Complex number equation

Quote:

Originally Posted by depecheSoul

Hello can you please help me to solve this equation.

$z^3+8=0$

(Wink) (Wink) (Wink)

An alternative...

$\displaystyle \begin{align*} z^3 + 8 &= 0 \\ z^3 &= -8 \\ z^3 &= 8e^{i\pi} \\ z &= \left(8e^{i\pi}\right)^{\frac{1}{3}} \\ z &= 2e^{\frac{i\pi}{3}} \end{align*}$

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 $\displaystyle 2e^{i\pi}$ .

Convert back to Cartesians.