Find all n complex roots of any complex number using De Moivre's theorem. Roots are evenly spaced on a circle of radius r1/n — see them all in polar form, rectangular form, and on the Argand diagram.
| k | Angle (degrees) | Angle (radians) | Rectangular form |
|---|
z^n = r·cis(θ) ⟹ exactly n solutions
The Fundamental Theorem of Algebra guarantees that every degree-n polynomial has exactly n roots in ℂ. The equation wn = z is a degree-n polynomial in w, so there are exactly n roots.
Geometrically, each root lies at the same distance r1/n from the origin — all on a circle of that radius. As k increases from 0 to n−1, adding 360°/n each time rotates around the full circle exactly once, hitting n equally spaced points before repeating.
If you used k = n, you'd get angle (θ + 360°·n)/n = θ/n + 360° — the same angle as k = 0. So adding more values of k just repeats roots already found.
z^(1/n) = r^(1/n) · cis((θ + 360°k)/n)
De Moivre's Theorem states: [r·cis(θ)]m = rm·cis(mθ). To find the nth root, we apply this with m = 1/n.
We write z in polar form as r·cis(θ). But cis(θ) = cis(θ + 360°k) for any integer k — because adding 360° to an angle gives the same point. Substituting:
w_k = [r · cis(θ + 360°k)]^(1/n)
= r^(1/n) · cis((θ + 360°k)/n)
For k = 0, 1, 2, …, n−1 this gives n distinct angles in [0°, 360°). The angular spacing 360°/n is the same between every consecutive pair of roots.
One-on-one Trigonometry tutoring makes De Moivre's theorem click — we work through polar form, Argand diagrams, and the geometry behind nth roots until it feels intuitive.