Expressing in polar form

Question

How can I express -1- i√(3) in polar form that is

z = r(cos θ + i sin θ) ?

Answer 1

The other answer (from Mark M.) is a little misleading (or perhaps incomplete). For a + bi in rectangular/Cartesian coordinates, you have:

r² = a² + b²

The usual convention is to use a non-negative value of r (but this isn't required, polar forms are not unique!), so take the positive square root. Assuming r > 0:

tan(θ) = b/a

...but this IS NOT EQUIVALENT to θ = arctan(b/a) in general; only when -π/2 < θ < π/2 [between -90 and 90, if working in degrees rather than radians]. If a < 0, which is the case here, you need to adjust θ by either adding or subtracting π [180 degrees] so that the final value θ is in Quadrant II (when b > 0) or Quadrant III (when b < 0).

Whether you should add or subtract depends on the problem/application; Mark's answer assumes θ from 0 to 2π (which is very common) but in some cases you want θ from -π to π (in which case you would instead subtract 180 degrees).

Answer 2

r² = (-1)² + (-√3)²

θ = arc tan (√3) +180°