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Complex exponential equation

  1. Sep 24, 2014 #1
    How can we show that the following equation has infinitely many solutions
    [tex]e^z-z^2=0[/tex].
    Thanks
     
  2. jcsd
  3. Sep 24, 2014 #2

    WWGD

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    Work within a fixed branch of logz and write : ## z^2=e^{2logz} ##

    Once you find a solution, you have infinitely-many, by periodicity of ##e^z##.
     
  4. Oct 7, 2014 #3
    Continuing WWGD's post: you get ##e^z=e^{2\log z}## and thus the equation ##e^{z-2log z}=1##. In other words you are looking at the solutions of each of the equations ##z-2\log z=2\pi n## for ##n\in\mathbb{Z}##.
     
  5. Oct 7, 2014 #4

    WWGD

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    It seems that something like Lambert's W function may be helpful in finding solutions to platetheduke's equation.
     
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