|k|=1$, and for some nonnegative integer $n$.

I was thinking to apply Blaschke product by breaking the number of zeros of $f(z)$ inside the circle and outside the circle. I also have to consider the cases $f(z)$ might have infinitely many zero in the whole complex plane or it might never vanish. I am getting confuse how to start.

Any help would be appreciated!

A detailed answer can be found here: https://math.stackexchange.com/questions/3408/characterizing-nonconstant-entire-functions-with-modulus-1-on-the-unit-circle — Lance, Apr 24 '17 at 01:37

score 0 · Answer 1 · answered Nov 14 '14 at 03:26

0

Hint: $f(z) = \dfrac{1}{\overline{f(1/\overline{z})}}$ on the circle, and therefore...

answered Nov 14 '14 at 03:26

Robert Israel

448,999

I guess it is same function again ,right? – Nov 14 '14 at 03:43
I did not see what you are trying to say. – Nov 14 '14 at 03:49
Therefore the only possible zero is at $z=0$. Also, after dividing by $z^n$ where $n$ is the order of that zero (if any), you get a bounded entire function. – Robert Israel Nov 14 '14 at 04:57
Thank you very much for you help. – Nov 14 '14 at 17:41

Entire function constant on $|z|=1$

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