Answer
In standard form,
$z_{1}z_{2}=(0+3i)(0-4i)=-12i^{2}=12$
In trigonometric form,
$z_{1}=0+3i=|z_{1}|(\cos90^{\circ}+isin90^{\circ})=3(\cos90^{\circ}+sin90^{\circ})$
$z_{2}=0-4i=4(\cos270^{\circ}+isin270^{\circ})= 4(\cos270^{\circ}+sin270^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$12 (cos360^{\circ}+isin360^{\circ})$
Check
$12(cos360^{\circ}+isin360^{\circ})=12(1+i0)=12$ which matches with the previous result of the standard form.
Work Step by Step
In standard form,
$z_{1}z_{2}=(0+3i)(0-4i)=-12i^{2}=12$
In trigonometric form,
$z_{1}=0+3i=|z_{1}|(\cos90^{\circ}+isin90^{\circ})=3(\cos90^{\circ}+sin90^{\circ})$
$z_{2}=0-4i=4(\cos270^{\circ}+isin270^{\circ})= 4(\cos270^{\circ}+sin270^{\circ})$
Therefore,
$z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$
=$12 (cos360^{\circ}+isin360^{\circ})$
Check
$12(cos360^{\circ}+isin360^{\circ})=12(1+i0)=12$ which matches with the previous result of the standard form.