AS & A Level · Practice questions · Summing series with de Moivre

Pair it, sum it, take a part.

Six original Cambridge-style questions on summing series with de Moivre's theorem: finite trigonometric sums as geometric series in \(e^{i\theta}\), binomial sums from \((1 + z)^{n}\), a convergent infinite series and two exact evaluations, with full model solutions.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

Pair, sum, simplify, take a part.

Pair the series with its partner and write \(C + iS = \sum a_r e^{ir\theta}\).
Sum it as a GP in \(z = e^{i\theta}\) (count the terms and the first term), or as \((1 + z)^{n}\) if there are binomial coefficients.
Simplify: \(e^{i\phi} - 1 = 2i\sin\tfrac{\phi}{2}\,e^{i\phi/2}\), \(1 + e^{i\phi} = 2\cos\tfrac{\phi}{2}\,e^{i\phi/2}\), or multiply by the conjugate of the denominator.
Take a part: real for \(C\), imaginary for \(S\), only once the answer is (real) \(\times\, e^{i\alpha}\) or has a real denominator.
Check: \(|w| \lt 1\) for an infinite series, and test the result with \(n = 1\) or \(n = 2\).

Recall

[3 marks]

Show that \(e^{i\theta} - 1 = 2i\sin\tfrac{\theta}{2}\,e^{i\theta/2}\), and deduce that \(\big|e^{i\theta} - 1\big| = 2\big|\sin\tfrac{\theta}{2}\big|\).

\(e^{i\theta} - 1 = e^{i\theta/2}\big(e^{i\theta/2} - e^{-i\theta/2}\big)\). ✓
\(e^{i\theta/2} - e^{-i\theta/2} = \big(\cos\tfrac{\theta}{2} + i\sin\tfrac{\theta}{2}\big) - \big(\cos\tfrac{\theta}{2} - i\sin\tfrac{\theta}{2}\big) = 2i\sin\tfrac{\theta}{2}\), so \(e^{i\theta} - 1 = 2i\sin\tfrac{\theta}{2}\,e^{i\theta/2}\). ✓
\(|2i| = 2\) and \(\big|e^{i\theta/2}\big| = 1\), so \(\big|e^{i\theta} - 1\big| = 2\big|\sin\tfrac{\theta}{2}\big|\). ✓

Calculation

[5 marks]

By considering \(\displaystyle\sum_{r=1}^{n}e^{i(2r - 1)\theta}\), show that, for \(\sin\theta \ne 0\), \[\sum_{r=1}^{n}\sin(2r - 1)\theta = \frac{\sin^{2}n\theta}{\sin\theta},\] and find \(\displaystyle\sum_{r=1}^{n}\cos(2r - 1)\theta\) in its simplest form.

\(C + iS = \sum_{r=1}^{n}e^{i(2r - 1)\theta} = e^{i\theta}\sum_{r=0}^{n-1}\big(e^{2i\theta}\big)^{r}\), a GP with ratio \(e^{2i\theta} \ne 1\). ✓
\[\begin{aligned} C + iS &= \frac{e^{i\theta}\big(e^{2in\theta} - 1\big)}{e^{2i\theta} - 1}\\ &= \frac{e^{i\theta}\cdot 2i\sin n\theta\,e^{in\theta}}{2i\sin\theta\,e^{i\theta}}\\ &= \frac{\sin n\theta}{\sin\theta}\,e^{in\theta}. \end{aligned}\] ✓
Imaginary part: \(S = \dfrac{\sin n\theta\,\sin n\theta}{\sin\theta} = \dfrac{\sin^{2}n\theta}{\sin\theta}\), as required. ✓
Real part: \(C = \dfrac{\sin n\theta\cos n\theta}{\sin\theta} = \dfrac{\sin 2n\theta}{2\sin\theta}\). ✓
Check \(n = 1\): \(\dfrac{\sin^{2}\theta}{\sin\theta} = \sin\theta\) and \(\dfrac{\sin 2\theta}{2\sin\theta} = \cos\theta\). ✓

Calculation

[5 marks]

(a) Show that \(\displaystyle\sum_{r=0}^{\infty}\big(\tfrac{2}{3}\big)^{r}\cos r\theta = \frac{9 - 6\cos\theta}{13 - 12\cos\theta}\).

(b) Hence find the exact value of \(\displaystyle\sum_{r=0}^{\infty}\big(\tfrac{2}{3}\big)^{r}\cos\frac{r\pi}{3}\).

(a) \(C + iS = \sum_{r=0}^{\infty}w^{r}\) with \(w = \tfrac{2}{3}e^{i\theta}\); \(|w| = \tfrac{2}{3} \lt 1\), so it converges to \(\dfrac{1}{1 - w} = \dfrac{3}{3 - 2e^{i\theta}}\). ✓
Multiply top and bottom by \(3 - 2e^{-i\theta}\): \[\begin{aligned} C + iS &= \frac{3\big(3 - 2e^{-i\theta}\big)}{9 - 6\big(e^{i\theta} + e^{-i\theta}\big) + 4}\\ &= \frac{9 - 6\cos\theta + 6i\sin\theta}{13 - 12\cos\theta}. \end{aligned}\] ✓
The denominator is real, so \(C = \dfrac{9 - 6\cos\theta}{13 - 12\cos\theta}\), as required. ✓
(b) \(\theta = \tfrac{\pi}{3}\), \(\cos\theta = \tfrac{1}{2}\): \(\dfrac{9 - 3}{13 - 6} = \dfrac{6}{7}\). ✓

Calculation

[6 marks]

(a) Show that \(\displaystyle\sum_{r=0}^{n}\binom{n}{r}\sin r\theta = 2^{n}\cos^{n}\tfrac{\theta}{2}\,\sin\tfrac{n\theta}{2}\).

(b) Hence evaluate \(\displaystyle\sum_{r=0}^{6}\binom{6}{r}\sin\frac{r\pi}{2}\), and check your answer by adding the terms.

(a) \(C + iS = \sum_{r=0}^{n}\binom{n}{r}e^{ir\theta} = \big(1 + e^{i\theta}\big)^{n}\) by the binomial theorem. ✓
\(1 + e^{i\theta} = e^{i\theta/2}\big(e^{-i\theta/2} + e^{i\theta/2}\big) = 2\cos\tfrac{\theta}{2}\,e^{i\theta/2}\). ✓
So \(C + iS = 2^{n}\cos^{n}\tfrac{\theta}{2}\,e^{in\theta/2}\) (de Moivre), and the imaginary part is \(2^{n}\cos^{n}\tfrac{\theta}{2}\sin\tfrac{n\theta}{2}\). ✓
(b) \(n = 6\), \(\theta = \tfrac{\pi}{2}\): \(64\big(\tfrac{\sqrt{2}}{2}\big)^{6}\sin\tfrac{3\pi}{2} = 64 \times \tfrac{1}{8} \times (-1) = -8\). ✓
Check: \(\sin\tfrac{r\pi}{2}\) is \(0, 1, 0, -1, 0, 1, 0\) for \(r = 0\) to 6, so the sum is \(6 - 20 + 6 = -8\). ✓

Analysis

[6 marks]

(a) Show that, for \(\sin\tfrac{\theta}{2} \ne 0\), \[\sum_{r=1}^{n}\cos r\theta = \frac{\cos\frac{(n + 1)\theta}{2}\,\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}.\]

(b) Hence show that \(\displaystyle\sum_{r=1}^{n}\cos\frac{2r\pi}{2n + 1} = -\frac{1}{2}\) for every positive integer \(n\).

(a) \(C + iS = \sum_{r=1}^{n}z^{r} = \dfrac{z(z^{n} - 1)}{z - 1}\) with \(z = e^{i\theta} \ne 1\). ✓
\[\begin{aligned} C + iS &= \frac{e^{i\theta}\cdot 2i\sin\frac{n\theta}{2}\,e^{in\theta/2}}{2i\sin\frac{\theta}{2}\,e^{i\theta/2}}\\ &= \frac{\sin\frac{n\theta}{2}}{\sin\frac{\theta}{2}}\,e^{i(n + 1)\theta/2}. \end{aligned}\] ✓
Taking the real part gives the result. ✓
(b) Put \(\theta = \dfrac{2\pi}{2n + 1}\) and let \(a = \dfrac{n\pi}{2n + 1}\), so \(\dfrac{n\theta}{2} = a\) and \(\dfrac{(n + 1)\theta}{2} = \dfrac{(n + 1)\pi}{2n + 1} = \pi - a\). ✓
\(\cos(\pi - a) = -\cos a\), so \(C = \dfrac{-\cos a\sin a}{\sin\frac{\pi}{2n + 1}} = \dfrac{-\sin 2a}{2\sin\frac{\pi}{2n + 1}}\). ✓
\(2a = \pi - \dfrac{\pi}{2n + 1}\), so \(\sin 2a = \sin\dfrac{\pi}{2n + 1}\) and \(C = -\tfrac{1}{2}\). Check \(n = 1\): \(\cos\tfrac{2\pi}{3} = -\tfrac{1}{2}\). ✓

Proof

[7 marks]

(a) Use de Moivre's theorem to show that, for \(\sin\theta \ne 0\), \[\sum_{r=0}^{n-1}\cos 2r\theta = \frac{\cos(n - 1)\theta\,\sin n\theta}{\sin\theta}.\]

(b) Hence find \(\displaystyle\sum_{r=0}^{n-1}\cos^{2}r\theta\) in terms of \(n\) and \(\theta\).

(a) \(C + iS = \sum_{r=0}^{n-1}w^{r}\) with \(w = e^{2i\theta} \ne 1\) (as \(\sin\theta \ne 0\)): \(C + iS = \dfrac{e^{2in\theta} - 1}{e^{2i\theta} - 1}\). ✓
\[\begin{aligned} C + iS &= \frac{2i\sin n\theta\,e^{in\theta}}{2i\sin\theta\,e^{i\theta}}\\ &= \frac{\sin n\theta}{\sin\theta}\,e^{i(n - 1)\theta}, \end{aligned}\] whose real part is \(\dfrac{\cos(n - 1)\theta\,\sin n\theta}{\sin\theta}\). ✓
(b) \(\cos^{2}r\theta = \tfrac{1}{2}(1 + \cos 2r\theta)\), and there are \(n\) terms, so \[\sum_{r=0}^{n-1}\cos^{2}r\theta = \frac{n}{2} + \frac{\cos(n - 1)\theta\,\sin n\theta}{2\sin\theta}.\] ✓
(c) Put \(\theta = \tfrac{\pi}{n}\). For \(n \ge 2\), \(0 \lt \theta \lt \pi\), so \(\sin\theta \ne 0\) and (b) applies. ✓
\(\sin n\theta = \sin\pi = 0\), so the second term vanishes and the sum is \(\dfrac{n}{2}\). ✓
Check \(n = 3\): \(\cos^{2}0 + \cos^{2}\tfrac{\pi}{3} + \cos^{2}\tfrac{2\pi}{3} = 1 + \tfrac{1}{4} + \tfrac{1}{4} = \tfrac{3}{2}\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.