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Solving equation involving trigonometry and complex numbers

  1. Mar 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the solutions of the equation

    [tex]2sin(z) + cos(z) = isin(z)[/tex]

    are given by

    [tex] z = (n\pi-\frac{\pi}{8}) - \frac{1}{4}iln2[/tex]

    2. Relevant equations

    [tex] e^{iz} = cos(z) + isin(z) [/tex]
    [tex] sinz = \frac{1}{2i}(e^{z}-e^{-z}) [/tex]

    [tex] z_{1}^{z_{2}} = e^{z_{2}lnz_{1}} [/tex]
    [tex] lnz = lnr + i(\theta + 2n\pi) [/tex]

    3. The attempt at a solution

    [tex] 2sinz + cosz = isinz [/tex]
    [tex] cosz - isinz = -2sinz [/tex]
    [tex] e^{-iz} = -\frac{1}{i}(e^z-e^{-z}) [/tex]

    [tex] e^{-iz} = i(e^z-e^{-z}) [/tex]

    [tex] e^{-iz} = e^{i\pi/2}(e^z-e^{-z}) [/tex]

    [tex] e^{-iz} = e^{i\pi/2+z}-e^{i\pi/2-z} [/tex]

    [tex] lne^{-iz + 2n\pi} = ln( e^{i\pi/2+z-i\pi/2+z}) [/tex]

    [tex] 2z = -iz + 2ni\pi [/tex]

    [tex] (2+i)z = 2ni\pi [/tex]

    [tex] z = \frac{2ni\pi}{2+i} = 2ni\pi\frac{2-i}{5} = \frac{4ni\pi}{5} + \frac{2n\pi}{5} [/tex]

    Don't know where to go from here
  2. jcsd
  3. Mar 22, 2010 #2


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    Science Advisor
    Homework Helper
    Gold Member

    You are missing [itex]i[/itex]'s in the exponents on the right-hand side.
  4. Mar 23, 2010 #3
    Thanks for pointing that out. I've retried it but the answer is still incorrect. I got z=2n(pi)/3
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