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Complex Logarithm Calculation

  1. Nov 4, 2014 #1
    Hello everyone,

    I am asked to calculate ##\log (e^{1+2i})##, and I would appreciate it if someone could verify my calculation..

    My textbook defines ##\log z## as ##\log z = \ln |z| + i \arg z##.

    ##\log (e^{1+2i}) = \ln |e^{1+2i}| + i \arg(e^{1+2i}) \iff##

    ##\log (e^{1 + 2i}) = \ln|e e^{2i}| + i \arg (e e^{2i}) \iff##

    ##\\log (e^{1+2i}) = \ln e + i \arg (e e^{2i}) \iff##

    ##\log (e^{1+2i}) = 1 + i \arg (e e^{2i})##

    Here is the step that I am not entirely certain about. I know that ##e^{2i}## is the exponential representation of the complex point whose angle is ##2##. But I am wondering, would the coefficient ##e## just represent the distance of the complex point from the origin? More generally, the complex exponential function ##e^z## is defined as ##e^z = e^x e^{iy}##, and so ##e^x## would represent the distance of the complex point ##e^{iy}##?

    If this is so, then the answer would seem to be

    ##\log (e^{1+2i}) = 1 + i(2 + 2 n \pi)##

    Is this correct?
     
    Last edited: Nov 4, 2014
  2. jcsd
  3. Nov 4, 2014 #2

    haruspex

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    Yes, but it is rather a long way about. You can straight away write down that one value of ln(ez) is z. The only question is what other values are there. If w is another value then w = ln(ez). Exponentiate both sides and see what develops.
     
  4. Nov 5, 2014 #3
    I am not certain I see where the quantity ##\ln e^z## arises. If recall correctly, my professor said that we do not use the natural logarithm when dealing with complex numbers.
     
  5. Nov 5, 2014 #4

    haruspex

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    OK, I see - you have to start with the definition you quoted, and your method is fine.
     
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