Nth-root branches

1. Jun 29, 2014

mahler1

The problem statement, all variables and given/known data.

Let $n \in \mathbb N$. If $\Omega \subset \mathbb C^*$ is open, we define a branch of the nth-root of $z$ on $\Omega$ to be any continuous function $f:\Omega \to \mathbb C$ such that ${f(z)}^n=z$ for all $z \in \Omega$. We will denote $\sqrt[n]{z}$ to $f(z)$.

(i) Prove that if $\Omega=\mathbb C \setminus \mathbb R_{\leq 0}$, there are exactly two branches of $\sqrt{z}$ on $\Omega$. Define them. Show that every branch of $\sqrt{z}$
is holomorphic.

(ii) If $\Omega$ is connected and $f$ is a branch of $\sqrt{z}$ on $\Omega$, then $f$ and $-f$ are all the branches.

The attempt at a solution

For $(i)$

By definition, $f(z)^2=e^{2\log(f(z))}$. This means $e^{2log(f(z))}=z$, So $2log(f(z))$ is a branch of the logarithm on $\Omega$. I am stuck at that point.

And I also don't know how to deduce that if $\Omega=\mathbb C \setminus \mathbb R_{\leq 0}$, then there are two functions $f_1$ and $f_2$ that satisfy the conditions required.

For $(ii)$ I have no idea where to start the problem, I would appreciate help and suggestions.

Last edited: Jun 29, 2014
2. Jun 29, 2014

haruspex

I guess you mean
We define a branch of the nth-root of $z$ on $\Omega$ to be any continuous function $f:\Omega \to \mathbb C$ such that ${f(z)}^n=z$ for all $z \in \Omega$.
Do you mean $f(z)^2=e^{2\ln(f(z))}$?

3. Jun 29, 2014

mahler1

Thanks for the corrections and sorry for my english. I've edited my original post.

4. Jul 6, 2014

haruspex

Sorry for the delay...
Suppose there are two different values w, w' of f(z). Both satisfy w2=z. What can you say about the relationship between them?