Answer
$16+16\sqrt{3}i$
Work Step by Step
We are given;
$z=1-\sqrt{3}i$
To find $r$ of a complex number $a+bi$, we use: $\sqrt{a^2+b^2}$:
$r=\sqrt{1^{2}+(-\sqrt{3})^{2}}=2$
To find $\theta$, we use $\tan{\theta}=\frac{b}{a}$:
$\displaystyle \tan\theta=\frac{-\sqrt{3}}{1}=-\sqrt{3}$
And since $(1,-\sqrt{3})$ is in the 4th quadrant:
$\displaystyle \theta=\frac{5\pi}{3} $
To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$:
$z=2(\displaystyle \cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})$ .
We apply the exponent to the polar form of the function and use De Moivre's Theorem:
$(1-\sqrt{3}i)^{5}=[2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})]^{5}=2^{5}(\cos\frac{5*5\pi}{3}+i\sin\frac{5*5\pi}{3})=2^{5}(\cos{(8\pi+\frac{\pi}{3})}+i\sin{(8\pi+\frac{\pi}{3})})=2^{5}(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3})
=32(\frac{1}{2}+\frac{\sqrt{3}}{2}i)=16+16\sqrt{3}i$
