How do you write the complex number in trigonometric form #4-4sqrt3i#?
1 Answer
Mar 18, 2017
Explanation:
#(x,y)tor(costheta+isintheta)#
#color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(r=sqrt(x^2+y^2))color(white)(2/2)|)))#
#"and " color(red)(bar(ul(|color(white)(2/2)color(black)(theta=tan^-1(y/x))color(white)(2/2)|)))#
#"here "x=4" and " y=-4sqrt3#
#rArrr=sqrt((4)^2+(-4sqrt3)^2)=sqrt64=8#
#"now " 4-4sqrt3# is in the 2nd quadrant so we must ensure that
#theta# is in the 2nd quadrant.
#rArrtan^-1((4sqrt3)/4)=tan^-1(sqrt3)=pi/3#
#rArrtheta=(pi-pi/3)=(2pi)/3larrcolor(red)" in 2nd quadrant"#
#rArr4-4sqrt3i=8[cos((2pi)/3)+isin((2pi)/3)]#
