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The complex exponential function

  1. Jul 26, 2010 #1

    I need to solve the equation
    [tex]e^z = -3[/tex]

    The problems arises when i set z to a+bi
    [tex]e^a(cos(b) + isin(b),~b = 0[/tex]

    Then im left with [tex]e^a = -3[/tex]

    However you're not allowed to take the log of a negative number.

    Also i know that [tex]cos(\pi) + isin(\pi) = -1[/tex]

    Obviously [tex]3e^{(\pi*i)}[/tex] is a solution, but it's not a solution for z.

    Any ideas?
  2. jcsd
  3. Jul 26, 2010 #2


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    Science Advisor

    No, if you set it in that form, [itex]e^z= e^{a+ bi}= e^ae^{bi}= e^a(cos(\theta)+ i sin(\theta))= -3[/itex] but now [itex]e^a[/itex] must be positive. We must have [itex]e^acos(\theta)= -3[/itex] and [itex]e^a sin(\theta)= 0[/itex]. Since [itex]e^a[/itex] is never 0, we must have [itex]sin(\theta)= 0[/itex] so that [itex]\theta= 0[/itex] or [itex]\pi[/itex]. If [itex]\theta= 0[/itex], [itex]cos(\theta)= 1[/itex] so [itex]e^a= -3[/itex] which, as I said, is impossible. Therefore, [itex]\theta= \pi[/itex]. Now continue.
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