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!


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

  • $\begingroup$ I guess it is same function again ,right? $\endgroup$ – user178061 Nov 14 '14 at 3:43
  • $\begingroup$ I did not see what you are trying to say. $\endgroup$ – user178061 Nov 14 '14 at 3:49
  • $\begingroup$ 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. $\endgroup$ – Robert Israel Nov 14 '14 at 4:57
  • $\begingroup$ Thank you very much for you help. $\endgroup$ – user178061 Nov 14 '14 at 17:41

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