Answer
$z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
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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$r=|z|=\sqrt{(-2)^{2}+(2)^{2}}=\sqrt{8}=2\sqrt{2}$
$\displaystyle \tan\theta=\frac{2}{-2}=-1$,
In quadrant I, $\displaystyle \tan\frac{\pi}{4}=1$ (reference),
so, by symmetry (on the unit circle),
$\theta$ is either $\displaystyle \frac{3\pi}{4}$ (quadrant II) or $\displaystyle \frac{7\pi}{4}$ (quadrant IV)
z=-2+2i
lies in quadrant II, so we select the appropriate argument:
$z=2\displaystyle \sqrt{2}(\cos\frac{3\pi}{4}+i\sin\frac{3\pi}{4})$
