Answer
$z=2\displaystyle \sqrt{2}(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$
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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$z=-\sqrt{6}+\sqrt{2}i$
$r=|z|=\sqrt{(\sqrt{6})^{2}+(\sqrt{2})^{2}}=\sqrt{8}=2\sqrt{2}$
$\tan\theta =\displaystyle \frac{\sqrt{2}}{-\sqrt{6}}=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$
In Q.I, $\displaystyle \tan\frac{\pi}{6} =\displaystyle \frac{\sqrt{3}}{3}$
By symmetry (on the unit circle),
$\theta$ can be $\displaystyle \frac{5\pi}{6}$ (Q.II) or $\displaystyle \frac{7\pi}{6}$ (Q.III).
z lies in Q.II,
so we select the appropriate argument:
$z=2\displaystyle \sqrt{2}(\cos\frac{5\pi}{6}+i\sin\frac{5\pi}{6})$
