Femtobytes
Back
Problem type

The n-th roots of unity

What it solves
Solve z to the power n equals 1. Find all n solutions, see why they are the corners of a regular polygon, why each is a power of the first, and prove that they add up to zero.
Subject
Complex Numbers
Topic
Length and angle
Also called
roots of unity, nth roots of 1, cyclotomic
Your problem

Find every complex number zz that satisfies this equation.

z11=1z^{11} = 1
Invent a new one:
Femtobytes
The 11 roots of unity: solve z^11 = 1, step by step
femtobytes.com/solve/roots-of-unity/11
The problem

Find every complex number zz that satisfies this equation.

z11=1z^{11} = 1
The result
ωk=cos⁡2kπ11+isin⁡2kπ11,k=0,1,…,10\omega_k = \cos \frac{2k\pi}{11} + i \sin \frac{2k\pi}{11}, \quad k = 0, 1, \ldots, 10

There are exactly 1111 solutions, equally spaced round the unit circle, and their sum is 00.

checked 22 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.
Picture 1 of 2Step 1: Say what is being asked, and what we already expect
0.50.5i-0.5-0.5i1i-1-ireal axisimaginary axis1
The number 1, where every root of unity must land after being multiplied by itself n times. Each root is a turn that, done n times, comes back exactly to here.
1

Say what is being asked, and what we already expect

Why this step

Among real numbers, z11=1z^{11} = 1 has one solution, 11. Yet an equation of degree 1111 is entitled to 1111 solutions. The missing ones are not on the number line at all. They are in the plane, and the only way to find them is to stop thinking of multiplication as repeated adding and start thinking of it as turning.

The number 11, as a point of the plane, has length 11 and points along the positive real axis, so its angle is 00, or equally 2π2\pi, 4π4\pi, and so on: 1=cos⁡(2kπ)+isin⁡(2kπ)1 = \cos(2k\pi) + i \sin(2k\pi) for every whole number kk. Here ii is the imaginary unit with i2=−1i^2 = -1, and angles are in radians, where a full turn is 2π2\pi.

Write an unknown solution as z=ρ(cos⁡φ+isin⁡φ)z = \rho \left( \cos \varphi + i \sin \varphi \right) with length ρ\rho (rho) and angle φ\varphi (phi). De Moivre's theorem says raising to the power 1111 raises the length to that power and multiplies the angle by 1111: z11=ρ11(cos⁡11φ+isin⁡11φ)z^{11} = \rho^{11} \left( \cos 11\varphi + i \sin 11\varphi \right).

ρ11=1and11φ=2kπ\rho^{11} = 1 \qquad \text{and} \qquad 11\varphi = 2k\pi
Equal complex numbers have equal lengths, and angles that agree up to whole turns.
0.50.5i-0.5-0.5i1i-1-ireal axisimaginary axis1
The number 1, where every root of unity must land after being multiplied by itself n times. Each root is a turn that, done n times, comes back exactly to here.
2

Solve for the length and the angles

Why this step

The length equation has no freedom in it: a length is a positive real number and the only positive real number whose power is 11 is 11 itself. So every solution lies on the unit circle. The angle equation is where the 1111 different solutions come from: a full turn, shared out 1111 ways, gives 1111 different directions before it starts repeating.

ρ=1,φk=2kπ11,k=0,1,…,10\rho = 1, \qquad \varphi_k = \frac{2k\pi}{11}, \quad k = 0, 1, \ldots, 10
Dividing 2kπ2k\pi by 1111. Taking k=11k = 11 would give 2π2\pi, a full turn, which is the same direction as k=0k = 0; so kk stops at 1010.
φ0=0\varphi_{0} = 0
With k=0k = 0. checked
φ1=2π11\varphi_{1} = \frac{2 \pi}{11}
With k=1k = 1: 2π11\frac{2 \pi}{11} in lowest terms, which is 36011∘\frac{360}{11}^\circ. checked
φ2=4π11\varphi_{2} = \frac{4 \pi}{11}
With k=2k = 2: 4π11\frac{4 \pi}{11} in lowest terms, which is 72011∘\frac{720}{11}^\circ. checked
φ3=6π11\varphi_{3} = \frac{6 \pi}{11}
With k=3k = 3: 6π11\frac{6 \pi}{11} in lowest terms, which is 108011∘\frac{1080}{11}^\circ. checked
φ4=8π11\varphi_{4} = \frac{8 \pi}{11}
With k=4k = 4: 8π11\frac{8 \pi}{11} in lowest terms, which is 144011∘\frac{1440}{11}^\circ. checked
φ5=10π11\varphi_{5} = \frac{10 \pi}{11}
With k=5k = 5: 10π11\frac{10 \pi}{11} in lowest terms, which is 180011∘\frac{1800}{11}^\circ. checked
φ6=12π11\varphi_{6} = \frac{12 \pi}{11}
With k=6k = 6: 12π11\frac{12 \pi}{11} in lowest terms, which is 216011∘\frac{2160}{11}^\circ. checked
φ7=14π11\varphi_{7} = \frac{14 \pi}{11}
With k=7k = 7: 14π11\frac{14 \pi}{11} in lowest terms, which is 252011∘\frac{2520}{11}^\circ. checked
φ8=16π11\varphi_{8} = \frac{16 \pi}{11}
With k=8k = 8: 16π11\frac{16 \pi}{11} in lowest terms, which is 288011∘\frac{2880}{11}^\circ. checked
φ9=18π11\varphi_{9} = \frac{18 \pi}{11}
With k=9k = 9: 18π11\frac{18 \pi}{11} in lowest terms, which is 324011∘\frac{3240}{11}^\circ. checked
φ10=20π11\varphi_{10} = \frac{20 \pi}{11}
With k=10k = 10: 20π11\frac{20 \pi}{11} in lowest terms, which is 360011∘\frac{3600}{11}^\circ. checked
0.50.5i-0.5-0.5i1i-1-ireal axisimaginary axis03C9₀03C9₁03C9₂03C9₃03C9₄03C9₅03C9₆03C9₇03C9₈03C9₉03C9₁₀
All the roots have length 1, so they sit on the unit circle, and their angles are the equal shares of a full turn.
3

Write out the roots

4

See that every root is a power of the first one

5

Prove that the roots add up to zero