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Solving equation involving trigonometry and complex numbers

  • #1

Homework Statement



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]


Homework 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]


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
 

Answers and Replies

  • #2
jbunniii
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[tex] e^{-iz} = -\frac{1}{i}(e^z-e^{-z}) [/tex]
You are missing [itex]i[/itex]'s in the exponents on the right-hand side.
 
  • #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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