# How do you find all the real and complex roots of (z-1)^3 =8i?

Dec 29, 2016

The roots are all complex:

${z}_{1} = 1 + \sqrt{3} + i$

${z}_{2} = 1 - \sqrt{3} + i$

${z}_{3} = 1 - 2 i$

#### Explanation:

Using de Moivre's formula, note that:

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

So one cube root of $i$ is:

$\cos \left(\frac{\pi}{6}\right) + i \sin \left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} + \frac{1}{2} i$

The other cube roots can be formed by adding $\frac{2 \pi}{3} = \frac{4 \pi}{6}$ to the angle a couple of times:

$\cos \left(\frac{5 \pi}{6}\right) + i \sin \left(\frac{5 \pi}{6}\right) = - \frac{\sqrt{3}}{2} + \frac{1}{2} i$

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

The cube roots of $8 i$ will be double these values since $8 = {2}^{3}$.

So the cube roots of $8 i$ are:

$\sqrt{3} + i$

$- \sqrt{3} + i$

$- 2 i$

We find the required values of $z$ by adding $1$ to each of these to give solutions:

${z}_{1} = 1 + \sqrt{3} + i$

${z}_{2} = 1 - \sqrt{3} + i$

${z}_{3} = 1 - 2 i$