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Homework Help: Whats wrong here?

  1. May 1, 2006 #1
    [tex]sin(z)=2[/tex]
    [tex]e^{iz}-e^{-iz} = 4i[/tex]
    [tex]e^{2iz}-4ie^{iz} = 1[/tex]
    [tex]iz \ln (e^{iz}-4i) = 0[/tex]
    [tex] z=0[/tex]

    when solving it by
    [tex]w = e^{iz}[/tex]
    [tex]w^2-4wi-1 = 0[/tex]
    i get one more solution, why is the first way not as good as the second way?
     
  2. jcsd
  3. May 1, 2006 #2

    HallsofIvy

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    Science Advisor

    ln(ab)= ln(a)+ ln(b), not ln(a)*ln(b).

    If [tex]e^{2iz}-4ie^{iz} = 1[/tex]
    then [tex]e^{iz}(e^{iz}- 4i)= 1[/tex]
    so [tex]iz+ ln(e^{iz}- 4i)= 0[/tex]
    NOT [tex]iz \ln (e^{iz}-4i) = 0[/tex]
     
  4. May 2, 2006 #3
    hehe, right, that was a dumb mistake :biggrin:
    thank you for pointing it out.
     
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