Question

USE DE MOIVRE'S THEOREM TO EVALUATE (SQUARE ROOT OF 3 MINUS 1)^5. EXPRESS YOUR ANSWER IN THE STANDARD FORM A+BI

Answer 1

De Moivre's theorem states

$$(\cos \theta +i\sin \theta)^n=\cos n\theta+i\sin n\theta$$

Let $z=\sqrt{3}-i$

$$|z|=\sqrt{(\sqrt{3})^2+(-1)^2}=2$$

$$\tan \theta=\dfrac{-1}{\sqrt{3}}, \theta=-\dfrac{\pi}{6}$$

$$z=\sqrt{3}-i=2(\cos (-\dfrac{\pi}{6}) +i\sin (-\dfrac{\pi}{6}))$$

Therefore

$$z^5=(\sqrt{3}-i)^5=(2(\cos (-\dfrac{\pi}{6}) +i\sin (-\dfrac{\pi}{6})))^5$$

$$=2^5(\cos (-\dfrac{\pi}{6}) +i\sin (-\dfrac{\pi}{6}))^5$$

$$=32(\cos (-\dfrac{5\pi}{6}) +i\sin (-\dfrac{5\pi}{6}))$$

$$=32(-\dfrac{\sqrt{3}}{2}-i\dfrac{1}{2})$$

$$=-16\sqrt{3}-16i$$