The problem statement, all variables and given/known data.
Let ##n \in \mathbb N##. If ##\Omega \subset \mathbb C^*## is open, 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##. 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^{2f(z)}##. This means ##e^{2f(z)}=z##, So ##2f(z)## is a branch of the logarithm on ##\Omega##. From here, ##f(z)=\dfrac{\log(z)}{2}## for a logarithm branch ##f## on ##\Omega##.
I don't know how to deduce from here 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. All I know is that ##f(z)=\dfrac{\log(z)}{2}##, would this mean there are only two branches of logarithm on this particular region ##\Omega##?
As for the other part of (i), I've found that ##f(z)=\dfrac{\log(z)}{2}##, as ##\log(z)## is holomorphic, from here I could deduce that ##f## is holomorphic.
For ##(ii)## I have no idea where to start the problem, I would appreciate help and suggestions.
Let ##n \in \mathbb N##. If ##\Omega \subset \mathbb C^*## is open, 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##. 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^{2f(z)}##. This means ##e^{2f(z)}=z##, So ##2f(z)## is a branch of the logarithm on ##\Omega##. From here, ##f(z)=\dfrac{\log(z)}{2}## for a logarithm branch ##f## on ##\Omega##.
I don't know how to deduce from here 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. All I know is that ##f(z)=\dfrac{\log(z)}{2}##, would this mean there are only two branches of logarithm on this particular region ##\Omega##?
As for the other part of (i), I've found that ##f(z)=\dfrac{\log(z)}{2}##, as ##\log(z)## is holomorphic, from here I could deduce that ##f## is holomorphic.
For ##(ii)## I have no idea where to start the problem, I would appreciate help and suggestions.
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