Square Root of Complex Number in Cartesian Form/Examples/-15-8i/Proof 1

Proof
We have that:
 * $-15 - 8 i = 17 \, \map \cis {\theta + 2 k \pi}$

where:
 * $\cos \theta = -\dfrac {15} {17}$
 * $\sin \theta = -\dfrac 8 {17}$

Then the square roots of $-15 - 8 i$ are:

We have that:

and:

As $\theta$ is in the third quadrant, $\dfrac \theta 2$ is in the second quadrant.

Hence:

Hence from $(1)$ and $(2)$: