# How do you find all the cube roots of z, where z = (sqrt3)/2 + 1/2i?

Jan 6, 2016

${r}_{1} = \cos \left(\frac{\pi}{18}\right) + i \sin \left(\frac{\pi}{18}\right)$

${r}_{2} = \cos \left(\frac{13 \pi}{18}\right) + i \sin \left(\frac{13 \pi}{18}\right)$

${r}_{3} = \cos \left(\frac{25 \pi}{18}\right) + i \sin \left(\frac{25 \pi}{18}\right)$

#### Explanation:

This is an example of "Casus Irreducibilis" - a Complex cube root that is not expressible in the form $a + b i$ where $a$ and $b$ are expressed in terms of Real $n$th roots.

You can express it in terms of trigonometric functions.

$z$ itself lies on the unit circle:

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

Use De Moivre's formula:

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

to deduce:

The cube roots also lie on the unit circle at intervals of $\frac{2 \pi}{3}$:

${r}_{1} = \cos \left(\frac{\pi}{18}\right) + i \sin \left(\frac{\pi}{18}\right)$

${r}_{2} = \cos \left(\frac{13 \pi}{18}\right) + i \sin \left(\frac{13 \pi}{18}\right)$

${r}_{3} = \cos \left(\frac{25 \pi}{18}\right) + i \sin \left(\frac{25 \pi}{18}\right)$