Femtobytes
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.

z8=1z^{8} = 1
Invent a new one:
The problem

Find every complex number zz that satisfies this equation.

z8=1z^{8} = 1
The result
ωk=cos⁡2kπ8+isin⁡2kπ8,k=0,1,…,7\omega_k = \cos \frac{2k\pi}{8} + i \sin \frac{2k\pi}{8}, \quad k = 0, 1, \ldots, 7

There are exactly 88 solutions, equally spaced round the unit circle, and their sum is 00.

checked 16 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, z8=1z^{8} = 1 has two solutions, 11 and −1-1. Yet an equation of degree 88 is entitled to 88 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 88 raises the length to that power and multiplies the angle by 88: z8=ρ8(cos⁡8φ+isin⁡8φ)z^{8} = \rho^{8} \left( \cos 8\varphi + i \sin 8\varphi \right).

ρ8=1and8φ=2kπ\rho^{8} = 1 \qquad \text{and} \qquad 8\varphi = 2k\pi
Equal complex numbers have equal lengths, and angles that agree up to whole turns.
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 88 different solutions come from: a full turn, shared out 88 ways, gives 88 different directions before it starts repeating.

ρ=1,φk=2kπ8,k=0,1,…,7\rho = 1, \qquad \varphi_k = \frac{2k\pi}{8}, \quad k = 0, 1, \ldots, 7
Dividing 2kπ2k\pi by 88. Taking k=8k = 8 would give 2π2\pi, a full turn, which is the same direction as k=0k = 0; so kk stops at 77.
φ0=0\varphi_{0} = 0
With k=0k = 0. checked
φ1=π4\varphi_{1} = \frac{\pi}{4}
With k=1k = 1: 2π8\frac{2 \pi}{8} in lowest terms, which is 45∘45^\circ. checked
φ2=π2\varphi_{2} = \frac{\pi}{2}
With k=2k = 2: 4π8\frac{4 \pi}{8} in lowest terms, which is 90∘90^\circ. checked
φ3=3π4\varphi_{3} = \frac{3 \pi}{4}
With k=3k = 3: 6π8\frac{6 \pi}{8} in lowest terms, which is 135∘135^\circ. checked
φ4=π\varphi_{4} = \pi
With k=4k = 4: 8π8\frac{8 \pi}{8} in lowest terms, which is 180∘180^\circ. checked
φ5=5π4\varphi_{5} = \frac{5 \pi}{4}
With k=5k = 5: 10π8\frac{10 \pi}{8} in lowest terms, which is 225∘225^\circ. checked
φ6=3π2\varphi_{6} = \frac{3 \pi}{2}
With k=6k = 6: 12π8\frac{12 \pi}{8} in lowest terms, which is 270∘270^\circ. checked
φ7=7π4\varphi_{7} = \frac{7 \pi}{4}
With k=7k = 7: 14π8\frac{14 \pi}{8} in lowest terms, which is 315∘315^\circ. checked
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