Solving equation involving trigonometry and complex numbers

[thepopasmurf]

Homework Statement

Show that the solutions of the equation

2sin(z) + cos(z) = isin(z)

are given by

z = (n\pi-\frac{\pi}{8}) - \frac{1}{4}iln2

Homework Equations

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

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

The Attempt at a Solution

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

e^{-iz} = i(e^z-e^{-z})

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

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

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

2z = -iz + 2ni\pi

(2+i)z = 2ni\pi

z = \frac{2ni\pi}{2+i} = 2ni\pi\frac{2-i}{5} = \frac{4ni\pi}{5} + \frac{2n\pi}{5}

Don't know where to go from here

 

[jbunniii]

e^{-iz} = -\frac{1}{i}(e^z-e^{-z})

You are missing i's in the exponents on the right-hand side.

 

[thepopasmurf]

Thanks for pointing that out. I've retried it but the answer is still incorrect. I got z=2n(pi)/3