# Find all the values of $\bigg(\large\frac{1}{2}$$-i\large\frac{\sqrt 3}{2}\bigg)^{\Large\frac{3}{4}} and hence prove that the product of the values is 1. ## 1 Answer Toolbox: • From De moivre's theorem we have • (i) (\cos\theta+i\sin\theta)^n=\cos n\theta+i\sin n\theta,n\in Q • (ii) (\cos\theta+i\sin\theta)^{-n}=\cos n\theta-i\sin n\theta • (iii) (\cos\theta-i\sin\theta)^n=\cos n\theta-i\sin n\theta • (iv) (\sin \theta+i\cos \theta)^n=[\cos(\large\frac{\pi}{2}$$-\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)$