Answer
In standard form,
$z_{1}z_{2}=(1+i\sqrt 3)(-\sqrt 3+i)=-\sqrt 3-3i+i+i^{2}\sqrt 3=-2\sqrt 3-2i$
In trigonometric form,
$z_{1}=1+i\sqrt 3=|z_{1}|(\cos60^{\circ}+isin60^{\circ})=2(\cos60^{\circ}+sin60^{\circ})$
$z_{2}=-\sqrt 3+i=2(\cos150^{\circ}+isin150^{\circ})= 2(\cos150^{\circ}+sin150^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$4 (cos210^{\circ}+isin210^{\circ})$
Check
$4(cos210^{\circ}+isin210^{\circ})=4(-\frac{\sqrt 3}{2}-i\frac{1}{2})=-2\sqrt 3-2i$ which matches with the previous result of the standard form.
Work Step by Step
In standard form,
$z_{1}z_{2}=(1+i\sqrt 3)(-\sqrt 3+i)=-\sqrt 3-3i+i+i^{2}\sqrt 3=-2\sqrt 3-2i$
In trigonometric form,
$z_{1}=1+i\sqrt 3=|z_{1}|(\cos60^{\circ}+isin60^{\circ})=2(\cos60^{\circ}+sin60^{\circ})$
$z_{2}=-\sqrt 3+i=2(\cos150^{\circ}+isin150^{\circ})= 2(\cos150^{\circ}+sin150^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$4 (cos210^{\circ}+isin210^{\circ})$
Check
$4(cos210^{\circ}+isin210^{\circ})=4(-\frac{\sqrt 3}{2}-i\frac{1}{2})=-2\sqrt 3-2i$ which matches with the previous result of the standard form.
