# Question #0c21c

Apr 11, 2017

$- 128 \left(1 + i \sqrt{3}\right) .$

#### Explanation:

Let $z = 1 - i \sqrt{3} = r \left(\cos \theta + i \sin \theta\right) = r c i s \theta , r > 0.$

$\therefore r \cos \theta = 1 , r \sin \theta = - \sqrt{3.}$

Squaring and adding, ${r}^{2} \left({\cos}^{2} \theta + {\sin}^{2} \theta\right) = 1 + 3.$

$\Rightarrow r = 2.$

$\therefore \cos \theta = \frac{1}{r} = \frac{1}{2} , \mathmr{and} , \sin \theta = - \frac{\sqrt{3}}{2.}$

$\therefore \theta = - \frac{\pi}{3.}$

$\Rightarrow z = r c i s \theta = 2 c i s \left(- \frac{\pi}{3}\right) .$

By De' Moivre's Theorem, then,

${\left(1 - i \sqrt{3}\right)}^{8} = {z}^{8} = {\left\{2 c i s \left(- \frac{\pi}{3}\right)\right\}}^{8} = {2}^{8} c i s \left\{8 \left(- \frac{\pi}{3}\right)\right\} .$

$= {2}^{8} \left\{\cos \left(- 8 \frac{\pi}{3}\right) + i \sin \left(- 8 \frac{\pi}{3}\right)\right\} ,$

$= {2}^{8} \left\{\cos \left(8 \frac{\pi}{3}\right) - i \sin \left(8 \frac{\pi}{3}\right)\right\} ,$

$= {2}^{8} \left\{\cos \left(3 \pi - \frac{\pi}{3}\right) - i \sin \left(3 \pi - \frac{\pi}{3}\right)\right\} ,$

$= {2}^{8} \left\{- \cos \left(\frac{\pi}{3}\right) - i \sin \left(\frac{\pi}{3}\right)\right\} ,$

$= - {2}^{8} \left(\frac{1}{2} + i \frac{\sqrt{3}}{2}\right) ,$

$= - {2}^{7} \left(1 + i \sqrt{3}\right) .$

Hence, the Reqd. Value=$- 128 \left(1 + i \sqrt{3}\right) .$

Enjoy Maths.!