Question

How do you divide `( 9i- 7) / (- 5 i -6 )` in trigonometric form?

Answer 1

`sqrt(130)/sqrt(61)cos(-91.931)+isin(-91.931)`

Explanation:

`z=a+bi` in trigonometric form is `z=r(cos\theta+isin\theta)`, where `r=sqrt(a^2+b^2)` and `\theta=tan^(-1)(b/a)`

So, we have `(-7+9i)/(-6-5i)=(sqrt((-7)^2+9^2)(cos(tan^(-1)(9/-7))+isin(tan^(-1)(9/-7))))/(sqrt((-6)^2+(-5)^2)(cos(tan^(-1)((-5)/-6))+isin(tan^(-1)((-5)/-6))))`
`=(sqrt(130)(cos(-52.125)+isin(-52.125)))/(sqrt(61)(cos(39.806)+isin(39.806)))`
`=sqrt(130)/sqrt(61)cos(-52.125-39.806)+isin(-52.125-39.806)`
`=sqrt(130)/sqrt(61)cos(-91.931)+isin(-91.931)`