Use de-Moivre's theorem to find the reciprocal of each number below. $$\sqrt 3 - i$$
Given $\sqrt{3}-i$ , we need to find the reciprocal of it using de-Moivre's theorem.
$$\frac{1}{\sqrt 3-i} $$
$$= \frac{1(\cos0^c + i\sin 0^c)}{2\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$
$$= \frac{1}{2}\cdot \frac{\cos\big(\frac{\pi}{6}\big) - i\sin\big(\frac{\pi}{6})}{\big(\frac{\sqrt3}{2}-\frac{i}{2}\big)}$$
$$ = \frac{1}{2}\cdot \cos\biggr(\frac{\pi}{6}\biggr) - i\sin\biggr(\frac{\pi}{6}\biggr)$$
$$ = \frac{1}{2} \cdot \biggr(\frac{\sqrt3}{2} +i \frac{1}{2}\biggr )$$
$$\boxed { = \frac{\sqrt3}{4}+\frac{i}{4}}$$
Does my assumption seem correct?
De Moivre's Theorem states that: $$[r(\cos \theta + i \sin \theta)]^n=r^n(\cos (n\theta)+ i \sin (n\theta))$$
To find the reciprocal, take $n=-1$. $$z=\sqrt{3}-i\to r=2, \theta=\tan^{-1}\bigg({\frac{-1}{\sqrt{3}}}\bigg)=\frac{-\pi}{6}$$ Hence $z^{-1}=2^{-1}(\cos{(-1)(\frac{-\pi}{6})}+ i\sin{(-1)(\frac{-\pi}{6})})=\frac{1}{2}(\cos \frac{\pi}{6}+i\sin\frac{\pi}{6})$
$$=\frac{\sqrt{3}}{4}+\frac{1}{4}i$$ as you achieved.