Question

How do you convert `sqrt(3)i - 1` to polar form?

Answer 1

`(2,(2pi)/3)`

Explanation:

Using the formulae that links Cartesian to Polar coordinates.

`•r=sqrt(x^2+y^2)" and " theta=tan^-1(y/x)`

now `sqrt3 i-1=-1+sqrt3 i`

here x = -1 and y =`sqrt3`

`rArrr=sqrt((-1)^2+(sqrt3)^2)=sqrt4=2`

and `theta=tan^-1(-sqrt3)=-pi/3`

`-1+sqrt3 i" is in the 2nd quadrant"`

so `theta" requires to be an angle in 2nd quadrant"`

`rArrtheta=(pi-pi/3)=(2pi)/3`

`rArr(-1,sqrt3)=(2,(2pi)/3)`