1. ## Complex number equation

$z^3+8=0$

2. ## Re: Complex number equation

Use the sum of two cubes: $a^3+b^3 = (a+b)(a^2-ab+b^2)$

3. ## Re: Complex number equation

Originally Posted by depecheSoul
$z^3+8=0$
\displaystyle \begin{align*} z^3 + 8 &= 0 \\ z^3 &= -8 \\ z^3 &= 8e^{i\pi} \\ z &= \left(8e^{i\pi}\right)^{\frac{1}{3}} \\ z &= 2e^{\frac{i\pi}{3}} \end{align*}
Since this is a cubic, there are three solutions, all evenly spaced around a circle. So their angles are separated by $\displaystyle \frac{2\pi}{3}$, which means the three solutions are $\displaystyle 2e^{-\frac{i\pi}{3}}, 2e^{\frac{i\pi}{3}}$ and $\displaystyle 2e^{i\pi}$.