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Nth-root branches

  1. Jun 29, 2014 #1
    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. jcsd
  3. Jun 29, 2014 #2

    haruspex

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    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))}##?
     
  4. Jun 29, 2014 #3
    Thanks for the corrections and sorry for my english. I've edited my original post.
     
  5. Jul 6, 2014 #4

    haruspex

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    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?
     
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