# Let $f$ be a non-constant entire function such that $\left \lvert f(z) \right\lvert=1$ for every $z$ with $\left \lvert z \right\lvert=1$.

I was thinking about the problem that says:

Let $f$ be a non-constant entire function such that $\left | f(z) \right |=1$ for every $z$ with $\left \lvert z \right \lvert =1$. Then which of the following option(s) is/are correct?

• (a) $f$ has a zero in the open unit disc,
• (b) $f$ always has a zero outside the closed unit disc,
• (c) $f$ need not have any zero,
• (d) any such $f$ has exactly one zero in the open unit disc.

If we take $f(z)=z^n$, then the given condition is satisfied and I think option (a) is correct. But I can not predict anything about the other options.

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Is it a CSIR problem? – Sugata Adhya Dec 14 '12 at 16:22

For (a), if it was not true, apply maximum modulus principle to $1/f$ and the unit disk.
(b) $f(z)=z$ gave a counter-example.
(d) Think about a factorization of $f$ (with elements of the form $f(z)=\prod\frac{z-\alpha_j}{1-\bar \alpha_jz}g(z)$, where $g$ doesn't vanish in the open unit disk. As $f$ needs to be entire, what can we say about $\alpha_j$?