Answer
$z=8(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
Work Step by Step
See p. 604,
A complex number $z=a+bi$ has the polar (or trigonometric) form
$z=r(\cos\theta+i\sin\theta)$
where $r=|z|=\sqrt{a^{2}+b^{2}}$ and $\tan\theta=b/a$.
The number $r$ is the modulus of $z$, and $\theta$ is an argument of $z$.
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$z=4(\sqrt{3}-i)=4\sqrt{3}-4i$
lies in Q.IV.
$r=|z|=\sqrt{(4\sqrt{3})^{2}+(-4)^{2}}=\sqrt{48+16}=8$
$\tan\theta =-\displaystyle \frac{4}{4\sqrt{3}}=-\frac{\sqrt{3}}{3}$
In Q.I, $\displaystyle \tan(\frac{\pi}{6}) =\displaystyle \frac{\sqrt{3}}{3},$
In Q.IV, $\displaystyle \tan(\frac{11\pi}{6}) =-\displaystyle \frac{\sqrt{3}}{3}$
So,
$z=8(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
