Finding nth roots of a complex number

Finding nth roots of a complex number

de Moivre’s formula can be used to obtain roots of complex numbers. Suppose n is a positive integer and a complex number ω is n th root of z denoted by z1/ n , then we have

ωn = z                     …………(1)

Let ω = ρ (cosϕ + i sinϕ ) and

z = r (cosθ + i sinθ ) = r (cos(θ + 2kπ ) + i sin (θ + 2kπ )),   k ∈ Z

Since w is the nth root of z , then

ωn = z

⇒ ρn (cosϕ + i sinϕ)n = r (cos(θ + 2kπ ) + i sin (θ + 2kπ )) ,   k ∈ Z

By de Moivre’s theorem,

⇒ ρn (cosnϕ + i sinnϕ) = r (cos (θ + 2kπ ) + i sin (θ + 2kπ )), k ∈ Z

Comparing the moduli and arguments, we get

Although there are infinitely many values of  k , the distinct values of  ω  are obtained when k = 0,1, 2, 3,K, n −1. When k = n, n +1, n + 2,K we get the same roots at regular intervals (cyclically). Therefore the nth roots of complex number z = r (cosθ + i sinθ ) are

If we set ω =  the formula for the  n th   roots of  a complex number has a nice geometric interpretation, as shown in Figure. Note that because | ω | = n√r the n roots all have the same modulus n√r they all lie on a circle of radius n√r with centre at the origin. Furthermore, the n roots are equally spaced along the circle, because successive n roots have arguments that differ by 2π/n .

Remark

(i) General form of de Moivre's Theorem

If x is rational, then cos xθ + i sin xθ is one of the values of (cosθ + i sinθ )x .

(ii) Polar form of unit circle

Let z  = eiθ  = cosθ + i sinθ . Then, we get

|z|2 = |cosθ + i sinθ|2

⇒  | x + iy|2 = cos2θ + sin2θ = 1

⇒ x2 + y2 = 1.

Therefore, |z| = 1 represents a unit circle (radius one) centre at the origin.