Maths › Complex numbers › Roots of unity and complex roots
Roots of unity and complex roots
Every number has exactly n distinct nth roots, arranged as a regular polygon on the Argand diagram: find one, and rotation finds the rest.
Builds on De Moivre's theorem and trigonometric identities.
IN THIS TOPIC
- Find all nth roots of unity and place them on the unit circle.
- Solve z to the n = w for any complex w, spacing the roots by 2π/n.
- Use the geometry: regular polygons, and roots of unity summing to zero.
WHAT YOU PROBABLY THINK
A number has one cube root, so z³ = 8i has exactly one solution.
The roots of unity
zn = 1 asks for numbers whose modulus powers to 1 and whose argument times n is a multiple of 2π. So every root has modulus 1 and argument 2πk/n for k = 0, 1, …, n − 1: exactly n distinct roots, equally spaced round the unit circle, the vertices of a regular n-gon with one vertex at 1.
Their sum is zero: the arrows balance by symmetry, since rotating the whole set by 2π/n permutes the roots yet must also rotate their sum. Only the zero vector survives both demands. That balancing act is a standard show-that mark in the exam.
Roots of any number
The same recipe opens every equation zn = w. Write w in modulus-argument form, take the real nth root of the modulus, divide the argument by n for one root, then space the remaining roots 2π/n apart. The opening claim undercounts badly: cube roots come in threes.
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: z = 2(cos π/6 + i sin π/6) = √3 + i.
The other two sit 2π/3 further round: arguments 5π/6 and 3π/2, giving −√3 + i and −2i.
Check the first: (√3 + i)³ has modulus 8 and argument π/2, which is 8i. Three roots, one equilateral triangle on a circle of radius 2.
YOUR TURN
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: 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, one corner on the positive real axis.
THE EXAM BIT
- Count before you finish: z to the n = w must produce exactly n roots.
- Give roots in the form asked for; exact surd form and modulus-argument form both appear.
- Space arguments by 2π/n from the first root, then reduce each into (−π, π].
- For 'show the roots sum to zero', the rotation-symmetry sentence earns the marks.
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
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nth roots: root the modulus, divide the argument by n, then space by 2π/n.
The n roots of unity form a regular n-gon on the unit circle and sum to zero.
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.
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 complex w, spacing the roots by 2π/n.
- Use the geometry: regular polygons, and roots of unity summing to zero.
Open the full revision checklist to see every objective in the course in one place.
No animated video for this topic yet; these notes stand alone.