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Complex logarithm rules

  1. Nov 2, 2008 #1

    I know that for real numbers log(z)=-log(1/z)

    is this also true in general for complex numbers?
  2. jcsd
  3. Nov 2, 2008 #2


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    yup. no promises about the log of 0 or [itex]\infty[/itex].
  4. Nov 2, 2008 #3
    But one might need to add [itex]2\pi i[/itex] somewhere sometimes because of some branch choosing issues.
  5. Nov 2, 2008 #4
    If you choose to use a branch

    \log(z) = \log(|z|) + i\textrm{arg}(z),\quad 0\leq \textrm{arg}(z) < 2\pi

    then for example

    \log(-1+i) = \log(\sqrt{2}) + \frac{3\pi i}{4}


    \log(\frac{1}{-1+i}) = \log(-\frac{1}{2}(1+i)) = \log(\frac{1}{\sqrt{2}}) + \frac{5\pi i}{4}.

    So you've got

    \log(-1+i) + \log(\frac{1}{-1+i}) = 2\pi i,

    in contradiction with your equation. But if you choose the branch so that

    -\pi < \textrm{arg}(z) \leq \pi,

    then you've got

    \log(-1+i) + \log(\frac{1}{-1+i}) = 0,

    as your equation stated. Even with this choice of branch still, for example,

    \log(-1) + \log(\frac{1}{-1}) = 2\pi i,

    so actually...

    for positive real numbers! :wink:
  6. Nov 2, 2008 #5


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    Science Advisor

    The essentail point is than ln(1)= [itex]2n\pi i[/itex] with n being any integer.
  7. Nov 2, 2008 #6
    Thank you for the answers and examples. I understand it much better now.
  8. Nov 2, 2008 #7
    You probably remember the trick where one does something like this:

    1 = \sqrt{1} = \cdots = -1

    with imaginary units. The examples I gave are very similar in nature. Most of the time, a blind use of familiar calculation rules might seem to work, but you never know when something tricky surprises you, if you are not careful.
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