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If $\{z_n\}$ is a sequence of distinct complex numbers in unit circle such that $z_n \rightarrow 0$ as $n \rightarrow \infty$, there exist an entire function $f$ such that $f(z_n)=z_n$ for all $n \in \mathbb{N}$ and $f(5)=0$. (T/F)

My work: Consider $g(z)=f(z)-z$, clearly $g(z_n)=0$. Thus by identity and uniqueness theorem $g(z)=0 $ for all $z$ in the unit circle, i.e $f(z)=z$, so $f(5)=5$, and the above statement is false.

CHECK the logic.

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  • $\begingroup$ Capitals are considered as rude and offensive $\endgroup$ – user312648 Jan 19 '18 at 13:58
  • $\begingroup$ How can the $z_n$'s be on the unit circle while $z_n \to 0$? $\endgroup$ – user517611 Jan 19 '18 at 14:02
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    $\begingroup$ If $f(z)=z$ for all $z$ in unit circle, how do you conclude $f(5)=5$?? $5$ does not belong to unit circle. $\endgroup$ – user312648 Jan 19 '18 at 14:03
  • $\begingroup$ @cello, yes ur point is correct, so is there any way to solve this problem? $\endgroup$ – 1256 Jan 19 '18 at 14:12
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    $\begingroup$ @orole rather, it is a rhetorical question, intended to get the OP fix the statement. $\endgroup$ – user517611 Jan 19 '18 at 14:26

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