Use De Moivre's Theorem to find ${(i - \sqrt{3} i)}^{8}$ .

Step 1

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

Step 2

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

Step 3

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

Step 4

Since $z$ lies in Quadrant IV, we choose $θ = a r c tan - \frac{\sqrt{3}}{1} + 2 π = a r c tan - \sqrt{3} + 2 π = \frac{5 π}{3}$

Step 5

The trigonometric form of a complex number then is.
$z = 2 (cos (5 \frac{π}{3}) + i sin (5 \frac{π}{3}))$

Step 6

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

Step 7

$= 256 [cos (\frac{40 π}{3} + i sin (40 \frac{π}{3}))] = 256 (- \frac{1}{2} + i \cdot - \frac{\sqrt{3}}{2}) = - 128 - 128 \sqrt{3} i$