Question

(sqrt(3)-i)^-10 Find the indicated power using De Moivre's Theorem.

Answer 1

You already know De Moivre's Theorem, right?
\([r(cos~ \theta +isin ~\theta)]^a=r^a(cos~ a.\theta+isin~ a.\theta)\)

Answer 2

First, try to convert \(\sqrt3 -i\) into trigonometric form.

\(\large \sqrt3 -i=2(\frac{\sqrt3}{2}-\frac{1}{2}i)=2(cos\frac{\pi}{6}-isin\frac{\pi}{6})\).

Now you can use the formula.

\(\large [2(cos\frac{\pi}{6}-isin \frac{\pi}{6})]^{-10}=2^{-10}[cos(-10.\frac{\pi}{6})-isin(-10.\frac{\pi}{6})]\)

\(=\large 2^{-10}[cos(-\frac{5\pi}{3})-isin(-\frac{5\pi}{3})]=2^{-10}(\frac{1}{2}+\frac{\sqrt3}{2}i)\)