# Show that a holomorph function $f:\mathbb{C}^{\times}\rightarrow\mathbb{C}$ doesn't exist which satisfies $f(z)^n=z$ for $n\in\mathbb{N}, n\gt1$.

I have to proof this with the help of the Riemann's theorem, which says:

Let $D\subset\mathbb{C}$ be an open subset of the complex plane, $a\in D$ a point of $D$ and $f$ a holomorphic function defined on the set $D\setminus\{a\}$. The following are equivalent:

1. $f$ is holomorphically extendable over $a$
2. $f$ is continuously extendable over $a$
3. There exists a neighborhood of $a$ on which $f$ is bounded
4. $\lim \limits_{z \to a}(z-a)f(z)=0$

I thought about assuming that $f$ satisfies $f(z)^n=z$ and then showing with this function $f$ that it doesn't satisfies Riemann's theorem, but i don't know how. I also thought about how $f$ has to look like in order to fulfill $f(z)^n=z$. The function $\tilde f:=\sqrt[n]{z}$ for example. Are there any other functions which would fulfill this equation? I'm assuming that we have to proof this for general function $f$, so do we even need a specific function?

The Riemann's theorem says that $f$ is a holomorphic function defined on the set $D\setminus\{a\}$. Our $f$ is defined on $\mathbb{C}^{\times}$ which is $\mathbb{C}\setminus\{0\}$ if I'm right. So our removable singularity $a$ should be $0$.

That's all I got, don't know how to proof this. Any help appreciated.

Assume that $f:\mathbb{C}^{\times}\rightarrow\mathbb{C}$ is a holomorphic function satisfying $f(z)^n = z$ for some integer $n \ge 2$.
Then $|f(z)| = |z|^{1/n}$, which implies $\lim_{z \to 0} f(z) = 0$, so that condition (2) of Riemann's theorem is fulfilled (at $a=0$).
It follows that $f$ can be extended to a holomorphic function on $\mathbb{C}$, and the (extended) function satisfies $f(z)^n = z$ for all $z \in \Bbb C$.
Now differentiate this identity $$n f(z)^{n-1} f'(z) = 1$$ and derive a contradiction at $z = 0$.
(Alternatively, assume that $f$ has a zero of multiplicity $k$ at $z=0$, and obtain a contradiction from $kn = 1$.)
For a slightly more general statement, see for example Show that there is no nth roots in $U$..