Question

Express `z_1=sqrt(​3)-i` in polar form?

Answer 1

`z_1=1angle(-pi/6)`
fourth quadrant

Explanation:

Given:
`z_1=sqrt3-i`
If `z=a+ib,`
`r=sqrt(a^2+b^2)`
and
`theta=tan^-1(b/a)`
Here,
`z=z_1, a=sqrt3, b=-1`
Thus,
`|z_1|=sqrt(sqrt3^2+(-1)^2`
`=sqrt(3+1)=sqrt4=2`
`theta_1=tan^-1((-1)/sqrt3)=-pi/6`

Hence, `z=r angletheta`

`z_1=1angle(-pi/6)`
fourth quadrant