Answer
In standard form,
$z_{1}z_{2}=(1+i)(2+2i)=2+2i^{2}+4i=2+2(\sqrt-1)^{2}+4i=4i$
In trigonometric form,
$z_{1}=1+i=|z_{1}|(\cos45^{\circ}+isin45^{\circ})=\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
$z_{2}=2+2i=|z_{2}|(\cos45^{\circ}+isin45^{\circ})=2\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$2\sqrt 2\sqrt 2 (cos90^{\circ}+isin90^{\circ})$=$4 (cos90^{\circ}+isin90^{\circ})$
Check
$4(cos90^{\circ}+isin90^{\circ})=4(0+1i)=4i$ which matches with the previous result of the standard form.
Work Step by Step
In standard form,
$z_{1}z_{2}=(1+i)(2+2i)=2+2i^{2}+4i=2+2(\sqrt-1)^{2}+4i=4i$
In trigonometric form,
$z_{1}=1+i=|z_{1}|(\cos45^{\circ}+isin45^{\circ})=\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
$z_{2}=2+2i=|z_{2}|(\cos45^{\circ}+isin45^{\circ})=2\sqrt 2(\cos45^{\circ}+sin45^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$2\sqrt 2\sqrt 2 (cos90^{\circ}+isin90^{\circ})$=$4 (cos90^{\circ}+isin90^{\circ})$
Check
$4(cos90^{\circ}+isin90^{\circ})=4(0+1i)=4i$ which matches with the previous result of the standard form.