# Existence of an entire function $f.$

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.

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