Answer
$z=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
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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$z=2\sqrt{3}-2i$
$r=|z|=\sqrt{(2\sqrt{3})^{2}+(-2)^{2}}=\sqrt{4\cdot 3+4}=4$
$\displaystyle \tan\theta=\frac{-2}{2\sqrt{3}}=-\frac{\sqrt{3}}{3}$,
In quadrant I, $\displaystyle \tan\frac{\pi}{6}=+\frac{\sqrt{3}}{3}$.
By symmetry (on the unit circle),
$\theta$ can be $\displaystyle \frac{5\pi}{6}$ (quadrant II) or $\displaystyle \frac{11\pi}{6}$ (quadrant IV)
$z=2\sqrt{3}-2i$
lies in quadrant IV, so we select the appropriate argument:
$z=4(\displaystyle \cos\frac{11\pi}{6}+i\sin\frac{11\pi}{6})$
