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Complex Root Calculator — nth Roots of a Complex Number

Enter any complex number a + bi and the root degree n to find all n complex roots. The roots are equally spaced around a circle of radius r^(1/n) in the complex plane, a beautiful consequence of De Moivre's theorem.
Coefficient of i in a + bi
Integer from 1 to 12 (e.g. 2 for square root, 3 for cube root)
Root magnitude
1

All n roots lie on a circle of this radius: r^(1/n)

Input magnitude |z|
1
Root angle spacing
120°
Number of roots
3
Input argument θ
z0 = 1 + 0i
|z0| = 1
z1 = -0.5 + 0.866i
|z1| = 1
z2 = -0.5 − 0.866i
|z2| = 1
z0z1z2
Step by step
  1. 1

    Input modulus |z|

    √(a² + b²) = 1
  2. 2

    Input argument θ

    atan2(b, a) =
  3. 3

    Angle spacing per root

    360° ÷ 3 = 120°
    All n roots are equally spaced on the circle at this angular interval.
  4. 4

    Root magnitude r^(1/n)

    1 ^ (1 ÷ 3) = 1
Results are estimates for general information only and are not professional advice — always verify important results independently before relying on them. Read the full disclaimer.
Quick answer

How does this calculator work?

The n roots of z = a + bi are z_k = r^(1/n)·cis((θ + 2πk)/n) for k = 0..n−1, where r = |z| and θ = atan2(b, a). All roots share magnitude r^(1/n) and are equally spaced by 360°/n on a circle in the complex plane.

Formula
z_k = r^(1/n) · [cos((θ + 2πk)/n) + i·sin((θ + 2πk)/n)] for k = 0, 1, …, n−1
How this is calculated

To find the nth roots of z = a + bi, first convert to polar form: r = |z| = √(a² + b²) and θ = atan2(b, a). By De Moivre's theorem, the n distinct nth roots of z are: z_k = r^(1/n) · [cos((θ + 2πk)/n) + i·sin((θ + 2πk)/n)] for k = 0, 1, …, n − 1.

All roots have the same modulus r^(1/n) but different arguments, equally spaced by 2π/n radians (360°/n) around a circle. This means the roots form a regular n-gon inscribed in a circle of radius r^(1/n) centred at the origin — which the plot illustrates.

For real numbers, this method also works: a real number x is treated as x + 0i. Note that negative real numbers have an argument of π (180°), so their square roots are pure imaginary (e.g. √(−4) = ±2i), as expected.

Frequently asked questions

Because e^(iθ) is periodic with period 2π, adding 2πk (for k = 0, 1, …, n − 1) to the argument before dividing by n gives n different angles, hence n different roots. Adding 2πn would repeat the same angle.

Only for odd n. A real positive number has one real positive nth root and n − 1 complex conjugate pairs. A real negative number has zero real nth roots for even n (all roots are complex) and one real negative root for odd n.

The principal nth root is the one with the smallest non-negative argument — corresponding to k = 0 in the formula. It is the root most commonly meant by the n-th root symbol ⁿ√.

APA

TG we-Calculate Editorial Team. (2026). Complex Root Calculator — nth Roots of a Complex Number [Online calculator]. TG we-Calculate. https://we-calculate.com/calculator/complex-root-calculator

Chicago

TG we-Calculate Editorial Team. "Complex Root Calculator — nth Roots of a Complex Number." TG we-Calculate. 2026. https://we-calculate.com/calculator/complex-root-calculator.

IEEE

TG we-Calculate Editorial Team, "Complex Root Calculator — nth Roots of a Complex Number," TG we-Calculate, 2026. [Online]. Available: https://we-calculate.com/calculator/complex-root-calculator

BibTeX

@misc{wecalculate_complex_root_calculator, title = {Complex Root Calculator — nth Roots of a Complex Number}, author = {{TG we-Calculate Editorial Team}}, howpublished = {\url{https://we-calculate.com/calculator/complex-root-calculator}}, year = {2026}, note = {TG we-Calculate} }

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