Question

How do you convert `-4 + 8i` to polar form?

Answer 1

In polar coordinates, the point will be `(r,theta)=(6.92i, -1.11i)`

Explanation:

For coordinates in polar form, `(r, theta)`, we need to find the value of `r` and the value of `theta`.

We will use Pythagoras' theorem to find `r` and the `tan` function to find `theta`.

`r=sqrt((-4)^2+8i^2)=sqrt(16-64) =sqrt(-48)=6.92i`

(since `8i^2=8ixx8i=8sqrt(-1)xx8sqrt(-1)=64xx-1=-64`)

`tantheta=(opp)/(adj)=(8i)/-4=-8/4i=-2i`

So `theta=tan^-1(-2i)=-1.11i`

Answer 2

`(4sqrt5,2.03)`

Explanation:

`"to convert from "color(blue)"cartesian to polar form"`

`"that is "(x,y)to(r,theta)" where"`

`•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=8`

`rArrr=sqrt((-4)^2+8^2)=sqrt80=4sqrt5`

`-4+8i" is in the second quadrant "`

`"so we must ensure that "theta" is in the second quadrant"`

`theta=tan^-1(2)=1.11larrcolor(red)" related acute angle"`

`rArrtheta=(pi-1.11)=2.03larrcolor(red)" in second quadrant"`

`rArr-4+8ito(-4,8)to(4sqrt5,2.03)`