Roots of Complex Number/Examples/z^6 + 1 = root 3 i

Theorem
The roots of the equation:
 * $z^6 + 1 = \sqrt 3 i$

are:
 * $\set {\sqrt [6] 2 \cis 20 \degrees, \sqrt [6] 2 \cis 80 \degrees, \sqrt [6] 2 \cis 140 \degrees, \sqrt [6] 2 \cis 200 \degrees, \sqrt [6] 2 \cis 260 \degrees, \sqrt [6] 2 \cis 320 \degrees}$

Proof
We have:


 * $z^6 = -1 + \sqrt 3 i$

and so this is equivalent to finding the complex $6$th roots of $-1 + \sqrt 3 i$:

We have that:
 * $z^6 = 2 \, \map \cis {\dfrac {2 \pi} 3 + 2 k \pi}$

Let $z = r \cis \theta$.

Then: