# Solving equation involving trigonometry and complex numbers

## 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

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jbunniii
$$e^{-iz} = -\frac{1}{i}(e^z-e^{-z})$$
You are missing $i$'s in the exponents on the right-hand side.