Factorising \(z^{2n} + 1\)

The hook

What is \(\cos\frac{\pi}{8}\cos\frac{3\pi}{8}\cos\frac{5\pi}{8}\cos\frac{7\pi}{8}\)? Each factor is an ugly number involving nested square roots, but the product is exactly \(\frac18\). Products of cosines at equally spaced angles keep collapsing to simple numbers like this, and the question explains why. The trick is to recognise the angles as arguments of roots of a polynomial, factorise the polynomial, and then substitute a cleverly chosen value.

Over the complex numbers, every polynomial factorises completely into linear factors. Over the real numbers, the non-real roots come in conjugate pairs, which combine into real quadratic factors. Both facts are on display here, and the second is how \(z^{2n} + 1\), which has no real roots at all, can be written as a product of real quadratics.

Before you start

They all have modulus \(1\), and their arguments \(\theta\) satisfy \(2n\theta = \pi + 2m\pi\), so \(\theta\) is an odd multiple of \(\frac{\pi}{2n}\). They form a regular \(2n\)-gon on the unit circle, rotated so that no vertex lies on the real axis (Figure 5.3.1). They pair up into conjugates \(e^{\pm\mathrm{i}\theta}\), mirror images in the real axis.

Working it through

Stage 1 Conjugate pairs give real quadratics

\((z - e^{\mathrm{i}\theta})(z - e^{-\mathrm{i}\theta}) = z^2 - (e^{\mathrm{i}\theta} + e^{-\mathrm{i}\theta})z + 1 = z^2 - 2z\cos\theta + 1\).■

Stage 2 The factorisation

The \(2n\)th roots of \(-1\) with \(-\pi < \theta \leq \pi\) are \(e^{\pm\mathrm{i}\theta_k}\) with \[\theta_k = \frac{(2k - 1)\pi}{2n}, \qquad k = 1, 2, \ldots, n.\] These are \(2n\) distinct numbers, all roots of the degree-\(2n\) polynomial \(z^{2n} + 1\), so they are all its roots and \[z^{2n} + 1 = \prod_{k=1}^{n}(z - e^{\mathrm{i}\theta_k})(z - e^{-\mathrm{i}\theta_k}) = \prod_{k=1}^{n}\bigl(z^2 - 2z\cos\theta_k + 1\bigr). \qed\]

Stage 3 Substituting \(z = \mathrm{i}\)

Each factor becomes \[\mathrm{i}^2 - 2\mathrm{i}\cos\theta_k + 1 = -2\mathrm{i}\cos\theta_k,\] so the right-hand side is \((-2\mathrm{i})^n\prod_k\cos\theta_k\). The left-hand side is \(\mathrm{i}^{2n} + 1 = (-1)^n + 1\), which is \(2\) when \(n\) is even. And for even \(n\), \[(-2\mathrm{i})^n = 2^n(-\mathrm{i})^n = 2^n\bigl((-\mathrm{i})^2\bigr)^{n/2} = 2^n(-1)^{n/2}.\] So \[\prod_{k=1}^{n}\cos\theta_k = \frac{2}{2^n(-1)^{n/2}} = (-1)^{\frac12 n}2^{1-n}. \qed\] For \(n = 4\) this is the product in the hook: \((-1)^2 2^{-3} = \frac18\).

Stage 4 When \(n\) is odd

When \(k = \frac{n+1}{2}\), \(\theta_k = \frac{n\pi}{2n} = \frac\pi2\), so that factor is \(z^2 + 1\), which vanishes at \(z = \mathrm{i}\); both sides are \(0\), and we learn nothing. The way round is to divide out that factor first, which is what the given identity does.

Dividing by the factor \(z^2 + 1\) (the term \(k = \frac{n+1}{2}\)), \[1 - z^2 + z^4 - \cdots + z^{2n-2} = \prod_{k \neq \frac{n+1}{2}}\bigl(z^2 - 2z\cos\theta_k + 1\bigr).\] At \(z = \mathrm{i}\), every term on the left is \((-z^2)^j = 1^j = 1\), and there are \(n\) terms, so the left-hand side is \(n\). The right-hand side is \((-2\mathrm{i})^{n-1}\prod_{k \neq \frac{n+1}{2}}\cos\theta_k\). Since \(n - 1\) is even, \((-2\mathrm{i})^{n-1} = 2^{n-1}(-1)^{(n-1)/2}\).

Now pair the angles: \(\theta_{n+1-k} = \pi - \theta_k\), so \(\cos\theta_{n+1-k} = -\cos\theta_k\). The \(n - 1\) remaining factors form \(\frac{n-1}{2}\) such pairs, with \(k = 1, \ldots, \frac{n-1}{2}\), and each pair contributes \(-\cos^2\theta_k\). So \[n = 2^{n-1}(-1)^{(n-1)/2}\cdot(-1)^{(n-1)/2}\prod_{k=1}^{(n-1)/2}\cos^2\theta_k = 2^{n-1}\prod_{k=1}^{(n-1)/2}\cos^2\theta_k,\] giving \(\prod\cos^2\theta_k = n2^{1-n}\). The last factor, with \(k = \frac{n-1}{2}\), is \(\cos^2\frac{(n-2)\pi}{2n}\), as in the question.■

Check with \(n = 3\): \(\cos^2\frac\pi6 = \frac34 = 3 \cdot 2^{-2}\).

Looking back

• Know the roots, know the polynomial. A polynomial is determined (up to a constant) by its roots, so listing the roots of \(z^{2n} + 1\) gave its factorisation.

• Substitute a clever value. Putting \(z = \mathrm{i}\) turned every quadratic factor into a multiple of a cosine.

• When a substitution gives \(0 = 0\), divide first. For odd \(n\) one factor vanished at \(z = \mathrm{i}\), so we removed it before substituting.

What’s really going on

Products of sines and cosines

The method gives a whole family of identities. Using \(z^n - 1\) in place of \(z^{2n} + 1\) leads to \[\prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{n}{2^{n-1}},\] which is proved, in its geometric form, in Section 5.4. Identities like these appear in the analysis of vibrating systems (the frequencies of \(n\) masses joined by springs involve \(\sin\frac{k\pi}{2(n+1)}\)) and in counting problems (the number of spanning trees of some graphs, and the number of domino tilings of a rectangle, are given by products of cosines). Kasteleyn’s 1961 formula for domino tilings of an \(m \times n\) rectangle is a double product of expressions in \(\cos^2\frac{j\pi}{m+1}\) and \(\cos^2\frac{k\pi}{n+1}\).

The thinking skill

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Further reading

• Tristan Needham, Visual Complex Analysis (Oxford University Press). Chapter 2 on roots of unity and polynomials, with beautiful pictures.

• Benjamin Fine & Gerhard Rosenberger, The Fundamental Theorem of Algebra (Springer). Several proofs of the theorem, from analysis, algebra and topology, written for undergraduates.

• Paul Nahin, An Imaginary Tale (Princeton University Press). A history of \(\sqrt{-1}\), including roots of unity and their uses.