Question

How do you convert `2 - 2i` to polar form?

Answer 1

`(2,-2)to(2sqrt2,-pi/4)`

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 = 2 and y = -2

`rArrr=sqrt(2^2+(-2)^2)=sqrt8=2sqrt2`

Now 2 - 2i is in the 4th quadrant, hence we must ensure that `theta` is in the 4th quadrant.

`theta=tan^-1((-2)/2)=tan^-1(-1)=-pi/4" in 4th quadrant"`

`rArr2-2i=(2,-2)to(2sqrt2,-pi/4)`