Question

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

Answer 1

I found: `2/_30^@`

Explanation:

Have a look:

Answer 2

`(2,pi/6)`

Explanation:

To convert cartesian coordinates (x ,y) to Polar coordinates `(r,theta)`
Use the formulae which link them.

`•r=sqrt(x^2+y^2)`

and since `sqrt3+i=(sqrt3,1)` is in the 1st quadrant , then.

`•theta=tan^-1(y/x)`

here x =`sqrt3" and "y=1`

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

and `theta=tan^-1(1/sqrt3)=pi/6" or "30^@`

`rArr(sqrt3,1)to(2,pi/6)`