How do you convert i(6 + 6i) i(6+6i) to polar form?
1 Answer
Explanation:
To convert from
color(blue)"cartesian to polar form"cartesian to polar form
color(orange)"Reminder"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"
