1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Existence of (complex) limit z->0 (z^a)

  1. Sep 22, 2009 #1
    1. The problem statement, all variables and given/known data
    Justify for which complex values of a does the principal value of [tex]z^a[/tex] have a limit as z tends to 0?


    2. Relevant equations

    [tex]z^a = e^{a log(z)} [/tex]

    [tex]log(z) = log|z| + (i) (arg(z)) [/tex]

    3. The attempt at a solution

    [tex]Lim_{z \rightarrow 0} z^a = Lim_{z \rightarrow 0} e^{(a) (log(z))} [/tex]

    [tex]=Lim_{|z| \rightarrow 0} e^{(a) (log|z|) + (i) (a) (arg(z))}[/tex]

    Let [tex]a = u + i v [/tex].

    [tex]=Lim_{|z| \rightarrow 0} e^{(u+iv) (log|z| + (i) (u+iv) (arg(z)))}[/tex]

    [tex]=Lim_{|z| \rightarrow 0} e^{(u) (log|z|) + (i) (v) (log|z|) + (i) (u) (arg(z)) - (v) (arg(z))}[/tex]

    [tex]=Lim_{|z| \rightarrow 0} e^{(u) (log|z|)} e^{(i) (v) (log|z|)} e^{(i) (u) (arg(z))} e^{-v (arg(z))}[/tex]

    [tex]=Lim_{|z| \rightarrow 0} |z|^u e^{(i) (v) (log|z|)} e^{((i) (u) - (v)) (arg(z))}[/tex]

    I just noticed a big mistake here, so I'm erasing this part. Any ideas?
     
    Last edited: Sep 22, 2009
  2. jcsd
  3. Sep 22, 2009 #2

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi Knissp! :wink:

    Isn't it easier just to go polar, and put z = re ? :smile:
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook