Question

How do you convert `-3-3sqrt(3)i` to polar form?

Answer 1

`(-3,-3sqrt3)to(6,-(2pi)/3)`

Explanation:

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

That is `(x,y)to(r,theta)`

`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)|)))`

here x = - 3 and `y=-3sqrt3`

`rArr r=sqrt((-3)^2+(-3sqrt3)^2)=sqrt(9+27)=6`

Now `-3-3sqrt3` is in the 3rd quadrant hence we must ensure that `theta` is in the 3rd quadrant.

`rArrtheta=tan^-1((-3sqrt3)/(-3))=tan^-1(sqrt3)=pi/3`

However, `pi/3` is in the 1st quadrant and we require the 3rd.

`color(orange)"Reminder " color(red)(|bar(ul(color(white)(a/a)color(black)(-pi < theta <= pi)color(white)(a/a)|)))`

`rArrtheta=-pi+pi/3=-(2pi)/3`

`rArr -3-3sqrt3 i=(-3,-3sqrt3)to(6,-(2pi)/3)`