Answer
$z=2\displaystyle \sqrt{5}(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})$
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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$z=-\sqrt{5}-\sqrt{15}i$
$r=|z|=\sqrt{(-\sqrt{5})^{2}+(-\sqrt{15})^{2}}=\sqrt{20}=2\sqrt{5}$
$\tan\theta =\displaystyle \frac{-\sqrt{15}}{-\sqrt{5}}=\sqrt{3}$
In Q.I, $\displaystyle \tan\frac{\pi}{3} =\displaystyle \frac{\sqrt{3}}{3}$
By symmetry (on the unit circle),
$\theta$ can also be $\displaystyle \frac{4\pi}{3}$ (Q.III).
z lies in Q.III,
so we select the appropriate argument:
$z=2\displaystyle \sqrt{5}(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3})$
