Complex number equation

**depecheSoul** · July 8th 2011, 03:08 AM

Hello can you please help me to solve this equation.

$z^3+8=0$

**e^(i*pi)** · July 8th 2011, 03:16 AM

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

**Prove It** · July 8th 2011, 03:38 AM

Originally Posted by depecheSoul

Hello can you please help me to solve this equation.

$z^3+8=0$

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.

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