AS & A Level · Practice questions · De Moivre's theorem

Turn, stretch, repeat.

Six original Cambridge-style questions on de Moivre's theorem: multiplying and dividing in modulus-argument form, the induction proof with a full conclusion, negative integer powers, exact powers in the form \(a + bi\), and the geometry of successive powers.

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

Modulus to the power, argument times the power.

Convert \(a + bi\) to \(r(\cos\theta + i\sin\theta)\); sketch the point to choose the quadrant for \(\theta\).
Check the form: same angle twice and a plus sign; rewrite \(\cos\theta - i\sin\theta\) as \(\cos(-\theta) + i\sin(-\theta)\).
Apply de Moivre: \(\big(r(\cos\theta + i\sin\theta)\big)^{n} = r^{n}(\cos n\theta + i\sin n\theta)\), for any integer \(n\).
Reduce the argument by multiples of \(2\pi\) into \(-\pi \lt \theta \le \pi\), then evaluate exactly.
Prove by induction: base case, assumption, step with the compound angle formulae, and the full concluding sentence.

Recall

[3 marks]

Given \(z = 4\left(\cos\frac{2\pi}{5} + i\sin\frac{2\pi}{5}\right)\) and \(w = 2\left(\cos\left(-\frac{\pi}{10}\right) + i\sin\left(-\frac{\pi}{10}\right)\right)\), find \(zw\) in modulus-argument form and \(\dfrac{z}{w}\) in the form \(a + bi\).

\(|zw| = 4 \times 2 = 8\) and \(\arg(zw) = \dfrac{2\pi}{5} - \dfrac{\pi}{10} = \dfrac{3\pi}{10}\), so \(zw = 8\left(\cos\frac{3\pi}{10} + i\sin\frac{3\pi}{10}\right)\). ✓
\(\left|\dfrac{z}{w}\right| = 2\) and \(\arg\dfrac{z}{w} = \dfrac{2\pi}{5} + \dfrac{\pi}{10} = \dfrac{\pi}{2}\). ✓
So \(\dfrac{z}{w} = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2i\). ✓

Proof

[5 marks]

Prove by mathematical induction that, for every positive integer \(n\), \[(\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta.\]

\(n = 1\): both sides equal \(\cos\theta + i\sin\theta\), so the result holds. ✓
Assume it holds for \(n = k\): \((\cos\theta + i\sin\theta)^{k} = \cos k\theta + i\sin k\theta\). ✓
\[(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)\] ✓
\[= \cos k\theta\cos\theta - \sin k\theta\sin\theta + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta)\] ✓
\(= \cos(k + 1)\theta + i\sin(k + 1)\theta\), by the compound angle formulae, so the result holds for \(n = k + 1\). ✓
The result is true for \(n = 1\), and if true for \(n = k\) it is true for \(n = k + 1\); hence by induction it is true for every positive integer \(n\). ✓

Proof

[3 marks]

Using the result for positive integers, show that \((\cos\theta + i\sin\theta)^{-n} = \cos n\theta - i\sin n\theta\) for every positive integer \(n\).

\(\dfrac{1}{\cos\theta + i\sin\theta} = \dfrac{\cos\theta - i\sin\theta}{\cos^{2}\theta + \sin^{2}\theta} = \cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)\). ✓
So \((\cos\theta + i\sin\theta)^{-n} = \big(\cos(-\theta) + i\sin(-\theta)\big)^{n} = \cos(-n\theta) + i\sin(-n\theta)\), by the positive-integer result. ✓
Cosine is even and sine is odd, so this is \(\cos n\theta - i\sin n\theta\), as required. ✓

Calculation

[4 marks]

Find \((\sqrt{3} + i)^{7}\) in the form \(a + bi\), where \(a\) and \(b\) are exact.

\(|\sqrt{3} + i| = 2\) and the point is in the first quadrant with \(\tan\theta = \dfrac{1}{\sqrt{3}}\), so \(\arg = \dfrac{\pi}{6}\). ✓
\[(\sqrt{3} + i)^{7} = 2^{7}\left(\cos\tfrac{7\pi}{6} + i\sin\tfrac{7\pi}{6}\right)\] ✓
\(\dfrac{7\pi}{6}\) is equivalent to \(-\dfrac{5\pi}{6}\): \(\cos = -\dfrac{\sqrt{3}}{2}\), \(\sin = -\dfrac{1}{2}\). ✓
\(128\left(-\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = -64\sqrt{3} - 64i\). ✓

Analysis

[4 marks]

(a) Find the smallest positive integer \(n\) for which \((1 + i\sqrt{3})^{n}\) is real, and state the value of \((1 + i\sqrt{3})^{n}\) for this \(n\).

(b) Find the smallest positive integer \(n\) for which \((1 + i\sqrt{3})^{n}\) is real and positive.

\(1 + i\sqrt{3} = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\), so \((1 + i\sqrt{3})^{n} = 2^{n}\left(\cos\frac{n\pi}{3} + i\sin\frac{n\pi}{3}\right)\). ✓
(a) It is real when \(\sin\dfrac{n\pi}{3} = 0\), that is when \(\dfrac{n\pi}{3}\) is a multiple of \(\pi\): smallest \(n = 3\). ✓
Then \((1 + i\sqrt{3})^{3} = 8(\cos\pi + i\sin\pi) = -8\). ✓
(b) Real and positive needs \(\dfrac{n\pi}{3}\) to be a multiple of \(2\pi\): smallest \(n = 6\), giving \(2^{6} = 64\). ✓

Analysis

[6 marks]

Let \(w = 1 + i\), and let \(P_n\) be the point representing \(w^{n}\) on an Argand diagram, where \(O\) is the origin.

(a) Use de Moivre's theorem to find \(w^{8}\).

(b) Find the smallest positive integer \(n\) for which \(P_n\) lies on the negative real axis.

\(w = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\), so \(w^{n} = 2^{n/2}\left(\cos\frac{n\pi}{4} + i\sin\frac{n\pi}{4}\right)\). ✓
(a) \(w^{8} = 2^{4}(\cos 2\pi + i\sin 2\pi) = 16\). ✓
(b) \(P_n\) is on the negative real axis when \(\dfrac{n\pi}{4} = \pi\) (plus a multiple of \(2\pi\)): smallest \(n = 4\), with \(w^{4} = -4\). ✓
(c) \(OP_n = |w|^{n} = 2^{n/2}\) and \(OP_{n+1} = 2^{(n+1)/2}\); multiplying by \(w\) turns \(P_n\) through \(\dfrac{\pi}{4}\), so \(\angle P_nOP_{n+1} = \dfrac{\pi}{4}\). ✓
\[\text{Area} = \tfrac{1}{2} \times 2^{n/2} \times 2^{(n+1)/2} \times \sin\tfrac{\pi}{4} = \tfrac{1}{2} \times 2^{n} \times \sqrt{2} \times \tfrac{\sqrt{2}}{2}\] ✓
\(= \tfrac{1}{2} \times 2^{n} = 2^{n-1}\), as required. ✓

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.