Answer
$z_{1}z_{2}=-3+3i$
Work Step by Step
$z_{1}z_{2}=(1+i)(3i)=3i+3i^{2}=3i-3=-3+3i$
$z_{1}$ in trigonometric form is
$1+i=\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})$
$z_{2}$ in trigonometric form is
$3i=3(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2})$
Applying the formula
$(r_{1}\,cis\,\theta_{1})(r_{2}\,cis\,\theta_{2})=r_{1}r_{2}\,cis\,(\theta_{1}+\theta_{2})$, we get
$z_{1}z_{2}=[\sqrt {2}(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4})][3(\cos \frac{\pi}{2}+i\sin\frac{\pi}{2})]$
$=\sqrt {2}\cdot3[\cos(\frac{\pi}{4}+\frac{\pi}{2})+i\sin(\frac{\pi}{4}+\frac{\pi}{2})]$
$=3\sqrt {2}(\cos \frac{3\pi}{4}+i\sin \frac{3\pi}{4})$
In standard form, our result is
$=3\sqrt {2}(-\frac{\sqrt {2}}{2}+i\cdot\frac{\sqrt 2}{2})=-3+3i$
