Question

What is the trigonometric form of `(4-2i)`?

Answer 1

`2sqrt5` `cis` `(-0.46)`

Explanation:

To find the trigonometric form, we have to know `r`, the distance of the point from the origin, and `theta`, the angle.

We can use the following formulas:

`r = sqrt(a^2+b^2)`

`tan theta = b/a`

`r = sqrt(4^2+(-2)^2)`
`r = sqrt(20)`
`r = 2sqrt(5)`

`tan theta = -2/4`
`theta = tan^-1(-2/4)`
`theta = -0.46`

So, the trigonometric form is `2sqrt5` `cis` `(-0.46)` or `2sqrt5 (cos (-0.46) + i sin (-0.46))`.

Answer 2

`2sqrt5(cos(0.46)-isin(0.46))`

Explanation:

`"to convert from "color(blue)"cartesian to trig. form"`

`"that is " (x,y)tor(costheta+isintheta)" use"`

`•color(white)(x)r=sqrt(x^2+y^2)`

`•color(white)(x)theta=tan^-1(y/x)color(white)(x);-pi < theta<=pi`

`"here "x=4" and " y=-2`

`rArrr=sqrt(4^2+(-2)^2)=sqrt20=2sqrt5`

4-2i is in the fourth quadrant so we must ensure that `theta` is in the fourth quadrant.

`theta=tan^-1(1/2)=0.46larrcolor(red)" related acute angle"`

`rArrtheta=-0.46larrcolor(red)" in fourth quadrant"`

`rArr4-2i=2sqrt5(cos(-0.46)+isin(-0.46))`

`rArr4-2i=2sqrt5(cos(0.46)-isin(0.46))`