Question

How do you divide `(-4+5i) / (2+i)` in trigonometric form?

Answer 1

`(-4+5i)/(2+i)`
in trigonometric form is
`(r1)/(r2)cis(theta1-theta2)`
where,
`r1 = sqrt(((-4)^2+5^2))`
`theta1 =tan^-1(5/(-4))`
and
`r2 = sqrt((2^2+1^2))`
`theta1 =tan^-1(1/2)`

Explanation:

Divide `(-4+5i)/(2+i)`
in trigonometric form
Expressing numerator and denominator as
`z = r cis theta`
For `z1=-4+5i`
`r1 = sqrt(((-4)^2+5^2))`
`theta1 =tan^-1(5/(-4))`
For `z2=2+i`
`r2 = sqrt((2^2+1^2))`
`theta1 =tan^-1(1/2)`
Thus,
`(-4+5i)/(2+i)=(r1cistheta1)/(r2cistheta2)`
`=(r1)/(r2)(cistheta1)/(cistheta2)`
By De-Moivre's theorem,
`(cistheta1)/(cistheta2)=cis(theta1-theta2)`
Now,
`(-4+5i)/(2+i)=(r1)/(r2)cis(theta1-theta2)`