How do you divide ( -i+8) / (2i +7 )i+82i+7 in trigonometric form?

1 Answer

1/53(54-23i)153(5423i)

Explanation:

We have

\frac{-i+8}{2i+7}i+82i+7

=\frac{8-i}{7+2i}=8i7+2i

=\frac{\sqrt{65}(\cos(-\tan^{-1}(1/8))+i\sin(-\tan^{-1}(1/8)))}{\sqrt{53}(\cos(\tan^{-1}(2/7))+i\sin(\tan^{-1}(2/7)))}=65(cos(tan1(18))+isin(tan1(18)))53(cos(tan1(27))+isin(tan1(27)))

=\sqrt{\frac{65}{53}}(\cos(-\tan^{-1}(1/8)-\tan^{-1}(2/7))+i\sin(-\tan^{-1}(1/8)-\tan^{-1}(2/7)))=6553(cos(tan1(18)tan1(27))+isin(tan1(18)tan1(27)))

=\sqrt{\frac{65}{53}}(\cos(-\tan^{-1}(23/54))+i\sin(-\tan^{-1}(23/54)))=6553(cos(tan1(2354))+isin(tan1(2354)))

=\sqrt{\frac{65}{53}}(54/\sqrt3445-i23/\sqrt{3445})=6553(543445i233445)

=\sqrt{\frac{65}{53\times 3445}}(54-23i)=6553×3445(5423i)

=1/53(54-23i)=153(5423i)