Answer
$2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})$
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 $z$ is in the 4th quadrant, we have:
$\displaystyle \theta=\frac{5\pi}{3}$
To put the number in polar form, we use $r(\cos{\theta}+i\sin{\theta})$:
$2(\cos\frac{5\pi}{3}+i\sin\frac{5\pi}{3})$
