Answer
$z=2(\displaystyle \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$
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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$r=|z|=\sqrt{(-\sqrt{2})^{2}+(-\sqrt{2})^{2}}=\sqrt{2+2}=2$
$\displaystyle \tan\theta=\frac{-\sqrt{2}}{-\sqrt{2}}=1$,
In quadrant I, $\displaystyle \tan\frac{\pi}{4}=1$.
By symmetry (on the unit circle),
$\theta$ can also be $\displaystyle \frac{5\pi}{4}$ (quadrant III)
$z=-\sqrt{2}-i\sqrt{2}$
lies in quadrant III, so we select the appropriate argument:
$z=2(\displaystyle \cos\frac{5\pi}{4}+i\sin\frac{5\pi}{4})$