4.02s - Using roots of unity

Roots of unity as repeated rotations

Why is a diagonal of a regular pentagon $(1+\sqrt5)/2$ times as long as a side? A pentagon does not look like a quadratic equation, but its vertices supply one.

Begin with the fifth roots of 1. One is 1 itself. The others are reached by repeatedly turning through $2\pi/5$ around the unit circle:

Their fifth powers are all 1, because their arguments become integer multiples of $2\pi$. Joining them in order gives a regular pentagon with centre $O$.

For roots of 1, the common modulus is 1 and the first angle is 0. It is useful to name the first anticlockwise step $\omega$.

The exponential list is supplied in the booklet.

With $\omega=e^{2\pi i/n}$, multiplication by $\omega$ moves one vertex anticlockwise to the next. Also $\omega^n=1$, so reduce powers modulo $n$: for fifth roots, $\omega^8=\omega^3$. The inverse and conjugate coincide, since $|\omega|=1$:

This relation is not separately printed in the formula booklet; it follows from $\omega\omega^*=|\omega|^2=1$. More generally, $(\omega^k)^*=\omega^{n-k}$, with exponents understood modulo $n$.

[DIAGRAM: asset_name: Fifth roots of unity form a pentagon; asset_slug: h245_4_02s_pentagon_unity; caption: Multiplication by omega moves one vertex anticlockwise around the regular pentagon; description: Equal-scale Argand axes labelled Re and Im with origin O and a unit circle. The five blue vertices are labelled 1, omega, omega squared, omega cubed and omega to the fourth. Sage dashed lines pair omega with omega to the fourth and omega squared with omega cubed across the real axis. One clay side from 1 to omega and one blue diagonal from 1 to omega squared are distinguished, and an arc marks the central angle 2pi/5]

Choose the step carefully: $e^{4\pi i/6}$ is a sixth root of unity, but its powers reach only three of the six vertices. The choice $e^{2\pi i/n}$ always reaches every vertex.

Reflection preserves a vertex's real part and reverses its imaginary part; it does not negate the whole vertex.

Sums of roots and their real parts

The complete set of roots of unity sums to zero when $n\ge2$, so the real coordinates sum to zero and the imaginary coordinates sum to zero.

Let $S=1+\omega+\cdots+\omega^{n-1}$, where $\omega=e^{2\pi i/n}\ne1$. This is a GP:

The denominator is non-zero because $n\ge2$. Geometrically, the vertex position vectors balance around $O$. Taking real and imaginary parts is legitimate because addition adds the coordinates separately.

The exception $n=1$ has only the root 1, whose sum is 1. Also, a partial list does not generally sum to zero: for example, $1+i\ne0$ although both terms are fourth roots of unity.

Worked example 2 changes only the number of vertices, from five to seven. There are now three conjugate pairs.

These exact cosine sums are consequences of the root sum, not separate formula-booklet entries. For even $n$, remember the real root $-1$ as well as 1 when pairing the remaining roots.

With nine vertices there are four conjugate pairs and the same single unpaired root 1.

Sides and diagonals as complex differences

The distance between vertices represented by $u$ and $v$ is $|v-u|$, so subtracting roots of unity measures sides and diagonals directly.

For the unit-circle points 1 and $e^{i\alpha}$, centre the difference:

Taking moduli gives $|e^{i\alpha}-1|=2|\sin(\alpha/2)|$, since $|i|=|e^{i\alpha/2}|=1$. This is the same chord length you get by splitting an isosceles triangle into two right-angled triangles.

For any pair of vertices $\omega^j$, $\omega^{j+k}$, factor out $\omega^j$. Its modulus is 1, so the distance is still $|1-\omega^k|$. The chord depends on the number of steps between vertices, not the starting vertex.

Worked example 4 changes only the step between vertices, from one step to two. The chord is now a diagonal.

The ratio is already simple:

This follows from the double-angle sine identity. It is not a separate formula-booklet result; the next part finds its exact value.

The same difference method works with another regular polygon; only the angular step changes.

Exact values from the pentagon

The fifth-root sum links two cosines; a double-angle identity turns that link into a quadratic equation.

From the second part,

Either cosine can be used as the unknown. The two values are the two roots of the same quadratic, and the diagram chooses the correct sign.

Worked example 6 changes only the chosen vertex from $\omega$ to $\omega^2$. The quadratic is unchanged, but the sign selection reverses.

These exact values are derived results, not printed in the formula booklet. To finish the opening question, $\pi/5$ is acute and

Therefore the pentagon's diagonal-to-side ratio is $(1+\sqrt5)/2$, the golden ratio. Scaling or rotating the pentagon does not change that ratio.

For exact lengths on a larger circle, squared distances avoid nested square roots: $|R-R\omega^k|^2=R^2(2-\omega^k-\omega^{-k})=2R^2(1-\cos(2k\pi/5))$. This is the squared chord formula, derived here rather than supplied separately in the booklet.

Quick Recap

• The nth roots of unity are $1,\omega,\ldots,\omega^{n-1}$, with $\omega=e^{2\pi i/n}$; they form a regular polygon when $n\ge3$.
• For $n\ge2$, all roots sum to zero. Taking real and imaginary parts gives trigonometric sums.
• Conjugate vertices have equal real parts; use their symmetry to pair cosine terms.
• Sides and diagonals are moduli of differences, and their lengths depend on the vertex step.
• A root sum combined with a multiple-angle identity can determine exact trigonometric values.

The opening pentagon has diagonal-to-side ratio $2\cos(\pi/5)=(1+\sqrt5)/2$, obtained from the geometry and algebra of its five roots.