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$f:D\rightarrow\mathbb{C}$ be analytic satisfying $f(1/n)=\frac{2n}{3n+1}$, $D$ is open unit disk,Then

1.$f(0)=2/3$

2.$f$ has a simple pole at $z=-3$

3.$f(3)=-3$

4.No such $f$ exist.

well, $1$ is correct due to continuity of $f$, define $g(z)=f(z)-\frac{2}{3+z}$ then $g(1/n)=0$ and $0$ is a limit point of set of zeroes of $g$ and hence $f(z)=\frac{2}{3+z}$ hence $2$ is also correct.am I right?

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Are $3$ and $-3$ inside your open unit disk? ;) – N. S. Jun 13 '12 at 15:54
    
:P :P :P :P :P :P extremely sorry.thnx for correcting :) – Un Chien Andalou Jun 13 '12 at 15:58
    
Ok so only $1$ is correct. – Un Chien Andalou Jun 13 '12 at 16:00
1  
Yup. What is true about (2) is that $f$ has an analityc continuation to $\mathbb{C}\backslash \{1 \}$, and this analytic continuation satisfies (2). But this is a new function... – N. S. Jun 13 '12 at 16:07
up vote 2 down vote accepted

What you did is correct. We have to consider the analytic extension to $\Bbb C\setminus \{-3\}$ after having proved it exists. So 1. and 2. are correct.

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