I have a complex analysis question that is giving me a lot of trouble. I think I have to argue by contradiction, saying S is not finite - but I don't know where to go from here. My friend said something about comparing it with the function $g(z)=\frac{z}{z+1}$. Here is the question;
Let $U\subseteq \Bbb{C}$ be an open set such that $R_2(0)\subset U$ where;
$R_2(0)=[-2,2]+i[-2,2]$$=${$z∈\Bbb{C}$:$\vert Re(z)\vert \le 2, \vert Im(z)\vert \le 2 $}
and suppose $f: U\to \Bbb{C}$ is a holomorphic function on $U$. Definte the (possibly empty) set $S\subseteq \Bbb{N}$ by;
$S =$ {$n∈\Bbb{N} : f(\frac{1}{n})=\frac{1}{n+1}$}
Show that the number of elements in $S$ is finite.
Thank you so much for the help!
Let $g(z)=(1+z)f(z)-z$. If $n \in S$ then $g(\frac 1 n )=0$. If $S$ has infinitely many points then 0 would be a limit point of zeros of $g$ which makes $g \equiv 0$. But then $z=(1+z)f(z)$ for all $z \in R_2 (0)$. You get a contradiction by putting $z=-1$.