## Trigonometry 7th Edition

In standard form, $z_{1}z_{2}=(0+2i)(0-5i)=-10i^{2}=10$ In trigonometric form, $z_{1}=0+2i=|z_{1}|(\cos90^{\circ}+isin90^{\circ})=2(\cos90^{\circ}+sin90^{\circ})$ $z_{2}=0-5i=5(\cos270^{\circ}+isin270^{\circ})= 5(\cos270^{\circ}+sin270^{\circ})$ Therefore, $z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ =$10 (cos360^{\circ}+isin360^{\circ})$ Check $10(cos360^{\circ}+isin360^{\circ})=10(1+i0)=10$ which matches with the previous result of the standard form.
In standard form, $z_{1}z_{2}=(0+2i)(0-5i)=-10i^{2}=10$ In trigonometric form, $z_{1}=0+2i=|z_{1}|(\cos90^{\circ}+isin90^{\circ})=2(\cos90^{\circ}+sin90^{\circ})$ $z_{2}=0-5i=5(\cos270^{\circ}+isin270^{\circ})= 5(\cos270^{\circ}+sin270^{\circ})$ Therefore, $z_{1}z_{2}=r_{1}r_{2}[\cos(\theta_{1}+\theta_{2})+i\sin(\theta_{1}+\theta_{2})]$ =$10 (cos360^{\circ}+isin360^{\circ})$ Check $10(cos360^{\circ}+isin360^{\circ})=10(1+i0)=10$ which matches with the previous result of the standard form.