How do you write the trigonometric form of #-4+2i#?

1 Answer
Jun 25, 2017

#2sqrt5(cos(2.678)+isin(2.678))#

Explanation:

#"to convert from "color(blue)"cartesian to polar form"#

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

#• r=sqrt(x^2+y^2)#

#• theta=tan^-1(y/x) ; -pi< theta <=pi#

#"here " x=4" and " y=2#

#rArrr=sqrt((-4)^2+2^2)=sqrt20=2sqrt5#

#"note " -4+2i" is in the second quadrant and so " theta" must"#
#"be in the second quadrant"#

#theta=tan^-1(1/2)~~0.464larrcolor(red)"related acute angle"#

#rArrtheta=(pi-0.464)=2.678larrcolor(red)" in second quadrant"#

#rArr-4+2ito2sqrt5(cos(2.678)+isin(2.678))#