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

##### 1 Answer

#### 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"#