How do you convert #(sqrt 5)-i# to polar form?
1 Answer
Aug 10, 2017
Explanation:
#"to convert from "color(blue)"cartesian to polar form"#
#"that is "(x,y)to(r,theta)" using"#
#•color(white)(x)r=sqrt(x^2+y^2)#
#•color(white)(x)theta=tan^-1(y/x)color(white)(x);-pi < theta<=pi#
#"here "x=sqrt5" and "y=-1#
#rArrr=sqrt((sqrt5)^2+(-1)^2)=sqrt6#
#sqrt5-i" is in fourth quadrant so we must ensure "theta# is in the fourth quadrant.
#theta=tan^-1(1/sqrt5)=0.42larrcolor(red)" related acute angle"#
#rArrtheta=-0.42larrcolor(red)" in fourth quadrant"#
#rArr(sqrt5,-1)to(sqrt6,-0.42)#
