$f:D\rightarrow\mathbb{C}$ 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?