Answer
See the answer below.
Work Step by Step
$z_{1}z_{2}=(-3)(\sqrt {3}+i)=-3\sqrt {3}-3i$
$z_{1}$ in trigonometric form is
$-3=3(\cos \pi+i\sin\pi)$
$z_{2}$ in trigonometric form is
$\sqrt {3}+i=2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})$
Applying the formula $(r_{1}\,cis\,\theta_{1})(r_{2}\,cis\,\theta_{2})=r_{1}r_{2}\,cis\,(\theta_{1}+\theta_{2})$, we get
$z_{1}z_{2}=[3(\cos\pi+i\sin\pi)][2(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6})]$
$=3\cdot2[\cos(\pi+\frac{\pi}{6})+i\sin(\pi+\frac{\pi}{6})]$
$=6(\cos\frac{7\pi}{6}+i\sin\frac{7\pi}{6})$
In standard form, our result is
$=6(-\frac{\sqrt 3}{2}+i\cdot -\frac{1}{2})=-3\sqrt {3}-3i$
We see that both the products are equal to $-3\sqrt {3}-3i$.
