# Complex number equation

• Jul 8th 2011, 03:08 AM
depecheSoul
Complex number equation
Hello can you please help me to solve this equation.

$\displaystyle z^3+8=0$

(Wink) (Wink) (Wink)
• Jul 8th 2011, 03:16 AM
e^(i*pi)
Re: Complex number equation
Use the sum of two cubes: $\displaystyle a^3+b^3 = (a+b)(a^2-ab+b^2)$
• Jul 8th 2011, 03:38 AM
Prove It
Re: Complex number equation
Quote:

Originally Posted by depecheSoul
Hello can you please help me to solve this equation.

$\displaystyle z^3+8=0$

(Wink) (Wink) (Wink)

An alternative...

\displaystyle \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 $\displaystyle \displaystyle \frac{2\pi}{3}$, which means the three solutions are $\displaystyle \displaystyle 2e^{-\frac{i\pi}{3}}, 2e^{\frac{i\pi}{3}}$ and $\displaystyle \displaystyle 2e^{i\pi}$.

Convert back to Cartesians.