Guided Example #1 - DeMoivre's Theorem and the n-th Roots of a Complex Number Given

First convert the complex number ${\displaystyle z=a+bi}$ from its standard form to its trigonometric form
${\displaystyle z=r(cos\theta +isin\theta )}$, by using ${\displaystyle r=\sqrt{a^2+b^2}}$ and ${\displaystyle tan\theta =\frac{b}{a}}$.

Here ${\displaystyle z=-1-\sqrt{3}i}$ . So comparing this with the standard form ${\displaystyle z=a+bi}$, we see that ${\displaystyle a=1}$ and ${\displaystyle b=-\sqrt{3}}$.

Thus, ${\displaystyle r=\sqrt{\left(1^2\right)+{(-\sqrt{3})}^2})=\sqrt{1+3}=\sqrt{4}}$=2 and ${\displaystyle tan\theta =\frac{-\sqrt{3}}{1}=-\sqrt{3}}$.

Since ${\displaystyle z}$ lies in Quadrant IV, we choose ${\displaystyle \theta =arctan-\frac{\sqrt{3}}{1}+2\pi =arctan-\sqrt{3}+2\pi =\frac{5\pi }{3}}$

The trigonometric form of a complex number then is.
${\displaystyle z=2(cos\left(5\frac{\pi }{3}\right)+isin\left(5\frac{\pi }{3}\right))}$

Then, by De Moivre's theorem, ${\displaystyle {(i -\sqrt{3}i)}^8=[2{(cos\left(5\frac{\pi }{3}\right)+isin\left(5\frac{\pi }{3}\right)]}^8=2^8(cos\left(8\frac{5\pi }{3}\right) +isin\left(8\frac{5\pi }{3}\right))}$

${\displaystyle =256\left[cos(\frac{40\pi }{3}+isin\left(40\frac{\pi }{3}\right))\right]=256(-\frac{1}{2}+i\cdot -\frac{\sqrt{3}}{2})=-128-128\sqrt{3}i}$