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How is this possible?

  1. Dec 10, 2008 #1
    How can this be true(z is complex)

    z^a = exp(a * ln(z))

    but in general ln(z^a) is not equal to a * ln(z)
     
  2. jcsd
  3. Dec 10, 2008 #2
    Are you mainly interested in the case where [itex]a\in\mathbb{R}[/itex]? (Answer would be simpler then)
     
  4. Dec 10, 2008 #3
    Well, anyway, I'll assume that [itex]a\in\mathbb{R}[/itex]. Arbitrary [itex]a\in\mathbb{C}[/itex] would result in more calculating, but I don't believe there would be anything highly different.

    This equation is usually true, because it is the definition of the left side. First you must decide what argument function [itex]\textrm{arg}:\mathbb{C}\to\mathbb{R}[/itex] you are using. This choice will fix the branch of the logarithm,

    [tex]
    \log(z) = \log |z| + i\textrm{arg}(z),
    [/tex]

    and then we denote

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

    The equation is not necessarily true, if you start using other branches of logarithm later, of course.

    This is not true always because of the branch cuts.

    [tex]
    \log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})
    [/tex]

    [tex]
    a\log(z) = a\log |z| + ia\textrm{arg}(z)
    [/tex]

    The equation

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

    is always true, but

    [tex]
    a\textrm{arg}(z) = \textrm{arg}(e^{ia\textrm{arg}(z)}) + 2\pi n
    [/tex]

    may need some [itex]n\in\mathbb{Z}[/itex] depending on how the argument function happens to behave with these [itex]a,z[/itex].
     
    Last edited: Dec 10, 2008
  5. Dec 10, 2008 #4

    So... basically it's not true due to the extra 2*pi*n factor at the end?

    No that can´t be true. Then the equation would be true for the principal value?

    No wait. It's the extra iarg factor in this

    [tex]
    \log (z^a) = \log(e^{a\log(z)}) = \log( \underbrace{e^{a\log |z|}}_{\in\mathbb{R}} \underbrace{e^{ia\textrm{arg}(z)}}_{\in S^1\subset\mathbb{C}} ) = \log(e^{a\log |z|}) \;+\; i\textrm{arg}(e^{ia\textrm{arg}(z)})
    [/tex]

    If there is no iarg, then there is only log(e^(a*log|z|)) which is |z|^a and this is true for the real numbers since |z| is really replaced by x
     
    Last edited: Dec 10, 2008
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