Question

How do you convert `-1+(sqrt3i)` to polar form?

Answer 1

`(r,theta)=(2,120^@)`

Explanation:

For a complex number in the form
`color(white)("XXX")a+bi`
the radius is given by the Pythagorean Theorem as
`color(white)("XXX")r=sqrt(a^2+b^2)`
In this case
`color(white)("XXX")r=sqrt((-1)^2_(sqrt(3))^2)=sqrt(1+3)=sqrt(4)=2`

The angle can be calculated using an inverse trig function and adjusting to place the point in the proper quadrant.

In this case `-1+sqrt(3)i` is in Quadrant 2 of the complex plane.

Using the `arctan` function gives an angle in `(-90^@,+90^@)`
so it will be necessary to add `180^@` to shift the angle into Q 2.
`color(white)("XXX")theta=arctan(-sqrt(3)/1)+180^@ =120^@`

(In this particular case, we have one of the standard triangles and the angle could easily be inferred with a quick sketch. )