Summing a Series with De Moivre

Two sums with no common ratio

Take $C = 1 + \cos \theta + \cos 2\theta + \dots + \cos n\theta$ and $S = \sin \theta + \sin 2\theta + \dots + \sin n\theta$. The angle goes up by $\theta$ each term, but neither sum is geometric: to be geometric, each term would have to be the one before it times the same number.

At $\theta = 30^\circ$, $\cos 60^\circ \div \cos 30^\circ \approx 0.577$, but cos 90° ÷ cos 60° = 0. The ratio changes from one term to the next, so the formula for a geometric sum cannot be used on C, or on S, alone.

Pair them into one complex sum

Add C to i times S, term by term. The terms pair off: C + iS $= 1 + \left(\cos \theta + i \sin \theta\right) + \left(\cos 2\theta + i \sin 2\theta\right) + \dots + \left(\cos n\theta + i \sin n\theta\right)$.

Let $z = \cos \theta + i \sin \theta$. By De Moivre’s theorem, $z^{k} = \cos k\theta + i \sin k\theta$ for every whole number k, so each bracket is a power of z, and C + iS $= 1 + z + z^{2} + \dots + z^{n}$. That sum is geometric: each term is the one before it times z.

The terms 1, z, $z^{2}$ and $z^{3}$ at $\theta = 30^\circ$, laid head to tail. Each has length 1 and turns 30° from the one before. The gold arrow is their sum, C + iS: its real part is $C = 1 + \cos 30^\circ + \cos 60^\circ + \cos 90^\circ \approx 2.366$, and its imaginary part is $S \approx 2.366$.

Sum it as a geometric series

A geometric series with first term 1 and ratio z has the sum $1 + z + z^{2} + \dots + z^{n} = \frac{{1 - z^{n+1}}}{{1 - z}}$, as long as $z \ne 1$. There are n + 1 terms, from $z^{0}$ up to $z^{n}$, so the power on top is n + 1.

So C + iS $= \frac{{1 - z^{n+1}}}{{1 - z}}$, and $z^{n+1} = \cos\left(\left(n + 1\right)\theta\right) + i \sin\left(\left(n + 1\right)\theta\right)$ by De Moivre again.

Split the quotient into its parts

C is the real part of the quotient and S is its imaginary part. To read them off, write the top and the bottom in polar form.

Use the double-angle formulas $1 - \cos \theta = 2\sin^{2}\left(\frac{\theta}{2}\right)$ and $\sin \theta = 2\sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)$. Then $1 - z = 2\sin^{2}\left(\frac{\theta}{2}\right) - 2i \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) = -2i \sin\left(\frac{\theta}{2}\right) \left[\cos\left(\frac{\theta}{2}\right) + i \sin\left(\frac{\theta}{2}\right)\right]$. Multiply out the right side to check: $-2i \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) - 2i^{2} \sin^{2}\left(\frac{\theta}{2}\right) = 2\sin^{2}\left(\frac{\theta}{2}\right) - 2i \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)$.

The top works the same way with $\left(n + 1\right)\theta$ in place of $\theta$: $1 - z^{n+1} = -2i \sin\left(\frac{\left(n + 1\right)\theta}{2}\right) \left[\cos\left(\frac{\left(n + 1\right)\theta}{2}\right) + i \sin\left(\frac{\left(n + 1\right)\theta}{2}\right)\right]$.

Divide. The −2i cancels, and dividing the two brackets subtracts their angles: $\frac{\left(n + 1\right)\theta}{2} - \frac{\theta}{2} = \frac{n\theta}{2}$. So C + iS $= \left[\sin\left(\frac{\left(n + 1\right)\theta}{2}\right) / \sin\left(\frac{\theta}{2}\right)\right] \times \left[\cos\left(\frac{n\theta}{2}\right) + i \sin\left(\frac{n\theta}{2}\right)\right]$.

The first factor is real, so the parts can be read off: $C = \sin\left(\frac{\left(n + 1\right)\theta}{2}\right) \cos\left(\frac{n\theta}{2}\right) / \sin\left(\frac{\theta}{2}\right)$ and $S = \sin\left(\frac{\left(n + 1\right)\theta}{2}\right) \sin\left(\frac{n\theta}{2}\right) / \sin\left(\frac{\theta}{2}\right)$.

Checks

At $\theta = 60^\circ$ with n = 2: $C = 1 + \cos 60^\circ + \cos 120^\circ = 1 + \frac{1}{2} - \frac{1}{2} = 1$. The formula gives $\sin 90^\circ \cos 60^\circ \div \sin 30^\circ = 1 \times \frac{1}{2} \div \frac{1}{2} = 1$. And $S = \sin 60^\circ + \sin 120^\circ = \sqrt{3}$, while the formula gives $\sin 90^\circ \sin 60^\circ \div \sin 30^\circ = \left(\frac{\sqrt{3}}{2}\right) / \left(\frac{1}{2}\right) = \sqrt{3}$.

At $\theta = 30^\circ$ with n = 3, the sum drawn above: $C = \sin 60^\circ \cos 45^\circ \div \sin 15^\circ \approx 0.866 \times 0.707 / 0.259 \approx 2.366$, and term by term $C = 1 + \frac{\sqrt{3}}{2} + \frac{1}{2} + 0 = \frac{{3 + \sqrt{3}}}{2} \approx 2.366$.

At $\theta = 90^\circ$ every term cancels

At $\theta = 90^\circ$, z = cos 90° + i sin 90° = i. With n = 3 the sum is $1 + i + i^{2} + i^{3} = 1 + i - 1 - i = 0$, and the formula agrees: $\frac{{1 - i^{4}}}{{1 - i}} = \frac{{1 - 1}}{{1 - i}} = 0$.

The real parts give C = 1 + cos 90° + cos 180° + cos 270° = 1 + 0 − 1 + 0 = 0, and the imaginary parts give S = 0 + 1 + 0 − 1 = 0. In the closed form, $\sin\left(\frac{\left(n + 1\right)\theta}{2}\right) = \sin 180^\circ = 0$, so both C and S are 0.

At $\theta = 90^\circ$ the terms 1, i, $i^{2}$ and $i^{3}$, laid head to tail, go round a square and end where they started, at 0.

When z = 1

The geometric formula needs $z \ne 1$. At $\theta = 0^\circ$, or any whole number of turns, z = 1 and 1 − z = 0. Then every term is 1, so C = n + 1 and S = 0, found by counting rather than by the formula.

At $\theta = 120^\circ$ with n = 2, z is the cube root of 1 called $\omega$, and the sum is $1 + \omega + \omega^{2} = 0$. The closed form gives sin 180° = 0 on top, so C = 0 and S = 0, as The Roots of Unity found.

The usual mistakes

Building the series from $z = \cos \theta$ alone. Its powers are $\cos^{2}\theta$, $\cos^{3}\theta$ and so on, and $\cos^{2}\theta$ is not $\cos 2\theta$: at $\theta = 30^\circ$, $\cos^{2}30^\circ = 0.75$ but cos 60° = 0.5. Only the whole of $\cos \theta + i \sin \theta$ has $z^{k} = \cos k\theta + i \sin k\theta$.

Taking C from the imaginary part. C has no i beside it in C + iS, so C is the real part and S is the imaginary part.

Putting $z^{n}$ on top. The sum runs from $z^{0}$ to $z^{n}$, which is n + 1 terms, so the top is $1 - z^{n+1}$.

Forgetting the leading 1. At $\theta = 90^\circ$, 1 + cos 90° + cos 180° + cos 270° is 0, not −1: the first term is cos 0° = 1, and it cancels cos 180° = −1.