# How to solve this equation e^(6z)+(1-i)e^(3z)-i=0, for z where z is a complex number?

Mar 13, 2017

See below.

#### Explanation:

Calling $w = {e}^{3 z}$ we have

${w}^{2} + \left(1 - i\right) w - i = 0$ solving for $w$ we have

$w = - 1 = {e}^{i \left(\pi + 2 k \pi\right)}$ and $w = i = {e}^{i \left(\frac{\pi}{2} + 2 k \pi\right)}$ or

1) ${e}^{3 z} = {e}^{i \left(\pi + 2 k \pi\right)}$ or

$3 z = i \left(\pi + 2 k \pi\right) = 3 \left(x + i y\right) \to \left\{\begin{matrix}x = 0 \\ y = \frac{\pi}{3} + \frac{2}{3} k \pi\end{matrix}\right.$

2) ${e}^{3 z} = {e}^{i \left(\frac{\pi}{2} + 2 k \pi\right)}$ or

$3 z = i \left(\frac{\pi}{2} + 2 k \pi\right) = 3 \left(x + i y\right) \to \left\{\begin{matrix}x = 0 \\ y = \frac{\pi}{6} + \frac{2}{3} k \pi\end{matrix}\right.$