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Nice!

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

Dec 28, 2018

#1
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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
+95171
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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
edited by Melody  Dec 28, 2018