Complex Roots and Radicals of Complex Numbers

Every nonzero complex number has exactly n different nth roots, which lie on a circle of radius ⁿ√r and are equally spaced at angles 2π/n: w_k = ⁿ√r (cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)), k = 0, 1, …, n − 1. In brief: the square roots of −4 are ±2i, the cube roots of 8 are 2, −1 + √3 i and −1 − √3 i, and the fourth roots of −16 are √2(±1 ± i). The roots are the vertices of a regular n-gon.

In the real numbers, a positive number has two square roots and a negative number has none. In the complex numbers, every number has n roots of order n, and the notation ⁿ√z is ambiguous without a choice. This guide derives the formula from de Moivre’s theorem, shows how to compute roots step by step, how to take square roots in algebraic form without the polar form and how the roots are placed in the complex plane, with a table of cube roots.

What is the formula for nth roots?

We want all w with wⁿ = z, where z = r(cos θ + i sin θ) ≠ 0. Write w = ρ(cos φ + i sin φ). By de Moivre’s theorem, wⁿ = ρⁿ(cos nφ + i sin nφ). Equating moduli, ρⁿ = r, so ρ = ⁿ√r (the positive real root). Equating arguments, nφ = θ + 2kπ, so φ = (θ + 2kπ)/n. Distinct values arise for k = 0, 1, …, n − 1, and then the values repeat. So:

w_k = ⁿ√r [cos((θ + 2kπ)/n) + i sin((θ + 2kπ)/n)], k = 0, 1, …, n − 1.

In exponential form, w_k = ⁿ√r e^{i(θ + 2kπ)/n}. The roots differ by the factor e^2πi/n, so each root is obtained from the previous one by a rotation by 2π/n.

How do you compute roots step by step?

1. Write z in polar form with the modulus r and an argument θ.
2. Compute the modulus of the roots: ⁿ√r.
3. Compute the arguments (θ + 2kπ)/n for k = 0, …, n − 1.
4. Write the n roots and, if needed, convert to algebraic form.
5. Check one root by raising it to the power n.

Example 1: cube roots of 8

8 = 8(cos 0 + i sin 0). The modulus is ∛8 = 2 and the arguments are 0, 2π/3 and 4π/3. The roots: 2; 2(cos 120° + i sin 120°) = −1 + √3 i; and 2(cos 240° + i sin 240°) = −1 − √3 i. Check: (−1 + √3 i)³: modulus 8 and argument 360°, so it equals 8 ✓.

Example 2: square roots of −4

−4 = 4(cos π + i sin π). The roots have the modulus 2 and the arguments π/2 and 3π/2: 2i and −2i. Indeed, (±2i)² = −4.

Example 3: fourth roots of −16

−16 = 16(cos 180° + i sin 180°). The modulus is 2 and the arguments are 45°, 135°, 225° and 315°. The roots: √2 + √2 i, −√2 + √2 i, −√2 − √2 i and √2 − √2 i, that is, √2(±1 ± i). Check: (√2 + √2 i)² = 2 + 4i − 2 = 4i, and (4i)² = −16 ✓.

Example 4: cube roots of 1 + i

1 + i = √2(cos 45° + i sin 45°). The modulus is (√2)^1/3 = 2^1/6 ≈ 1.1225. The arguments: 15°, 135° and 255°. The roots are 1.1225(cos 15° + i sin 15°) ≈ 1.0842 + 0.2905i, 1.1225(cos 135° + i sin 135°) ≈ −0.7937 + 0.7937i and 1.1225(cos 255° + i sin 255°) ≈ −0.2905 − 1.0842i.

Example 5: cube roots of −27i

−27i = 27(cos 270° + i sin 270°). The modulus is 3 and the arguments: 90°, 210° and 330°. The roots: 3i; 3(cos 210° + i sin 210°) = −(3√3/2) − (3/2)i; and 3(cos 330° + i sin 330°) = (3√3/2) − (3/2)i. The sum of the three roots is 0, a general property for n ≥ 2.

How do you find square roots in algebraic form?

To find √(a + bi), set (x + yi)² = a + bi, so x² − y² = a and 2xy = b. Also x² + y² = |z| = √(a² + b²), so x² = (|z| + a)/2 and y² = (|z| − a)/2, with the sign of x and y chosen so that 2xy has the sign of b.

Example 6: √(3 + 4i)

|z| = 5. x² = (5 + 3)/2 = 4 and y² = (5 − 3)/2 = 1. Since b = 4 > 0, x and y have the same sign: the roots are ±(2 + i). Check: (2 + i)² = 4 + 4i − 1 = 3 + 4i ✓.

Example 7: √(−5 + 12i)

|z| = 13. x² = (13 − 5)/2 = 4 and y² = (13 + 5)/2 = 9. With b > 0: the roots are ±(2 + 3i). Check: (2 + 3i)² = 4 + 12i − 9 = −5 + 12i ✓.

Example 8: a quadratic equation with complex coefficients

z² − (3 + i)z + (2 + i) = 0. D = (3 + i)² − 4(2 + i) = 9 + 6i − 1 − 8 − 4i = 2i. √(2i) = ±(1 + i) (since (1 + i)² = 2i). So z = (3 + i ± (1 + i))/2, giving z = 2 + i and z = 1. Check by Vieta’s formulas: the sum 3 + i ✓ and the product 2 + i ✓.

What do the roots look like in the plane?

All n roots lie on the circle of radius ⁿ√r centered at the origin, and the angle between neighbouring roots is 2π/n. So they form the vertices of a regular n-gon. For the cube roots of 8: an equilateral triangle with vertices 2, −1 + √3 i, −1 − √3 i. For the fourth roots of −16: a square with vertices (±√2, ±√2).

Table of cube roots

Where does the root formula come from?

The need for complex roots arose with Cardano’s formula for cubic equations in the sixteenth century, which involved cube roots of complex numbers even when all solutions of the cubic are real (the casus irreducibilis). De Moivre gave the general formula for the nth roots in 1730, and the geometric picture of the roots as the vertices of a regular polygon is due to Gauss and others in the nineteenth century.

Where are complex roots used today?

The solutions of differential equations with constant coefficients involve roots of characteristic polynomials, which are often complex and give oscillating solutions. The Fast Fourier Transform relies on the roots of unity. In geometry, constructing a regular polygon in the complex plane means finding roots.

Which mistakes are common?

Giving only one root

A complex number has n distinct nth roots. List all of them, if “all roots” is asked.

Using the wrong argument for the starting number

An error in θ affects all roots. Check the quadrant.

Not dividing the whole angle by n

The argument of the roots is (θ + 2kπ)/n, so 2kπ is also divided by n.

Assuming that √(z₁z₂) = √z₁√z₂

With several values the rule needs care. For example √(−1)√(−1) = i · i = −1, but √((−1)(−1)) = √1 = 1.