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

 Dec 28, 2018
 #1
avatar+844 
+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+103697 
+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

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