Question

How do you convert `i(6 + 6i)` to polar form?

Answer 1

`(6sqrt2,(3pi)/4)`

Explanation:

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

`color(orange)"Reminder"`

`color(red)(|bar(ul(color(white)(a/a)color(black)(r=sqrt(x^2+y^2))color(white)(a/a)|)))" and "`

`color(red)(|bar(ul(color(white)(a/a)color(black)(theta=tan^-1(y/x))color(white)(a/a)|)))`

The first step, however, is to distribute the bracket.

`rArri(6+6i)=6i+6i^2" and " color(red)(|bar(ul(color(white)(a/a)color(black)(i^2=(sqrt(-1))^2=-1)color(white)(a/a)|)))`

`rArr6i+6i^2=-6+6i`

here x = -6 and y = 6

`rArrr=sqrt((-6)^2+6^2)=sqrt72=6sqrt2`

Now - 6 + 6i is in the 2nd quadrant so we must ensure that `theta` is in the 2nd quadrant.

`theta=tan^-1((6)/-6)=tan^-1(-1)=-pi/4" in 4th quadrant"`

`rArrtheta=(pi-pi/4)=(3pi)/4" in 2nd quadrant"`

`rArr-6+6i=(-6,6)to(6sqrt2,(3pi)/4)" in polar form"`