# How do you find all solutions to x^6-64i=0?

Aug 20, 2016

Use de Moivre's theorem to find all $6$ Complex roots.

#### Explanation:

Given:

${x}^{6} - 64 i = 0$

Add $64 i$ to both sides to get:

${x}^{6} = 64 i$

$\textcolor{w h i t e}{{x}^{6}} = {2}^{6} \left(0 + 1 i\right)$

$\textcolor{w h i t e}{{x}^{6}} = {2}^{6} \left(\cos \left(\frac{\pi}{2}\right) + i \sin \left(\frac{\pi}{2}\right)\right)$

From de Moivre's theorem, we have:

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

Hence principal root:

${x}_{1} = \sqrt[6]{64 i} = 2 \left(\cos \left(\frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{12}\right)\right)$

The other $5$ roots can be found by multiplying by the primitive Complex $6$th root of $1$, i.e. $\left(\cos \left(\frac{\pi}{3}\right) + i \sin \left(\frac{\pi}{3}\right)\right)$ to find:

${x}_{2} = 2 \left(\cos \left(\frac{5 \pi}{12}\right) + i \sin \left(\frac{5 \pi}{12}\right)\right)$

${x}_{3} = 2 \left(\cos \left(\frac{9 \pi}{12}\right) + i \sin \left(\frac{9 \pi}{12}\right)\right)$

${x}_{4} = 2 \left(\cos \left(\frac{13 \pi}{12}\right) + i \sin \left(\frac{13 \pi}{12}\right)\right)$

${x}_{5} = 2 \left(\cos \left(\frac{17 \pi}{12}\right) + i \sin \left(\frac{17 \pi}{12}\right)\right)$

${x}_{6} = 2 \left(\cos \left(\frac{21 \pi}{12}\right) + i \sin \left(\frac{21 \pi}{12}\right)\right)$

These trigonometric values can be expressed in terms of square roots:

$\cos \left(\frac{\pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right) \text{ } \sin \left(\frac{\pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

$\cos \left(\frac{5 \pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right) \text{ } \sin \left(\frac{5 \pi}{12}\right) = \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

$\cos \left(\frac{9 \pi}{12}\right) = - \frac{\sqrt{2}}{2} \text{ } \sin \left(\frac{9 \pi}{12}\right) = \frac{\sqrt{2}}{2}$

$\cos \left(\frac{13 \pi}{12}\right) = - \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right) \text{ } \sin \left(\frac{13 \pi}{12}\right) = - \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right)$

$\cos \left(\frac{17 \pi}{12}\right) = - \frac{1}{4} \left(\sqrt{6} - \sqrt{2}\right) \text{ } \sin \left(\frac{17 \pi}{12}\right) = - \frac{1}{4} \left(\sqrt{6} + \sqrt{2}\right)$

$\cos \left(\frac{21 \pi}{12}\right) = \frac{\sqrt{2}}{2} \text{ } \sin \left(\frac{21 \pi}{12}\right) = - \frac{\sqrt{2}}{2}$

So here are the $6$ roots in the Complex plane:

graph{((x-1/2(sqrt(6)+sqrt(2)))^2 + (y-1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x-1/2(sqrt(6)-sqrt(2)))^2 + (y-1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x+sqrt(2))^2+(y-sqrt(2))^2-0.01)((x+1/2(sqrt(6)+sqrt(2)))^2 + (y+1/2(sqrt(6)-sqrt(2)))^2 - 0.01)((x+1/2(sqrt(6)-sqrt(2)))^2 + (y+1/2(sqrt(6)+sqrt(2)))^2 - 0.01)((x-sqrt(2))^2+(y+sqrt(2))^2-0.01) = 0 [-5, 5, -2.5, 2.5]}