🔬 Grade 12 · Lesson 11 of 13
Unity Roots: The nth Roots of 1 in the Complex Plane
Asking “what number, raised to the nth power, gives 1?” has not one answer but n — and they arrange themselves into a perfect regular polygon on the unit circle.
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The big idea: Over the real numbers, only 1 (and sometimes −1) solves z<sup>n</sup> = 1. Over the complex numbers there are always exactly n solutions, the <strong>nth roots of unity</strong>. They all sit on the unit circle, spaced evenly at angles of 2π/n, forming the corners of a regular n-gon. This is the clearest picture in mathematics of why complex numbers make rotation and multiplication the same thing.
More answers than you expect
Over the real numbers, x² = 1 has two solutions and x³ = 1 has just one. Something feels lopsided. In the complex plane the pattern becomes beautifully regular: zn = 1 always has exactly n solutions, called the nth roots of unity.
They are the answer to “which complex numbers, raised to the nth power, land exactly on 1?” And the answer, drawn out, is a piece of pure symmetry.
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Multiplying is rotating
A complex number on the unit circle can be written as an angle: cos θ + i sin θ. Multiplying two of them adds their angles. So raising to the nth power multiplies the angle by n. To land back on 1 (angle 0, or any full turn 2πk), the starting angle must be a whole fraction of a full circle.
\[ z_k = \cos\!\left(\dfrac{2\pi k}{n}\right) + i\,\sin\!\left(\dfrac{2\pi k}{n}\right), \qquad k = 0, 1, 2, \dots, n-1 \]
The n roots of unity. Each has modulus 1 (so it sits on the unit circle) and an argument that is a multiple of 2π/n.
Evenly spaced around the circle
Look at the angles: 0, 2π/n, 4π/n, and so on. Each root is one step of 2π/n radians further around than the last. Starting at 1 (angle 0), the roots march around the unit circle at equal spacing and return to the start after n steps.
Connect them and you get a regular n-gon inscribed in the unit circle, with one vertex always pinned at 1.
🎮 Roots of Unity LIVE
The nth roots of 1 sit evenly around the unit circle.
📝 Worked example: Find all the fourth roots of unity (the solutions of z⁴ = 1).
- There are n = 4 roots, spaced 2π/4 = 90° apart, starting at angle 0.
- k = 0: angle 0° → cos0 + i sin0 = 1.
- k = 1: angle 90° → cos90° + i sin90° = i.
- k = 2: angle 180° → cos180° + i sin180° = −1.
- k = 3: angle 270° → cos270° + i sin270° = −i.
✓ The fourth roots of unity are <strong>1, i, −1, and −i</strong> — the corners of a square on the unit circle.
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They always add to zero
For any n ≥ 2, the nth roots of unity sum to 0. By symmetry each root has a mirror image pulling the opposite way, so all the arrows cancel. Check it on the fourth roots: 1 + i + (−1) + (−i) = 0. This fact is the seed of the discrete Fourier transform.
\[ 1 + z_1 + z_2 + \cdots + z_{n-1} = 0 \quad (n \ge 2) \]
The sum of all nth roots of unity vanishes — the evenly spaced vectors cancel out perfectly.
📝 Worked example: Find the three cube roots of unity (the solutions of z³ = 1).
- There are n = 3 roots, spaced 2π/3 = 120° apart, starting at angle 0.
- k = 0: angle 0° → 1.
- k = 1: angle 120° → cos120° + i sin120° = \(-\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}\,i\).
- k = 2: angle 240° → cos240° + i sin240° = \(-\tfrac{1}{2} - \tfrac{\sqrt{3}}{2}\,i\).
✓ The cube roots of unity are <strong>1, −½ + (√3/2)i, and −½ − (√3/2)i</strong> — an equilateral triangle on the unit circle.
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Real thinking misses most of them
If you only look for real solutions, zn = 1 seems to have just one or two answers. That is the trap: the equation has exactly n complex roots, most of them off the real axis. All n share modulus 1 — they differ only in angle, never in distance from the origin.
Check your understanding
1. How many distinct fifth roots of unity are there (solutions of z⁵ = 1)?
zⁿ = 1 always has exactly n complex solutions, so z⁵ = 1 has 5 of them, evenly spaced on the unit circle.
2. The fourth roots of unity are…
Spaced 90° apart from angle 0: 1 (0°), i (90°), −1 (180°), −i (270°) — the corners of a square.
3. What is the angle between two consecutive sixth roots of unity?
Consecutive roots are 2π/n apart. For n = 6 that is 360°/6 = 60°.
4. What is the sum of all n roots of unity when n ≥ 2?
The evenly spaced roots are symmetric about the origin, so they cancel in pairs and sum to exactly 0.
5. Where in the complex plane do all the nth roots of unity lie?
Every nth root of unity has modulus 1, so all lie on the unit circle, evenly spaced as the vertices of a regular n-gon.
✅ Key takeaways
- The equation z^n = 1 has exactly n complex solutions, the nth roots of unity.
- Each root is cos(2πk/n) + i sin(2πk/n) for k = 0 to n−1 — modulus 1, angle a multiple of 2π/n.
- The roots sit on the unit circle spaced 2π/n apart, forming a regular n-gon with a vertex at 1.
- For n ≥ 2 the nth roots of unity always sum to zero, because they cancel by symmetry.
- Raising a unit-circle number to a power multiplies its angle, which is why rotation and multiplication coincide.