Proving Nth-Root Branches on Open Sets

[mahler1]

Homework Statement .

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.

 

[haruspex]

we call a branch of the nth-root of $z$ on $\Omega$ to every continuous function $f:\Omega \to \mathbb C$ such that ${f(z)}^n=z$ for all $z \in \Omega$.

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 $z \in \Omega$.

By definition, $f(z)^2=e^{2f(z)}$.

Do you mean $f(z)^2=e^{2\ln(f(z))}$?

 

[mahler1]

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 $f(z)^2=e^{2\ln(f(z))}$?

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

 

[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?