Question

How do you convert `(sqrt 5)-i` to polar form?

Answer 1

`(sqrt6,-0.42)`

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)`