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Formulas/maths/M3 De Moivre/nth Roots of a Complex Number

nth Roots of a Complex Number

Every non-zero complex number z = r·cis θ has exactly n distinct nth roots. They lie on a circle of radius r^(1/n) centred at the origin, spaced equally at angular intervals of 2π/n — forming a regular n-gon. In Euler form: r^(1/n) · e^(i(θ+2kπ)/n).
Derivation

Finding the nnth roots of a complex number is one of the most geometric results in the chapter. The algebra gives a formula; the geometry reveals why there are exactly nn roots and where they sit.

Setup

Let z=reiθz = re^{i\theta} (with r=z>0r = |z| > 0 and θ=arg(z)\theta = \arg(z)). We want all ww such that wn=zw^n = z.

Write w=ρeiϕw = \rho e^{i\phi}. Then:

wn=ρneinϕ=reiθw^n = \rho^n e^{in\phi} = re^{i\theta}

Equating moduli and arguments separately:

ρn=r    ρ=r1/n(unique positive real root)\rho^n = r \implies \rho = r^{1/n} \quad \text{(unique positive real root)} nϕ=θ+2kπ,kZ    ϕ=θ+2kπnn\phi = \theta + 2k\pi, \quad k \in \mathbb{Z} \implies \phi = \frac{\theta + 2k\pi}{n}

The n Distinct Roots

wk=r1/n(cosθ+2kπn+isinθ+2kπn),k=0,1,,n1w_k = r^{1/n} \left(\cos\frac{\theta + 2k\pi}{n} + i\sin\frac{\theta + 2k\pi}{n}\right), \quad k = 0, 1, \ldots, n-1

For knk \geq n, the angle (θ+2kπ)/n(\theta + 2k\pi)/n differs from some wjw_j with 0jn10 \leq j \leq n-1 by a multiple of 2π2\pi — giving the same point. So there are exactly nn distinct roots.

The Geometric Picture

All nn roots:

  • lie on a circle of radius r1/nr^{1/n} centred at the origin
  • are separated by equal angular gaps of 2π/n\mathbf{2\pi/n}
  • form the vertices of a regular nn-gon

The first root w0=r1/neiθ/nw_0 = r^{1/n} e^{i\theta/n} anchors the polygon. Every subsequent root is obtained by rotating w0w_0 by 2π/n2\pi/n.

Special Case: Roots of Unity

When z=1z = 1 (r=1r = 1, θ=0\theta = 0):

wk=e2πik/n=cos2kπn+isin2kπn,k=0,1,,n1w_k = e^{2\pi ik/n} = \cos\frac{2k\pi}{n} + i\sin\frac{2k\pi}{n}, \quad k = 0, 1, \ldots, n-1

A regular nn-gon inscribed in the unit circle with one vertex always at (1,0)(1, 0).

Why the Modulus Has a Unique Value

The equation ρn=r\rho^n = r with ρ,r>0\rho, r > 0 has exactly one positive real solution ρ=r1/n\rho = r^{1/n}. All nn roots share this modulus — they lie on a single circle. The only freedom is in the argument, which produces the nn distinct angular positions.