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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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    What are your tools? Is it from a book or a homework?2012-09-07
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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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    I do not understand the connection between non-existance of such sequence and finitely many $0$'s.2012-09-07
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    @miosaki: Try thinking about the contrapositive of the first sentence of this answer: If there are infinitely many zeros, then can you find such a sequence?2012-09-07
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    @JonasMeyer If there are infinitely many zeros of $f$ in the open unit disk then by the theorem "every bounded infinite set of $\mathbb{R}^n$ has a limit point in $\mathbb{R}^n$" and hence by Identity Theorem of analytic function we get $f\equiv 0$ thats what I can say. But here limit point may not be in the open unit disk right? Then I can not say $f\equiv 0$2012-09-07
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    contrapositive statement of the 1st line of the answer: $f$ has infinitely many zeros $\rightarrow$ there exist such sequence. but still I do not get.2012-09-07
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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