Find the solutions to \(z^3 = -8.\) Enter the solutions, separated by commas.

mathtoo Dec 28, 2018

#1**+1 **

if the question is find z because im not sure,

z^3 = -8

(-2)^3 = -8

therefore z = -2

YEEEEEET Dec 28, 2018

#2**+3 **

That is correct Zeet.

The only real solution is z=-2

but I think mathtoo wants the complex solutions too. It is a cubic so there are 3 solutions altogether.

There will be equally spaced at \(\frac{2\pi}{3}\) radians apart on the complex plane. So the others are at \(\frac{\pi}{3}\;\;and \;\;\frac{-\pi}{3}\)

The modulus or absolute value is 2

so

\(z=-2,\;\;2e^{(\pi/3)i},\;\;2e^{(-\pi/3)i}\)

\(2e^{(\frac{\pi}{3})i}\\=2[cos(\pi/3)+isin(\pi/3)]\\ =2[\frac{1}{2}+i\frac{\sqrt3}{2}]\\ =1+i\sqrt3\\~\\ 2e^{(\frac{\pi}{3})i}\\ =1-i\sqrt3\)

so

\(z=-2,\;\;1+i\sqrt3,\;\;1-i\sqrt3\)

Here is the diagramatic representation. Copied from Wolfram|Alpha.

Melody
Dec 28, 2018