Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Branching points

  1. Jul 21, 2015 #1
    ##f(z)=z^{\frac{1}{2}}## has brancing point at ##z=0##.
    ##z=re^{i\varphi}=re^{i(\varphi+2n\pi)}##
    From that
    [tex]z^{\frac{1}{2}}=r^{\frac{1}{2}}e^{i (\frac{\varphi}{2}+n\pi)}[/tex]
    For ##n=0##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
    For ##n=1##, ##z^{\frac{1}{2}}=e^{i (\frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
    For ##n=2##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
    For ##n=3##, ##z^{\frac{1}{2}}=e^{i( \frac{\varphi}{2}+\pi)}=-e^{i \frac{\varphi}{2}}##
    For ##n=4##, ##z^{\frac{1}{2}}=e^{i \frac{\varphi}{2}}##
    ##...##
    I see that this is multivalued function, but I am not sure why only problem or why only ##z=0## is branching point? Could you please explain me this?
     
  2. jcsd
  3. Jul 21, 2015 #2
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Branching points
  1. Help on Branch Points (Replies: 2)

  2. Branch points (Replies: 5)

Loading...