+0  
 
0
38
2
avatar+809 

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

 Dec 28, 2018
 #1
avatar+622 
+1

if the question is find z because im not sure,

z^3 = -8 

(-2)^3 = -8

therefore z = -2

 Dec 28, 2018
edited by YEEEEEET  Dec 28, 2018
edited by YEEEEEET  Dec 28, 2018
 #2
avatar+95171 
+3

That is correct Zeet.    laugh

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
edited by Melody  Dec 28, 2018

26 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.