How do you write the complex number in trigonometric form #4-4sqrt3i#?

1 Answer
Mar 18, 2017

Answer:

#z=8[cos((2pi)/3)+isin((2pi)/3)]#

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)]#