Answer
$z=2\displaystyle \sqrt{2}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
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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$z=i(\sqrt{2}-\sqrt{6}i)=i\sqrt{2}-\sqrt{6}i^{2}=\sqrt{6}+i\sqrt{2}$
lies in Q.I.
$r=|z|=\sqrt{(\sqrt{6})^{2}+(\sqrt{2})^{2}}=\sqrt{8}=2\sqrt{2}$
$\tan\theta =\displaystyle \frac{\sqrt{2}}{\sqrt{6}}=\frac{\sqrt{3}}{3}$
In Q.I,$\quad \displaystyle \tan(\frac{\pi}{6}) =\displaystyle \frac{\sqrt{3}}{3},$
So,
$z=2\displaystyle \sqrt{2}(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
