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Homework Help: Complex cosine equation (complex analysis)

  1. Jun 14, 2007 #1

    malawi_glenn

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    1. The problem statement, all variables and given/known data

    Solve [tex]cosz = 2i , z\in \mathbb{C} [/tex]

    3. The attempt at a solution


    [tex]e^{iz}+e^{-iz} = 4i[/tex]

    [tex]t=e^{-z} [/tex]

    [tex]t+t^{-1}=4i \Rightarrow t^{2}-4it+1=0 [/tex]

    [tex]t = (2 \pm \sqrt{5})i[/tex]

    [tex]log(e^{-z}) = logt [/tex]

    [tex]z = x + yi;x,y \in \mathbb{R}[/tex]

    [tex]log(e^{-z}) = log(e^{-y+ix}) = -y +xi + 2\pi ni; n\in\mathbb{Z}[/tex]

    [tex]logt = log((2 \pm \sqrt{5})i) = ln|(2 \pm \sqrt{5})| \pm \frac{\pi}{2} + 2\pi hi, h \in \mathbb{Z}[/tex]

    [tex]\Rightarrow y= -ln|(2 \pm \sqrt{5})| [/tex]

    [tex]\Rightarrow x= \pm \frac{\pi}{2} + 2\pi p, p \in \mathbb{Z}[/tex]

    [tex]z= \pm \frac{\pi}{2} + 2\pi p -iln|(2 \pm \sqrt{5})|[/tex]

    answer in book:
    [tex]z = \pm\lbrace \frac{\pi}{2} -iln(2 + \sqrt{5})\rbrace+2\pi n, n\in \mathbb{Z}
    [/tex]

    Were did I do wrong? :S
     
  2. jcsd
  3. Jun 14, 2007 #2

    Dick

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    You didn't go wrong. The book just found a sneakier way to write the answer. Since (sqrt(5)+2)*(sqrt(5)-2)=1, they are actually reciprocals. So their logs are negatives of each other. Thus you can sneak the log under the same +/- as the pi/2.
     
  4. Jun 14, 2007 #3

    malawi_glenn

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    aa, it is almost always the most simplest of solutions :) I shall check it out tomorrow. It is very late now in Sweden hehe. This is a summer course I am taking now, Complex analysis with applications.

    thanx!
     
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