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# Find all complex roots of z^3= -8i?

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#### Explanation

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#### Explanation:

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3
Dec 14, 2017

The roots are: $2 i$, $- \sqrt{3} - i$ and $\sqrt{3} - i$

#### Explanation:

In trigonometric form:

$- 8 i = 8 \left(\cos \left(- \frac{\pi}{2}\right) + i \sin \left(- \frac{\pi}{2}\right)\right)$

By de Moivre we have:

${\left(\cos \theta + i \sin \theta\right)}^{n} = \cos n \theta + i \sin n \theta$

Hence one of the cube roots of $- 8 i$ is:

$2 \left(\cos \left(- \frac{\pi}{6}\right) + i \sin \left(- \frac{\pi}{6}\right)\right) = \sqrt{3} - i$

The others can be found by adding multiples of $\frac{2 \pi}{3}$...

$2 \left(\cos \left(- \frac{\pi}{6} + \frac{2 \pi}{3}\right) + i \sin \left(- \frac{\pi}{6} + \frac{2 \pi}{3}\right)\right) = 2 \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right) = 2 i$

$2 \left(\cos \left(- \frac{\pi}{6} + \frac{4 \pi}{3}\right) + i \sin \left(- \frac{\pi}{6} + \frac{4 \pi}{3}\right)\right) = 2 \left(\cos \left(\frac{7 \pi}{6}\right) + i \sin \left(\frac{7 \pi}{6}\right)\right) = - \sqrt{3} - i$

Here are the three roots in the complex plane...

graph{(x^2+(y-2)^2-0.005)((x-sqrt(3))^2+(y+1)^2-0.005)((x+sqrt(3))^2+(y+1)^2-0.005) = 0 [-5, 5, -2.5, 2.5]}

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