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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.

z12=1z^{12} = 1
Invent a new one:
Femtobytes
The 12 roots of unity: solve z^12 = 1, step by step
femtobytes.com/solve/roots-of-unity/12
The problem

Find every complex number zz that satisfies this equation.

z12=1z^{12} = 1
The result
ωk=cos⁡2kπ12+isin⁡2kπ12,k=0,1,…,11\omega_k = \cos \frac{2k\pi}{12} + i \sin \frac{2k\pi}{12}, \quad k = 0, 1, \ldots, 11

There are exactly 1212 solutions, equally spaced round the unit circle, and their sum is 00.

checked 24 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, z12=1z^{12} = 1 has two solutions, 11 and −1-1. Yet an equation of degree 1212 is entitled to 1212 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 1212 raises the length to that power and multiplies the angle by 1212: z12=ρ12(cos⁡12φ+isin⁡12φ)z^{12} = \rho^{12} \left( \cos 12\varphi + i \sin 12\varphi \right).

ρ12=1and12φ=2kπ\rho^{12} = 1 \qquad \text{and} \qquad 12\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 1212 different solutions come from: a full turn, shared out 1212 ways, gives 1212 different directions before it starts repeating.

ρ=1,φk=2kπ12,k=0,1,…,11\rho = 1, \qquad \varphi_k = \frac{2k\pi}{12}, \quad k = 0, 1, \ldots, 11
Dividing 2kπ2k\pi by 1212. Taking k=12k = 12 would give 2π2\pi, a full turn, which is the same direction as k=0k = 0; so kk stops at 1111.
φ0=0\varphi_{0} = 0
With k=0k = 0. checked
φ1=π6\varphi_{1} = \frac{\pi}{6}
With k=1k = 1: 2π12\frac{2 \pi}{12} in lowest terms, which is 30∘30^\circ. checked
φ2=π3\varphi_{2} = \frac{\pi}{3}
With k=2k = 2: 4π12\frac{4 \pi}{12} in lowest terms, which is 60∘60^\circ. checked
φ3=π2\varphi_{3} = \frac{\pi}{2}
With k=3k = 3: 6π12\frac{6 \pi}{12} in lowest terms, which is 90∘90^\circ. checked
φ4=2π3\varphi_{4} = \frac{2 \pi}{3}
With k=4k = 4: 8π12\frac{8 \pi}{12} in lowest terms, which is 120∘120^\circ. checked
φ5=5π6\varphi_{5} = \frac{5 \pi}{6}
With k=5k = 5: 10π12\frac{10 \pi}{12} in lowest terms, which is 150∘150^\circ. checked
φ6=π\varphi_{6} = \pi
With k=6k = 6: 12π12\frac{12 \pi}{12} in lowest terms, which is 180∘180^\circ. checked
φ7=7π6\varphi_{7} = \frac{7 \pi}{6}
With k=7k = 7: 14π12\frac{14 \pi}{12} in lowest terms, which is 210∘210^\circ. checked
φ8=4π3\varphi_{8} = \frac{4 \pi}{3}
With k=8k = 8: 16π12\frac{16 \pi}{12} in lowest terms, which is 240∘240^\circ. checked
φ9=3π2\varphi_{9} = \frac{3 \pi}{2}
With k=9k = 9: 18π12\frac{18 \pi}{12} in lowest terms, which is 270∘270^\circ. checked
φ10=5π3\varphi_{10} = \frac{5 \pi}{3}
With k=10k = 10: 20π12\frac{20 \pi}{12} in lowest terms, which is 300∘300^\circ. checked
φ11=11π6\varphi_{11} = \frac{11 \pi}{6}
With k=11k = 11: 22π12\frac{22 \pi}{12} in lowest terms, which is 330∘330^\circ. checked
0.50.5i-0.5-0.5i1i-1-ireal axisimaginary axis03C9₀03C9₁03C9₂03C9₃03C9₄03C9₅03C9₆03C9₇03C9₈03C9₉03C9₁₀03C9₁₁
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