Maths › Complex numbers › Roots of unity and complex roots
Roots of unity and complex roots
Every non-zero number has exactly n distinct nth roots, arranged as a regular polygon on the Argand diagram. Once one root is found, rotating it about the origin in steps of 2π/n gives the rest.
Builds on De Moivre's theorem and trigonometric identities and Radians, arcs and small angles.
IN THIS TOPIC
- Find all nth roots of unity and place them on the unit circle.
- Solve z to the n = w for any non-zero complex w, spacing the roots by 2π/n, and know that w = 0 gives z = 0 with multiplicity n.
- Use the geometry, including regular polygons and roots summing to zero.
COMMON MISCONCEPTION
A number has one cube root, so z³ = 8i has exactly one solution.
A non-zero complex number has exactly n distinct nth roots. Take the real positive nth root of the modulus, divide the argument by n, then space the rest 2π/n apart: 8i has modulus 8 and argument π/2, so the cube roots are √3 + i, −√3 + i and −2i, an equilateral triangle on the circle of radius 2. Only zn = 0 has a single root.
The roots of unity
zn = 1 asks for numbers whose modulus has nth power 1 and whose argument times n is a whole number of turns. Every root therefore has modulus 1 and argument 2πk/n for k = 0, 1, …, n − 1, giving exactly n distinct roots equally spaced round the unit circle, the vertices of a regular n-gon with one vertex at 1.
Write ω = e2πi/n; the set of roots is then 1, ω, ω², …, ωn−1, the powers of a single root. For n ≥ 2 their sum is zero; for n = 1 the only root is 1 and the sum is 1. Proof for n ≥ 2: multiplying every root by ω permutes the set, so the sum S satisfies ωS = S, that is (ω − 1)S = 0. Since ω ≠ 1 when n ≥ 2, S = 0.
Roots of any non-zero number
The same method solves every equation zn = w with w ≠ 0. Write w in modulus-argument form, take the real positive nth root of the modulus, divide the argument by n to obtain one root, then space the remaining roots 2π/n apart. Multiplying that first root by each nth root of unity gives the same set. Either way, a non-zero complex number has exactly n distinct nth roots.
For square roots there is an alternative method, preferable when the argument is not a standard angle. Instead of halving such an argument, square x + yi and compare real and imaginary parts, as with 3 + 4i in the complex arithmetic lesson. This keeps the answer exact rather than leaving an inverse tangent in it.
Zero is the exception. zn = 0 forces |z| = 0, so z = 0 is the only solution, a single root of multiplicity n rather than n distinct ones. Zero has no argument to divide, and there is no polygon to draw.
WORKED EXAMPLE
The three cube roots of 8i
Solve z³ = 8i.
8i has modulus 8 and argument π/2, so one root has modulus 2 and argument π/6, namely z = 2(cos π/6 + i sin π/6) = √3 + i.
The other two sit 2π/3 further round, at arguments 5π/6 and −π/2, giving −√3 + i and −2i.
Check the first. (√3 + i)³ has modulus 8 and argument π/2, which is 8i. The three roots are the vertices of an equilateral triangle on a circle of radius 2.
GUIDED PRACTICE
Fourth roots, read from a square
Solve z⁴ = 16 and describe the roots' arrangement.
Show the working
16 has modulus 16 and argument 0, so one root is 2 and the rest sit π/2 apart, giving 2, 2i, −2, −2i.
Each powers to 16. For 2i, (2i)⁴ = 16i⁴ = 16.
The four roots are the corners of a square of circumradius 2, with one corner on the positive real axis.
ASSESSMENT FOCUS
- Check the count: zⁿ = w with w non-zero has exactly n distinct roots.
- Space arguments by 2π/n from the first root, then reduce each into (−π, π].
- Give roots in the form asked for; exact surd form and r eiθ form both come up.
- For 'show the roots sum to zero', give the argument in full: multiplying the set by ω permutes it, so ωS = S and hence S = 0 when ω ≠ 1.
CHECK YOURSELF
Solve z³ = 27, giving all three roots exactly, and state the shape they make on the Argand diagram.
Show a hint
One root is real; the others sit 2π/3 either side.
Show the answer
27 has modulus 27 and argument 0, so the roots have modulus 3 and arguments 0, 2π/3, −2π/3. That gives z = 3, and 3(−1/2 ± (√3/2)i) = −3/2 ± (3√3/2)i. They form an equilateral triangle of circumradius 3 with one vertex at 3.
nth roots of a non-zero number: take the real positive nth root of the modulus, divide the argument by n, then space the roots by 2π/n.
The n roots of unity form a regular n-gon on the unit circle and, for n ≥ 2, sum to zero.
Every root of zⁿ = w is one root times a root of unity, and zⁿ = 0 has only z = 0.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the roots of unity and complex roots questions page.
CHECK YOUR PROGRESS
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- Find all nth roots of unity and place them on the unit circle.
- Solve z to the n = w for any non-zero complex w, spacing the roots by 2π/n, and know that w = 0 gives z = 0 with multiplicity n.
- Use the geometry, including regular polygons and roots summing to zero.
Open the full revision checklist to see every objective in the course in one place.