Extracting the square root of a complex number
A square root of z = r(cos φ + i sin φ) is a complex number w = ρ(cos ψ + i sin ψ) whose square is z. From w² = z we get ρ² = r and 2ψ = φ + 2πn, so ρ = √r and ψ = (φ + 2πn)/2. For n = 0 and n = 1 we obtain two different roots, w₀ = √r(cos φ/2 + i sin φ/2) and w₁ = √r(cos(φ/2 + π) + i sin(φ/2 + π)) = −w₀; for n = 2, 3, … the roots repeat. So every non-zero complex number has two square roots, and they are opposite numbers. In the same way an nth root has modulus ⁿ√r and arguments (φ + 2πk)/n, k = 0, 1, …, n − 1: there are n roots and they lie at the vertices of a regular n-gon inscribed in the circle with centre at the origin and radius ⁿ√r. It helps to check the result by w² = z.
In pairs draw a circle and mark the three cube roots of z = 1 (arguments 0°, 120°, 240°); check with a protractor and ruler that they form a regular triangle.