How do you write #-3-2i# in trigonometric form?

1 Answer
Apr 23, 2017

#sqrt13(cos(2.55)-isin(2.55))#

Explanation:

#"to convert into "color(blue)"trigonometric form"#

#"that is " x+yitor(costheta+isintheta)# where

#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);-pi< theta<=pi)color(white)(2/2)|)))#

#"here " x=-3" and " y=-2#

#rArrr=sqrt((-3)^2+(-2)^2)=sqrt13#

#"now " -3-2i" is in the 3rd quadrant so must ensure that"#

#theta" is in the 3rd quadrant"#

#theta=tan^-1(2/3)=0.588larrcolor(red)" related acute angle"#

#rArrtheta=-pi+0.588~~-2.55larrcolor(red)" in 3rd quadrant"#

#rArr-3-2i=sqrt13(cos(-2.55)+isin(-2.55))#

#• color(orange)" Reminder" cos(-theta)=costheta ; sin(-theta)=-sintheta#

#rArr-3-2i=sqrt13(cos(2.55)-isin(2.55))#