Home / A-Level Further Maths 9231 / Complex numbers / Summing series with de Moivre
Cambridge International AS & A Level · Topic 2.5
A-Level Further Maths 9231 / Topic 2.5 / AS & A Level

Sum a trig series as a geometric one.

A sum such as \(\cos\theta + \cos 2\theta + \dots + \cos n\theta\) looks hopeless term by term. Pair it with the matching sine series, write \(C + iS\) as a geometric series in \(z = e^{i\theta}\), sum it, and read off the real or imaginary part. The same idea handles binomial sums and infinite series.

Topic 2.5All complex numbers lessons →
The key idea

To sum \(C = \sum a_r\cos r\theta\), form the partner \(S = \sum a_r\sin r\theta\) and write \(C + iS = \sum a_r e^{ir\theta}\). With \(z = e^{i\theta}\) this is usually a geometric series (or, with binomial coefficients, the expansion of \((1 + z)^{n}\)). Sum it, then make the answer look like (real) \(\times\, e^{i\alpha}\) using the half-angle factors \(e^{i\phi} - 1 = 2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\) and \(1 + e^{i\phi} = 2\cos\tfrac{\phi}{2}\,e^{i\phi/2}\). Then \(C\) is the real part and \(S\) the imaginary part. Always test the result with a small \(n\).

Re Im O θ 1 z z² z³ z⁴ C + iS terms 1, z, …, z⁴ z = cos θ + i sin θ, θ = 40° their sum C + iS length sin(5θ/2) ÷ sin(θ/2) argument 2θ, half of 4θ one circle every vertex lies on it each step turns by θ, so C is the horizontal and S the vertical reach
Fig. 1: the five unit steps 1, z, z², z³, z⁴ (each turned θ = 40° from the last) placed head to tail. The arrow from O to the end is C + iS: its real part is cos 0 + cos θ + ... + cos 4θ and its imaginary part is the matching sine sum. Its length sin(5θ/2)/sin(θ/2) and argument 2θ are exactly what the half-angle method predicts
Section 01

C + iS as a geometric series.

By de Moivre, \(\cos r\theta + i\sin r\theta = e^{ir\theta} = z^{r}\) with \(z = e^{i\theta}\). So if \(C = \sum\cos r\theta\) and \(S = \sum\sin r\theta\) over the same values of \(r\), then \(C + iS = \sum z^{r}\), a geometric series with ratio \(z\). Sum it with \(\dfrac{a(z^{n} - 1)}{z - 1}\) (valid when \(z \ne 1\), that is \(\theta\) not a multiple of \(2\pi\)). The answer is a quotient of complex numbers, so you cannot yet see its real part. The trick is to take out a half angle from each bracket, which is the same as multiplying top and bottom by \(e^{-i\phi/2}\).

For any real \(\phi\):

\[\begin{aligned} e^{i\phi} - 1 &= e^{i\phi/2}\big(e^{i\phi/2} - e^{-i\phi/2}\big)\\ &= 2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}, \end{aligned}\]

and equally \(1 - e^{i\phi} = -2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\). Use it on the top (\(\phi = n\theta\)) and on the bottom (\(\phi = \theta\)): the factors \(2i\) cancel and what is left is a REAL number times one exponential, whose real and imaginary parts you can read straight off.

SumClosed form
\(\displaystyle\sum_{r=0}^{n-1}\cos r\theta\)\(\dfrac{\cos\frac{(n-1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\)
\(\displaystyle\sum_{r=0}^{n-1}\sin r\theta\)\(\dfrac{\sin\frac{(n-1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\)
\(\displaystyle\sum_{r=1}^{n}\cos r\theta\)\(\dfrac{\cos\frac{(n+1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\)

Do not learn the table: derive each sum in an exam. Then check it with a small \(n\) (one or two terms), which catches almost every slip in the limits or the half angles.

Worked example A finite cosine sum
Show that, for \(\theta\) not a multiple of \(2\pi\), \[\sum_{r=0}^{n-1}\cos r\theta = \frac{\cos\frac{(n-1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}.\]
  1. Let \(C\) be the sum and \(S = \sum_{r=0}^{n-1}\sin r\theta\). Then \(C + iS = \sum_{r=0}^{n-1}z^{r}\) with \(z = e^{i\theta}\): a GP with first term 1, ratio \(z\), \(n\) terms.
  2. \(z \ne 1\), so \(C + iS = \dfrac{z^{n} - 1}{z - 1} = \dfrac{e^{in\theta} - 1}{e^{i\theta} - 1}\).
  3. Take out half angles: \(e^{in\theta} - 1 = 2i\sin\tfrac{n\theta}{2}\,e^{in\theta/2}\) and \(e^{i\theta} - 1 = 2i\sin\tfrac{\theta}{2}\,e^{i\theta/2}\).
  4. \[\begin{aligned} C + iS &= \frac{2i\sin\frac{n\theta}{2}\,e^{in\theta/2}}{2i\sin\frac{\theta}{2}\,e^{i\theta/2}}\\ &= \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\,e^{i(n-1)\theta/2}. \end{aligned}\]
  5. The fraction is real, so the real part is \(\dfrac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\cos\dfrac{(n-1)\theta}{2}\).
  6. Check \(n = 2\): the sum is \(1 + \cos\theta\); the formula gives \(\dfrac{\cos\frac{\theta}{2}\sin\theta}{\sin\frac{\theta}{2}} = 2\cos^{2}\tfrac{\theta}{2} = 1 + \cos\theta\).
Answer: \(\displaystyle\sum_{r=0}^{n-1}\cos r\theta = \frac{\cos\frac{(n-1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\), as required
Worked example A sum of sin 2rθ
Find \(\displaystyle\sum_{r=1}^{n}\sin 2r\theta\) in closed form, where \(\sin\theta \ne 0\).
  1. Pair it with \(C = \sum_{r=1}^{n}\cos 2r\theta\). Then \(C + iS = \sum_{r=1}^{n}w^{r}\) with \(w = e^{2i\theta}\): first term \(w\), ratio \(w\), \(n\) terms.
  2. \(C + iS = \dfrac{w(1 - w^{n})}{1 - w} = \dfrac{e^{2i\theta}\big(1 - e^{2in\theta}\big)}{1 - e^{2i\theta}}\).
  3. Half angles, with \(1 - e^{i\phi} = -2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\): the top bracket is \(-2i\sin n\theta\,e^{in\theta}\), the bottom is \(-2i\sin\theta\,e^{i\theta}\).
  4. \[\begin{aligned} C + iS &= \frac{e^{2i\theta}\sin n\theta\,e^{in\theta}}{\sin\theta\,e^{i\theta}}\\ &= \frac{\sin n\theta}{\sin\theta}\,e^{i(n+1)\theta}. \end{aligned}\]
  5. Take the imaginary part: \(S = \dfrac{\sin n\theta\,\sin(n+1)\theta}{\sin\theta}\).
  6. Check \(n = 1\): \(\dfrac{\sin\theta\sin 2\theta}{\sin\theta} = \sin 2\theta\), the single term.
Answer: \(\displaystyle\sum_{r=1}^{n}\sin 2r\theta = \frac{\sin n\theta\,\sin(n+1)\theta}{\sin\theta}\)
Your turn
Use the result of the first example to write down \(\displaystyle\sum_{r=0}^{n-1}\sin r\theta\), and check it for \(n = 2\).
Examiner trap

The real part of a quotient is NOT the quotient of the real parts: \(\operatorname{Re}\dfrac{e^{in\theta} - 1}{e^{i\theta} - 1} \ne \dfrac{\cos n\theta - 1}{\cos\theta - 1}\). Make the denominator real first (half angles or the conjugate). And count the terms: \(r = 0\) to \(n - 1\) and \(r = 1\) to \(n\) give different answers, so test your formula with \(n = 1\) or \(n = 2\).

Section 02

Binomial sums from (1 + z)n.

When each term carries a binomial coefficient, \(C + iS = \sum_{r=0}^{n}\binom{n}{r}z^{r}\) is not a GP: it is the binomial expansion of \((1 + z)^{n}\). The half-angle idea now uses a plus sign: \(1 + e^{i\phi} = e^{i\phi/2}\big(e^{-i\phi/2} + e^{i\phi/2}\big) = 2\cos\tfrac{\phi}{2}\,e^{i\phi/2}\). Raise it to the power \(n\) with de Moivre and read off the parts.

With \(z = e^{i\theta}\):

\[\begin{aligned} (1 + z)^{n} &= \big(2\cos\tfrac{\theta}{2}\big)^{n}\big(e^{i\theta/2}\big)^{n}\\ &= 2^{n}\cos^{n}\tfrac{\theta}{2}\,e^{in\theta/2}, \end{aligned}\]
SumClosed form
\(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\cos r\theta\)\(2^{n}\cos^{n}\tfrac{\theta}{2}\,\cos\tfrac{n\theta}{2}\)
\(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\sin r\theta\)\(2^{n}\cos^{n}\tfrac{\theta}{2}\,\sin\tfrac{n\theta}{2}\)

\(\cos\tfrac{\theta}{2}\) may be negative; that does not matter, because \(2^{n}\cos^{n}\tfrac{\theta}{2}\) is still a real number multiplying \(e^{in\theta/2}\). If the angle in each term is \(2r\theta\) or \((2r + 1)\theta\), adjust \(z\) and pull out any common factor such as \(e^{i\theta}\) first.

Worked example A binomial sum with cos 2rθ
Show that \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\cos 2r\theta = 2^{n}\cos^{n}\theta\,\cos n\theta\), and hence evaluate \(\displaystyle\sum_{r=0}^{4}\binom{4}{r}\cos\frac{r\pi}{3}\).
  1. \(C + iS = \sum_{r=0}^{n}\binom{n}{r}\big(e^{2i\theta}\big)^{r} = \big(1 + e^{2i\theta}\big)^{n}\).
  2. \(1 + e^{2i\theta} = e^{i\theta}\big(e^{-i\theta} + e^{i\theta}\big) = 2\cos\theta\,e^{i\theta}\).
  3. So \(C + iS = 2^{n}\cos^{n}\theta\,e^{in\theta}\), and the real part is \(2^{n}\cos^{n}\theta\cos n\theta\).
  4. \(\cos\frac{r\pi}{3} = \cos 2r\theta\) with \(\theta = \frac{\pi}{6}\), \(n = 4\): \(16\big(\tfrac{\sqrt{3}}{2}\big)^{4}\cos\tfrac{2\pi}{3} = 16 \times \tfrac{9}{16} \times \big(-\tfrac{1}{2}\big) = -\tfrac{9}{2}\).
  5. Check term by term: \(1 + 4\big(\tfrac{1}{2}\big) + 6\big(-\tfrac{1}{2}\big) + 4(-1) + 1\big(-\tfrac{1}{2}\big) = 1 + 2 - 3 - 4 - \tfrac{1}{2} = -\tfrac{9}{2}\).
Answer: \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\cos 2r\theta = 2^{n}\cos^{n}\theta\cos n\theta\); the value is \(-\tfrac{9}{2}\)
Worked example Odd multiples, with a factor taken out
Find \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\sin(2r + 1)\theta\) in terms of \(n\) and \(\theta\).
  1. \(C + iS = \sum_{r=0}^{n}\binom{n}{r}e^{i(2r + 1)\theta} = e^{i\theta}\sum_{r=0}^{n}\binom{n}{r}\big(e^{2i\theta}\big)^{r}\).
  2. \(= e^{i\theta}\big(1 + e^{2i\theta}\big)^{n} = e^{i\theta}\big(2\cos\theta\,e^{i\theta}\big)^{n}\).
  3. \(= 2^{n}\cos^{n}\theta\,e^{i(n + 1)\theta}\).
  4. Imaginary part: \(2^{n}\cos^{n}\theta\,\sin(n + 1)\theta\).
  5. Check \(n = 1\): \(\sin\theta + \sin 3\theta = 2\sin 2\theta\cos\theta\) (sum to product), and the formula gives \(2\cos\theta\sin 2\theta\).
Answer: \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\sin(2r + 1)\theta = 2^{n}\cos^{n}\theta\,\sin(n + 1)\theta\)
Your turn
Show that \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\cos r\theta = 2^{n}\cos^{n}\tfrac{\theta}{2}\cos\tfrac{n\theta}{2}\), and use it to find \(\displaystyle\sum_{r=0}^{3}\binom{3}{r}\cos\frac{2r\pi}{3}\).
Examiner trap

Write \(1 + e^{i\theta} = 2\cos\tfrac{\theta}{2}\,e^{i\theta/2}\), with HALF angles in both places, not \(2\cos\theta\,e^{i\theta/2}\). Then raise BOTH factors to the power \(n\): forgetting the \(2^{n}\), or writing \(\cos\tfrac{n\theta}{2}\) where \(\cos^{n}\tfrac{\theta}{2}\) belongs, are the usual slips.

Stage 1 · Learn

Check your understanding

Four quick checks. Each correct answer earns XP and records this skill on your profile.

Quick check+10 XP

Let \(C = \sum_{r=0}^{n-1}\cos r\theta\) and \(S = \sum_{r=0}^{n-1}\sin r\theta\). Which expression equals \(C + iS\) (for \(e^{i\theta} \ne 1\))?

Quick check+10 XP

Which expression equals \(1 - e^{i\theta}\)?

Quick check+10 XP

Which expression equals \((1 + e^{i\theta})^{n}\)?

Quick check+10 XP

Find the sum of the infinite series \(\displaystyle\sum_{r=0}^{\infty}\big(\tfrac{1}{2}e^{i\theta}\big)^{r}\).

Section 03

Infinite series that converge.

An infinite geometric series \(\sum w^{r}\) converges only when \(|w| \lt 1\), and then \(\sum_{r=0}^{\infty}w^{r} = \dfrac{1}{1 - w}\) and \(\sum_{r=1}^{\infty}w^{r} = \dfrac{w}{1 - w}\). For \(\sum k^{r}\cos r\theta\) the ratio is \(w = ke^{i\theta}\) with \(|w| = |k|\), so state \(|k| \lt 1\) before you use the formula. To find the real part, multiply top and bottom by the conjugate of the denominator, using \(\big(a - e^{i\theta}\big)\big(a - e^{-i\theta}\big) = a^{2} - 2a\cos\theta + 1\).

Worked example Powers of one third
Show that \(\displaystyle\sum_{r=1}^{\infty}\frac{\sin r\theta}{3^{r}} = \frac{3\sin\theta}{10 - 6\cos\theta}\).
  1. \(C + iS = \sum_{r=1}^{\infty}w^{r}\) with \(w = \tfrac{1}{3}e^{i\theta}\); \(|w| = \tfrac{1}{3} \lt 1\), so the series converges.
  2. \(C + iS = \dfrac{w}{1 - w} = \dfrac{e^{i\theta}}{3 - e^{i\theta}}\) (multiply top and bottom by 3).
  3. Multiply top and bottom by \(3 - e^{-i\theta}\): \[\begin{aligned} C + iS &= \frac{3e^{i\theta} - 1}{9 - 3\big(e^{i\theta} + e^{-i\theta}\big) + 1}\\ &= \frac{3\cos\theta - 1 + 3i\sin\theta}{10 - 6\cos\theta}. \end{aligned}\]
  4. The denominator is real, so \(S = \dfrac{3\sin\theta}{10 - 6\cos\theta}\).
  5. Check \(\theta = \tfrac{\pi}{2}\): the series is \(\tfrac{1}{3} - \tfrac{1}{27} + \tfrac{1}{243} - \dots = \dfrac{1/3}{1 + 1/9} = \tfrac{3}{10}\), and the formula gives \(\tfrac{3}{10}\).
Answer: \(\displaystyle\sum_{r=1}^{\infty}\frac{\sin r\theta}{3^{r}} = \frac{3\sin\theta}{10 - 6\cos\theta}\) (and the cosine series is \(\dfrac{3\cos\theta - 1}{10 - 6\cos\theta}\))
Worked example A ratio that depends on θ
For \(0 \lt \theta \lt \pi\), find \(\displaystyle\sum_{r=0}^{\infty}\cos^{r}\theta\,\cos r\theta\) and \(\displaystyle\sum_{r=0}^{\infty}\cos^{r}\theta\,\sin r\theta\).
  1. \(C + iS = \sum_{r=0}^{\infty}w^{r}\) with \(w = \cos\theta\,e^{i\theta}\). Since \(0 \lt \theta \lt \pi\), \(|w| = |\cos\theta| \lt 1\): it converges.
  2. \(1 - w = 1 - \cos^{2}\theta - i\sin\theta\cos\theta = \sin^{2}\theta - i\sin\theta\cos\theta\).
  3. Factor: \(1 - w = -i\sin\theta\,(\cos\theta + i\sin\theta) = -i\sin\theta\,e^{i\theta}\).
  4. \[\begin{aligned} C + iS &= \frac{1}{-i\sin\theta\,e^{i\theta}} = \frac{i\,e^{-i\theta}}{\sin\theta}\\ &= \frac{\sin\theta + i\cos\theta}{\sin\theta} = 1 + i\cot\theta. \end{aligned}\]
  5. So \(C = 1\) and \(S = \cot\theta\). Check \(\theta = \tfrac{\pi}{3}\) against the next exercise's formula: \(\sum\big(\tfrac{1}{2}\big)^{r}\cos\tfrac{r\pi}{3} = \dfrac{4 - 1}{5 - 2} = 1\).
Answer: \(\displaystyle\sum_{r=0}^{\infty}\cos^{r}\theta\cos r\theta = 1\) and \(\displaystyle\sum_{r=0}^{\infty}\cos^{r}\theta\sin r\theta = \cot\theta\)
Your turn
Show that \(\displaystyle\sum_{r=0}^{\infty}\frac{\cos r\theta}{2^{r}} = \frac{4 - 2\cos\theta}{5 - 4\cos\theta}\) and write down the matching sine series.
Examiner trap

Check convergence first: \(|w| \lt 1\) must be stated, and a ratio such as \(\cos\theta\,e^{i\theta}\) fails when \(\theta\) is a multiple of \(\pi\). Then watch the first term: a sum from \(r = 1\) is \(\dfrac{w}{1 - w}\), not \(\dfrac{1}{1 - w}\). Finally, \(|3 - e^{i\theta}|^{2} = 10 - 6\cos\theta\), not \(9 - \cos^{2}\theta\) or \(10 + 6\cos\theta\).

Stage 2 · Exam

Exam-style questions

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Exam style+20 XP

Use \(C + iS\) to find \(\displaystyle\sum_{r=1}^{n}\sin 2r\theta\), where \(\sin\theta \ne 0\).

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Evaluate \(\displaystyle\sum_{r=0}^{6}\binom{6}{r}\cos\frac{r\pi}{3}\).

Exam style+20 XP

Find \(\displaystyle\sum_{r=0}^{\infty}\frac{\cos r\theta}{3^{r}}\).

Exam style+20 XP

For an integer \(n \ge 2\), find \(\displaystyle\sum_{r=0}^{n-1}\cos\frac{2\pi r}{n}\).

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Series paired, summed as a GP, and checked with a small n.

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Six original Cambridge-style questions
The half-angle factor proved, finite sums of \(\sin(2r - 1)\theta\) and \(\cos r\theta\), a binomial sine sum, an infinite series in powers of \(\tfrac{2}{3}\), and \(\sum\cos^{2}\tfrac{r\pi}{n} = \tfrac{n}{2}\) by the \(C + iS\) method.