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WASpecialist MathematicsSyllabus dot point

Why does every nonzero complex number have exactly n distinct nth roots, evenly spaced on a circle?

Find the nth roots of a complex number and the nth roots of unity, and represent them on the Argand plane

WACE Specialist Unit 3 roots of complex numbers: solving z to the n equals w with the general root formula, the nth roots of unity, equal spacing on a circle of radius the nth root of the modulus, and their sum, with a worked example.

Generated by Claude Opus 4.76 min answer

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  1. What this dot point is asking
  2. The general root formula
  3. The nth roots of unity
  4. Plotting strategy

What this dot point is asking

SCSA wants you to solve zn=wz^n = w for all roots, identify the roots of unity, and plot them as a regular polygon on the Argand plane.

The general root formula

To solve zn=wz^n = w, write w=Rcisϕw = R\,\text{cis}\,\phi and let z=ρcisαz = \rho\,\text{cis}\,\alpha. By de Moivre zn=ρncis(nα)z^n = \rho^n\,\text{cis}(n\alpha), so matching moduli gives ρ=R1/n\rho = R^{1/n} and matching arguments gives nα=ϕ+2πkn\alpha = \phi + 2\pi k for integer kk. Hence

Taking k=0k = 0 to n1n-1 gives nn distinct roots; further values of kk repeat them. All roots share the modulus R1/nR^{1/n}, so they lie on a circle of that radius, spaced 2πn\tfrac{2\pi}{n} apart. This is why they form a regular nn-gon.

The nth roots of unity

Setting w=1=cis0w = 1 = \text{cis}\,0 gives the nnth roots of unity:

ωk=cis ⁣(2πkn),k=0,1,,n1.\omega_k = \text{cis}\!\left(\frac{2\pi k}{n}\right), \qquad k = 0, 1, \dots, n-1.

They all lie on the unit circle, include z=1z = 1, and are equally spaced. Writing ω=cis2πn\omega = \text{cis}\tfrac{2\pi}{n}, every root is a power ωk\omega^k.

Plotting strategy

Find one root (usually k=0k = 0), then rotate by 2πn\tfrac{2\pi}{n} repeatedly to get the rest. You never recompute the modulus.