How do you convert #-3-3sqrt(3)i# to polar form?
1 Answer
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)#
