We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

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