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Let $f$ be analytic in open unit disk, we need to show there exist $\{z_n\}$ with $|z_n|<1$ and $|z_n|\rightarrow 1$ then $f(z_n)$ is bounded.

could any one give me Hints for this one?

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    "then" doesn't make sense there. Do you mean "and"?2012-09-07

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Hint: If no such sequence exists, $f$ has only finitely many zeros in the open unit disk. Use the Maximum Modulus principle on $p(z)/f(z)$ for a suitable polynomial $p$.

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    If there are infinitely many zeros, they have a limit point. If that limit point is in the open disk, $f = 0$ everywhere. Otherwise it's on the circle $|z|=1$, and you take $z_n$ to be a sequence of those zeros.2012-09-07