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,
$z_{1}=-1+i\sqrt 3=|z_{1}|(\cos120^{\circ}+isin120^{\circ})=2(\cos120^{\circ}+sin120^{\circ})$
$z_{2}=\sqrt 3+i=2(\cos30^{\circ}+isin30^{\circ})= 2(\cos30^{\circ}+sin30^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$4 (cos150^{\circ}+isin150^{\circ})$
Check
$4(cos150^{\circ}+isin150^{\circ})=4(-\frac{\sqrt 3}{2}+i\frac{1}{2})=-2\sqrt 3+2i$ 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,
$z_{1}=-1+i\sqrt 3=|z_{1}|(\cos120^{\circ}+isin120^{\circ})=2(\cos120^{\circ}+sin120^{\circ})$
$z_{2}=\sqrt 3+i=2(\cos30^{\circ}+isin30^{\circ})= 2(\cos30^{\circ}+sin30^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$4 (cos150^{\circ}+isin150^{\circ})$
Check
$4(cos150^{\circ}+isin150^{\circ})=4(-\frac{\sqrt 3}{2}+i\frac{1}{2})=-2\sqrt 3+2i$ which matches with the previous result of the standard form.