# 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)#