Question

How do you write the following in trigonometric form and perform the operation given `(4i)/(-4+2i)`?

Answer 1

From Cartesian, `a + bi`, to trig `r(cos(theta)+isin(theta))`:

`r = sqrt(a^2+b^2)`
`theta = {(tan^-1(b/a);a>0,b>0),(pi/2; a=0,b>0),(pi+tan^-1(b/a),a<0),((3pi)/2;a=0,b<0), (2pi+tan^-1(b/a);a>0,b<0):}`

Explanation:

For the numerator, #a = 0, b = 4:

`r = sqrt(0^2+4^2)`

`r = 4`
`theta = (pi)/2`

The trig form of the numerator: `4(cos(pi/2)+isin(pi/2))`

For the denominator, #a = -4, b = 2:

`r = sqrt((-4)^2+2^2)`

`r = sqrt(16+4)`

`r = sqrt(20)`

`r = 2sqrt(5)`

`theta = pi + tan(2/-4)`

`theta ~~ 2.68`

The trig form of the numerator: `2sqrt(5)(cos(2.68)+isin(2.68))`

To divide you subtract the angle of the denominator from the angle of the numerator:

`theta = pi/2-2.68`

`theta ~~ -1.1`

And divide the magnitudes:

`r = 4/(2sqrt(5))`

`r = 2sqrt(5)/5`

The trig form is:

`2sqrt(5)/5(cos(-1.1)+isin(-1.1))`