How do you write the trigonometric form of #-1+sqrt3i#?

1 Answer
May 30, 2017

Answer:

#2(cos((2pi)/3)+isin((2pi)/3))#

Explanation:

#"to convert from "color(blue)"cartesian to trig. form"#

#"that is " (x,y)tor(costheta+isintheta)#

#• r=sqrt(x^2+y^2)#

#• theta=tan^-1(y/x)#

#"here " x=-1" and " y=sqrt3#

#rArrr=sqrt((-1)^2+(sqrt3)^2)=2#

#-1+isqrt3" is in the second quadrant so we must ensure "#
#"that " theta" is in the second quadrant"#

#theta=tan^-1(sqrt3)=pi/3larrcolor(red)" related acute angle"#

#rArrtheta=(pi-pi/3)=(2pi)/3larr" in second quadrant"#

#rArr-1+isqrt3=2(cos((2pi)/3)+isin((2pi)/3))#