Differential-Equations · Unit 6 · Video 4 · Interactive Practice

Why One Has Five Fifth Roots: Zeta on the Unit Circle

IKey Formulas

FormulaNameWhat it says
∣zw∣=∣z∣ ∣w∣,arg⁡(zw)=arg⁡z+arg⁡w|zw| = |z|\,|w|, \quad \arg(zw) = \arg z + \arg wPolar multiplicationMultiply the moduli, add the angles
ζ=e2πi/n,α=2πn\zeta = e^{2\pi i/n}, \quad \alpha = \dfrac{2\pi}{n}Primitive nn-th rootModulus 11, angle one nn-th of a full turn
ζn=e2πi=e0=1\zeta^n = e^{2\pi i} = e^{0} = 1Check with no pictureAngles 2π2\pi and 00 are the same angle
zn=1  ⟺  z∈{1,ζ,ζ2,…,ζn−1}z^n = 1 \iff z \in \{1, \zeta, \zeta^2, \dots, \zeta^{n-1}\}All nn-th roots of 11(ζk)n=(ζn)k=1k=1(\zeta^k)^n = (\zeta^n)^k = 1^k = 1

Key Insight: Over the reals zn=1z^n = 1 has one root (nn odd) or two (nn even); over the complex numbers it always has exactly nn, the corners of a regular nn-gon on the unit circle.

IIVisualization 1 — Which Points Have z5=1z^5 = 1?

Raising z=reiθz = re^{i\theta} to the fifth gives modulus r5r^5 and angle 5θ5\theta, so z5=1z^5 = 1 forces r=1r = 1 and 5θ5\theta a full-turn multiple.

💡 Challenge: find all five points whose fifth power is exactly 11 — only four of them are new.

IIIVisualization 2 — Each Multiplication by ζ\zeta Adds 2π/n2\pi/n

Repeated multiplication by ζ=e2πi/n\zeta = e^{2\pi i/n} walks around the unit circle and closes the turn after exactly nn steps.

IVVisualization 3 — Where Does a High Power of ζ\zeta Land?

Because ζn=1\zeta^n = 1, only the remainder of kk on division by nn decides where ζk\zeta^k sits.

VQuiz Questions

Problem 1 · Spacing of the Eighth Roots

Given: the eight solutions of z8=1z^8 = 1, equally spaced on the unit circle — find the angle α\alpha between neighbouring roots.

✅ Correct! A full turn 2π2\pi shared among 88 equal gaps gives α=2π/8=π/4\alpha = 2\pi/8 = \pi/4.
❌ Close, but… a full turn is 2π2\pi radians, not π\pi: the gap is 2π/82\pi/8.
❌ Close, but… the root 11 is one of the eight points, so the circle is cut into 88 gaps, not 77.
❌ Not quite. π/2\pi/2 is the spacing of the fourth roots 1,i,−1,−i1, i, -1, -i; eight points are twice as dense.
Show solution

The eight roots are 1,ζ,ζ2,…,ζ71, \zeta, \zeta^2, \dots, \zeta^7 with ζ=e2πi/8\zeta = e^{2\pi i/8}. Consecutive roots differ by one factor of ζ\zeta, which adds its angle:

α=arg⁡ζ=2π8=π4.\alpha = \arg\zeta = \frac{2\pi}{8} = \frac{\pi}{4}.

Check: eight gaps of π/4\pi/4 make 8⋅π4=2π8 \cdot \tfrac{\pi}{4} = 2\pi, one full turn.

Problem 2 · A Power Beyond the Fifth

Given: ζ=e2πi/5\zeta = e^{2\pi i/5} — find ζ13\zeta^{13} as one of the five fifth roots of 11.

✅ Correct! ζ13=(ζ5)2ζ3=ζ3=e6πi/5\zeta^{13} = (\zeta^5)^2\zeta^3 = \zeta^3 = e^{6\pi i/5}.
❌ Not quite. Only exponents that are multiples of 55 give 11; 1313 leaves remainder 33.
❌ Close, but… the angle of ζ\zeta is 2π/52\pi/5, not π/5\pi/5: thirteen copies give 26π/526\pi/5.
❌ Close, but… 26π/5=4π+6π/526\pi/5 = 4\pi + 6\pi/5: subtracting whole turns leaves +6π/5+6\pi/5, not −6π/5-6\pi/5.
❌ Not quite. Write 13=2⋅5+313 = 2\cdot 5 + 3 and use ζ5=1\zeta^5 = 1.
Show solution

Split the exponent by whole turns: 13=2⋅5+313 = 2 \cdot 5 + 3, so

ζ13=(ζ5)2⋅ζ3=12⋅ζ3=ζ3=e6πi/5.\zeta^{13} = (\zeta^5)^2 \cdot \zeta^3 = 1^2 \cdot \zeta^3 = \zeta^3 = e^{6\pi i/5}.

Equivalently, by angles: 13⋅2π5=26π5=4π+6π513 \cdot \tfrac{2\pi}{5} = \tfrac{26\pi}{5} = 4\pi + \tfrac{6\pi}{5}, and the two full turns 4π4\pi change nothing.

Problem 3 · The Sixth Roots of One

Given: ζ=e2πi/6\zeta = e^{2\pi i/6}, so the sixth roots of 11 are 1,ζ,…,ζ51, \zeta, \dots, \zeta^5 — find which power equals −1-1, and write ζ\zeta in the form a+bia + bi.

Which power of ζ equals −1?

What is ζ in the form a + bi?

✅ Correct! ζ=eiπ/3=12+32i\zeta = e^{i\pi/3} = \tfrac12 + \tfrac{\sqrt3}{2}i, and ζ3=eiπ=−1\zeta^3 = e^{i\pi} = -1: since 66 is even, −1-1 is a root.
❌ Check the power. −1-1 sits at angle π\pi; each factor of ζ\zeta adds 2π/6=π/32\pi/6 = \pi/3.
❌ Not quite. (−1)6=1(-1)^6 = 1: for even nn, −1-1 is always one of the nn-th roots.
❌ Close, but… that is the point at angle π/6\pi/6; ζ\zeta is at π/3\pi/3, where cos⁡=12\cos = \tfrac12 and sin⁡=32\sin = \tfrac{\sqrt3}{2}.
❌ Check ζ. Use eiθ=cos⁡θ+isin⁡θe^{i\theta} = \cos\theta + i\sin\theta with θ=2π/6\theta = 2\pi/6.
Show solution

Step 1: the power. arg⁡ζk=k⋅π3\arg\zeta^k = k \cdot \tfrac{\pi}{3}. We need angle π\pi, so k=3k = 3:

ζ3=e3⋅2πi/6=eπi=−1.\zeta^3 = e^{3 \cdot 2\pi i/6} = e^{\pi i} = -1.

Step 2: Cartesian form. With θ=2π/6=π/3\theta = 2\pi/6 = \pi/3:

ζ=cos⁡π3+isin⁡π3=12+32 i.\zeta = \cos\tfrac{\pi}{3} + i\sin\tfrac{\pi}{3} = \tfrac{1}{2} + \tfrac{\sqrt{3}}{2}\,i.

In the second group, option c is ζ2=e2πi/3\zeta^2 = e^{2\pi i/3}, the next root along; option d is the eighth root eiπ/4e^{i\pi/4}.

Problem 4 · Adding All Five Roots

Given: ζ=e2πi/5\zeta = e^{2\pi i/5} and S=1+ζ+ζ2+ζ3+ζ4S = 1 + \zeta + \zeta^2 + \zeta^3 + \zeta^4 — find SS. (Compare ζS\zeta S with SS.)

✅ Correct! ζS=S\zeta S = S with ζ≠1\zeta \ne 1 forces S=0S = 0: the pentagon's corners balance at the centre.
❌ Not quite. 55 is the sum of the moduli; complex numbers add as vectors, and these five point in different directions.
❌ Not quite. 11 is the value of each root raised to the fifth, not of their sum.
❌ Not quite. −1-1 is ζ+ζ2+ζ3+ζ4\zeta + \zeta^2 + \zeta^3 + \zeta^4 alone; the full sum also includes the root 11.
Show solution

Multiply by ζ\zeta; every term moves to the next root and ζ5=1\zeta^5 = 1 wraps around:

ζS=ζ+ζ2+ζ3+ζ4+ζ5=ζ+ζ2+ζ3+ζ4+1=S.\zeta S = \zeta + \zeta^2 + \zeta^3 + \zeta^4 + \zeta^5 = \zeta + \zeta^2 + \zeta^3 + \zeta^4 + 1 = S.

So (ζ−1)S=0(\zeta - 1)S = 0. Since ζ≠1\zeta \ne 1, S=0S = 0.

Geometrically, rotating the regular pentagon by 2π/52\pi/5 maps it onto itself, so its vector sum is unchanged by the rotation — the only such vector is 00.

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