Question

How do you divide `( 9i- 2) / (- 5 i+ 2 )` in trigonometric form?

Answer 1

`=1/29 (41+8i)`

Explanation:

Rewrite 9i-2 as -2+9i. Now square and the numbers -2 and 9, which would be 85. Its square root is`sqrt85`. Multiply and divide the expression -2 +9i with `sqrt85` as follows:

`sqrt85 (-2/sqrt85 +i 9/sqrt85)`

If `theta` is some angle then let `cos theta= -2/sqrt85 and sin theta=9/sqrt85`

Thus `-2 +9i= sqrt85(cos theta +isin theta)`

Like wise -5i +2 would be 2 -5i =`sqrt29( 2/sqrt 29-i5/sqrt29)`

If `phi` is some angle, then let `cos phi= 2/sqrt29 and sin phi= -5/sqrt29`

Thus `2-5i=sqrt29(cos phi -sin phi)`

`(9i-2)/(-5i+2)= sqrt(85/29) (cos theta +isin theta)/(cosphi-isin phi)=sqrt(85/29) e^(itheta)/e^(-iphi) =sqrt(85/29) e^(i(theta +phi)`

`=sqrt(85/29)( cos(theta+phi)+i sin (theta+phi) )`

`=sqrt(85/29)[ (cos theta cos phi- sin theta sin phi) +i(sin theta cos phi + cos theta sin phi)]`

`=sqrt(85/29)[(-2)/sqrt85 * 2/sqrt29 -9/sqrt85 * (-5)/sqrt29 +i(9/sqrt85 * 2/sqrt29 +(-2)/sqrt85 * (-5)/sqrt29)]`

`sqrt(85/29) (41/(sqrt85 sqrt29)+i 28/(sqrt 85 sqrt29))`
`=1/29 (41+28i)`