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?