Holomorphic function such that $f(\frac{1}{n})=\frac{1}{n+1}$.

Let $U$ be an open connected neighborhood of $0$ in $\Bbb{C}$ and $f$ a holomorphic function on $U$.

The exercise is to prove that if $f(\frac{1}{n})=\frac{1}{n+1}$ for all $n\ge 1$ then $f(z)=\frac{z}{z+1}$ on $U$.

So I did it using the function $g:\Bbb{C} \setminus\{-1\}\to \Bbb{C}$ such that $g(z)=\frac{z}{1+z}$ so $g$ is holomorphic on $\Bbb{C} \setminus\{-1\}$, now as $U$ is open we can find a disc $D(0,r)\subset U$ for $r$ so that $r<1$.

Therefore, using analytic continuation we have $f=g$ on $D(0,r)$.

My main problem is now to prove that $U\subset\Bbb{C} \setminus\{-1\}$. Perhaps, I can say that if $-1\in U$ we can find a path from $0$ to $-1$ but not sure how can I continue.

Solution 1:

Hint If you assume by contradiction that $-1 \in U$, then prove first that $f(z)=g(z)$ on $U \backslash \{-1\}$.

Then use the fact that $f$ is continuous at $-1$, and hence $$f(-1)=\lim_{z \to -1} \frac{z}{z+1}=\infty$$