Practice Maths

← Back to Further Complex Numbers›Specialist Maths

Roots of Complex Numbers

Key Terms

nth roots of a complex number
If z = reiθ, the n distinct nth roots are zk = r1/n ei(θ+2πk)/n, k = 0, 1, …, n−1.
Equally spaced
The nth roots lie on a circle of radius r1/n centred at the origin, equally spaced by angles of 2π/n.
Roots of unity
The nth roots of unity satisfy zn = 1 and are ωk = e2πik/n, k = 0, 1, …, n−1. They always sum to 0 (for n ≥ 2).
Regular polygon
The n roots form the vertices of a regular n-gon in the Argand plane.
Strategy
Convert to polar form, add 2πk to the argument, take the nth root of the modulus, divide the argument by n. List all k = 0, 1, …, n−1.
Primitive root
ω = e2πi/n is the primitive nth root of unity; all nth roots of unity are powers of ω.
Key Formula:
  • If z = r(cosθ + i sinθ), the nth roots are: zk = r1/n(cos((θ+2πk)/n) + i sin((θ+2πk)/n)), k = 0, 1, …, n−1
  • Modulus of each root: |zk| = r1/n
  • Angular spacing: 2π/n radians between consecutive roots
  • Sum of nth roots of unity: ∑k=0n−1 ωk = 0
Hot Tip Always write the polar form with the general argument θ + 2πk before dividing by n. If you forget the + 2πk step, you find only one root instead of n. Check your answers by verifying each root raised to the nth power gives the original number, and that consecutive roots are separated by exactly 2π/n.

Worked Example 1 — Cube roots of 8

Find all cube roots of 8.

Write 8 = 8(cos 0 + i sin 0) = 8ei·0. So r = 8, θ = 0, n = 3.

zk = 81/3(cos(2πk/3) + i sin(2πk/3)) = 2(cos(2πk/3) + i sin(2πk/3)), for k = 0, 1, 2.

k = 0: z0 = 2(cos 0 + i sin 0) = 2

k = 1: z1 = 2(cos(2π/3) + i sin(2π/3)) = 2(−1/2 + i√3/2) = −1 + i√3

k = 2: z2 = 2(cos(4π/3) + i sin(4π/3)) = 2(−1/2 − i√3/2) = −1 − i√3

These three roots form an equilateral triangle inscribed in the circle |z| = 2.

Worked Example 2 — 4th roots of −1

Find all 4th roots of −1.

Write −1 = cosπ + i sinπ = eiπ. So r = 1, θ = π, n = 4.

zk = 11/4(cos((π+2πk)/4) + i sin((π+2πk)/4)), k = 0, 1, 2, 3.

k = 0: cos(π/4) + i sin(π/4) = √2/2 + i√2/2 = (1+i)/√2

k = 1: cos(3π/4) + i sin(3π/4) = −√2/2 + i√2/2 = (−1+i)/√2

k = 2: cos(5π/4) + i sin(5π/4) = −√2/2 − i√2/2 = (−1−i)/√2

k = 3: cos(7π/4) + i sin(7π/4) = √2/2 − i√2/2 = (1−i)/√2

These form a square inscribed in the unit circle, rotated π/4 from the axes.

Deriving the Formula for nth Roots

Suppose we want to solve zn = w, where w = reiθ is a given complex number. Write z in polar form as z = ρeiφ. Then zn = ρneinφ. For this to equal w = reiθ, we need:

  • ρn = r   ⇒   ρ = r1/n (unique positive real solution)
  • nφ = θ + 2πk for some integer k   ⇒   φ = (θ + 2πk)/n

Since the complex exponential has period 2π, different values of k give different arguments only for k = 0, 1, …, n−1. For k = n, we get φ = (θ + 2πn)/n = θ/n + 2π, which is the same angle as k = 0. So there are exactly n distinct roots.

Geometric Meaning: Regular Polygons

All n roots have the same modulus r1/n, so they all lie on a circle of radius r1/n. The arguments are (θ/n), (θ/n + 2π/n), (θ/n + 4π/n), …, each separated by 2π/n radians. These n equally spaced points on a circle form the vertices of a regular n-gon.

For n = 3: equilateral triangle. For n = 4: square. For n = 6: regular hexagon. The first root (k=0) has argument θ/n, and the polygon is “rotated” relative to the standard orientation by this amount.

Roots of Unity

The nth roots of unity are the solutions to zn = 1. Since 1 = ei·0, the roots are ωk = e2πik/n for k = 0, 1, …, n−1, where ω = e2πi/n is the primitive nth root of unity.

For n = 3: the cube roots of unity are 1, ω = e2πi/3 = −1/2 + i√3/2, and ω2 = e4πi/3 = −1/2 − i√3/2. These form an equilateral triangle.

For n = 4: the 4th roots of unity are 1, i, −1, −i — a square aligned with the axes.

Sum of roots of unity = 0: The nth roots of unity form a geometric series: 1 + ω + ω2 + … + ωn−1 = (1 − ωn)/(1 − ω) = 0/(1 − ω) = 0, since ωn = 1 and ω ≠ 1. Geometrically, the roots are symmetrically placed around the circle, so their centroid is the origin.

Connection to Polynomial Equations

Solving zn = w is equivalent to finding the roots of the polynomial zn − w = 0. By the fundamental theorem of algebra, this degree-n polynomial has exactly n roots (counting multiplicity) in ℂ. The formula gives all n distinct roots explicitly.

For example, z3 = 8 is equivalent to z3 − 8 = (z−2)(z2+2z+4) = 0. The three roots are z = 2 and z = (−2 ± i√12)/2 = −1 ± i√3, consistent with the formula.

Exam Tip: To show roots form a regular n-gon: state that all roots have equal modulus r1/n (so lie on a circle) and are equally spaced with angular separation 2π/n. Then name the polygon.
Exam Tip: The sum of all nth roots of any complex number w is 0. This is because the nth roots of w are w1/n times the nth roots of unity, and the roots of unity sum to 0. Use this as a quick check or shortcut.

Mastery Practice

  1. Find all cube roots of 8, expressing each in Cartesian form a + bi. Fluency

  2. Find all 4th roots of −16, expressing each in Cartesian form. Fluency

  3. Find all 6th roots of unity. Express each in exact Cartesian form and state what regular polygon they form. Fluency

  4. Solve z3 = 1 + i√3, expressing the roots in polar form reiθ. Fluency

  5. Show that the three cube roots of any nonzero complex number form the vertices of an equilateral triangle in the Argand plane. Understanding

  6. Let ω = e2πi/5 be a primitive 5th root of unity. Show that 1 + ω + ω2 + ω3 + ω4 = 0. Understanding

  7. Find all solutions to z4 + 1 = 0 in polar and Cartesian form. Understanding

  8. Find all primitive 6th roots of unity. Explain what “primitive” means and why not all 6th roots of unity are primitive. Understanding

  9. Find all complex numbers z such that |z| = 2 and z3 is a negative real number. Express your answers in Cartesian form. Problem Solving

  10. Let the four 4th roots of a complex number w be z0, z1, z2, z3. Problem Solving

    1. (a) Show that the sum z0 + z1 + z2 + z3 = 0 for any nonzero w.
    2. (b) If w = 1 + i, find the product z0z1z2z3 in Cartesian form.
See Answers ➔