**Toolbox:**

- From De moivre's theorem we have
- (i) $(\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q$
- (ii) $(\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta$
- (iii) $(\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta$
- (iv) $(\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\theta)+i\sin(\large\frac{\pi}{2}$$-\theta)]^n=\cos n(\large\frac{\pi}{2}$$-\theta)+i\sin n(\large\frac{\pi}{2}$$-\theta)$
- $e^{i\theta}=\cos\theta+i\sin\theta$
- $e^{-i\theta}=\cos\theta-i\sin\theta$,also written as $\cos\theta$ and $\cos(-\theta)$

Step 1:

Let $x=\big(\large\frac{1}{2}-i\large\frac{\sqrt 3}{2}\big)^{\Large\frac{3}{4}}$

Now $\big|\large\frac{1}{2}$$-i\large\frac{\sqrt 3}{2}\big|=\frac{1}{4}+\frac{3}{4}$

$\qquad\qquad\quad\quad\;\;=1$

Step 2:

We have $\large\frac{1}{2}-$$i\frac{\sqrt 3}{2}$$=\cos\big(\large\frac{-\pi}{3}\big)$$+i\sin\big(\large\frac{-\pi}{3}\big)$

Therefore $x=[\cos\big(\large\frac{-\pi}{3}\big)$$+i\sin\big(\large\frac{-\pi}{3}\big)]^{\Large\frac{3}{4}}$

$\qquad\qquad\;\;=[\cos(-\pi)+i\sin(-\pi)]^{\large\frac{1}{4}}$

$\qquad\qquad\;\;=[\cos(2k\pi-\pi)+i\sin(2k\pi-\pi)]^{\large\frac{1}{4}}\;\;k\in z$

$\qquad\qquad\;\;=[\cos(2k-1)\pi+i\sin(2k-1)\pi]^{\large\frac{1}{4}}\;\;k\in z$

$\qquad\qquad\;\;=[\cos(2k-1)\large\frac{\pi}{4}$$+i\sin(2k-1)\large\frac{\pi}{4}$$]\;\;\;k=0,1,2,3$

Step 3:

The four values are $cis\big(\large\frac{-\pi}{4}\big),$$cis \;0,cis\large\frac{3\pi}{4},$$cis\big(\large\frac{5\pi}{4}\big)$

$\Rightarrow cis\big(\large\frac{-\pi}{4}\big),$$cis \large\frac{\pi}{4},$$\;cis\large\frac{3\pi}{4},$$cis\big(\large\frac{-3\pi}{4}\big)$