Answer
In standard form,
$z_{1}z_{2}=(1+i)(-1+i)=i^{2}-1=(\sqrt-1)^{2}-1=-2.$
In trigonometric form,
$z_{1}=1+i=|z_{1}|(\cos45^{\circ}+isin45^{\circ})=\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
$z_{2}=-1+i=|z_{2}|(\cos135^{\circ}+isin135^{\circ})=\sqrt 2(\cos135^{\circ}+sin135^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$\sqrt 2\sqrt 2 (cos180^{\circ}+isin180^{\circ})$
Check
$\sqrt 2\sqrt 2 (cos180^{\circ}+isin180^{\circ})=2(-1+0)=-2$ which matches with the previous result of the standard form.
Work Step by Step
In standard form,
$z_{1}z_{2}=(1+i)(-1+i)=i^{2}-1=(\sqrt-1)^{2}-1=-2.$
In trigonometric form,
$z_{1}=1+i=|z_{1}|(\cos45^{\circ}+isin45^{\circ})=\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
$z_{2}=-1+i=|z_{2}|(\cos135^{\circ}+isin135^{\circ})=\sqrt 2(\cos135^{\circ}+sin135^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$\sqrt 2\sqrt 2 (cos180^{\circ}+isin180^{\circ})$
Check
$\sqrt 2\sqrt 2 (cos180^{\circ}+isin180^{\circ})=2(-1+0)=-2$ which matches with the previous result of the standard form.