A-Level Further Maths 9231 / Topic 2.5 / AS & A Level

Every power is a rotation.

Multiplying complex numbers multiplies their moduli and adds their arguments, so raising \(\cos\theta + i\sin\theta\) to the power \(n\) simply turns it through \(n\theta\). Prove de Moivre's theorem by induction, extend it to negative powers, and use it to evaluate powers such as \((1 + i\sqrt{3})^{8}\) exactly in a few lines.

Topic 2.5All complex numbers lessons →

The key idea

If \(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) and \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\), then \(|z_1z_2| = r_1r_2\) and \(\arg(z_1z_2) = \theta_1 + \theta_2\); for \(\dfrac{z_1}{z_2}\) divide the moduli and subtract the arguments. Repeating this gives de Moivre's theorem: \((\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta\) for every integer \(n\), positive, negative or zero. So \(\big(r(\cos\theta + i\sin\theta)\big)^{n} = r^{n}(\cos n\theta + i\sin n\theta)\): raise the modulus to the power \(n\), multiply the argument by \(n\).

Fig. 1: z = 1.2(cos 40° + i sin 40°) and its powers on an Argand diagram. Each multiplication by z turns the point through another 40° about O and stretches its distance from O by a factor 1.2, so 1, z, z², z³ lie on a spiral (dashed). If |z| = 1 the powers stay on the unit circle and the multiplication is a pure rotation

Section 01

Multiply moduli, add arguments.

Write \(z_1 = r_1(\cos\theta_1 + i\sin\theta_1)\) and \(z_2 = r_2(\cos\theta_2 + i\sin\theta_2)\) and multiply out. The real part is \(r_1r_2(\cos\theta_1\cos\theta_2 - \sin\theta_1\sin\theta_2) = r_1r_2\cos(\theta_1 + \theta_2)\) and the imaginary part is \(r_1r_2(\sin\theta_1\cos\theta_2 + \cos\theta_1\sin\theta_2) = r_1r_2\sin(\theta_1 + \theta_2)\). So multiplying complex numbers multiplies their moduli and adds their arguments; dividing divides the moduli and subtracts the arguments.

Operation	Modulus	Argument	Effect on the Argand diagram
\(z_1z_2\)	\(r_1r_2\)	\(\theta_1 + \theta_2\)	enlarge \(z_1\) by scale factor \(r_2\) about \(O\) and rotate it through \(\theta_2\) anticlockwise
\(\dfrac{z_1}{z_2}\)	\(\dfrac{r_1}{r_2}\)	\(\theta_1 - \theta_2\)	enlarge \(z_1\) by \(\dfrac{1}{r_2}\) and rotate it through \(\theta_2\) clockwise
\(z^{n}\)	\(r^{n}\)	\(n\theta\)	\(n\) turns of \(\theta\) from the point 1, scaling by \(r\) each time (Fig. 1)

Multiplying by a number of modulus 1, \(\cos\alpha + i\sin\alpha\), is a pure rotation through \(\alpha\) about \(O\); multiplying by \(i = \cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\) turns every point through a right angle. Repeating the product rule \(n\) times on \(z = \cos\theta + i\sin\theta\) adds \(\theta\) to the argument \(n\) times, which is the content of de Moivre's theorem.

Worked example Product and quotient in modulus-argument form

Given \(z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)\) and \(w = 3\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)\), write \(zw\) and \(\dfrac{z}{w}\) in modulus-argument form.

\(|zw| = 2 \times 3 = 6\) and \(\arg(zw) = \dfrac{\pi}{3} + \dfrac{\pi}{4} = \dfrac{7\pi}{12}\).
\(\left|\dfrac{z}{w}\right| = \dfrac{2}{3}\) and \(\arg\dfrac{z}{w} = \dfrac{\pi}{3} - \dfrac{\pi}{4} = \dfrac{\pi}{12}\).
Both arguments already lie in \(-\pi \lt \theta \le \pi\), so no adjustment is needed.

Answer: \(zw = 6\left(\cos\frac{7\pi}{12} + i\sin\frac{7\pi}{12}\right)\), \(\dfrac{z}{w} = \dfrac{2}{3}\left(\cos\frac{\pi}{12} + i\sin\frac{\pi}{12}\right)\)

Worked example A rotation done by multiplication

The point \(P\) represents \(3 + i\). \(P\) is rotated through \(\dfrac{\pi}{3}\) anticlockwise about the origin. Find the complex number representing its new position, in exact form \(a + bi\).

A rotation through \(\dfrac{\pi}{3}\) is multiplication by \(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3} = \dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\), a number of modulus 1.
\[(3 + i)\left(\tfrac{1}{2} + \tfrac{\sqrt{3}}{2}i\right) = \tfrac{3}{2} + \tfrac{3\sqrt{3}}{2}i + \tfrac{1}{2}i + \tfrac{\sqrt{3}}{2}i^{2}\]
With \(i^{2} = -1\): real part \(\dfrac{3 - \sqrt{3}}{2}\), imaginary part \(\dfrac{3\sqrt{3} + 1}{2}\).
Check: the modulus is unchanged, \(\sqrt{10}\), as a rotation must keep the distance from \(O\).

Answer: \(\dfrac{3 - \sqrt{3}}{2} + \dfrac{3\sqrt{3} + 1}{2}\,i\)

Your turn

Given \(z = 2\left(\cos\frac{5\pi}{6} + i\sin\frac{5\pi}{6}\right)\) and \(w = \cos\left(-\frac{\pi}{3}\right) + i\sin\left(-\frac{\pi}{3}\right)\), find \(zw\) in the form \(a + bi\).

Examiner trap

Arguments ADD when you multiply; they do not multiply. And after adding or subtracting, bring the answer back into the principal range \(-\pi \lt \theta \le \pi\): \(\dfrac{2\pi}{3} + \dfrac{3\pi}{4} = \dfrac{17\pi}{12}\) is given as \(-\dfrac{7\pi}{12}\).

Section 02

Proof by induction, then negative powers.

For a positive integer \(n\), de Moivre's theorem is proved by induction, using the product rule as the inductive step. For a negative integer, the key fact is that the reciprocal of \(\cos\theta + i\sin\theta\) is \(\cos\theta - i\sin\theta\), which is \(\cos(-\theta) + i\sin(-\theta)\): a turn of \(\theta\) the other way.

Rational exponents. The theorem is also used with a fraction for the power: \(\cos\frac{p\theta}{q} + i\sin\frac{p\theta}{q}\) is one value of \((\cos\theta + i\sin\theta)^{p/q}\), because raising it to the power \(q\) gives \(\cos p\theta + i\sin p\theta\). It is not the only value: for example \((\cos\theta + i\sin\theta)^{1/2}\) is \(\pm\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right)\), since both square to \(\cos\theta + i\sin\theta\). Finding all the values is the work of the lesson on \(n\)th roots.

The theorem needs the form \(\cos\theta + i\sin\theta\) with the SAME angle in both places and a plus sign. Convert anything else first: \(\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)\), and \(\sin\theta + i\cos\theta = \cos\left(\frac{\pi}{2} - \theta\right) + i\sin\left(\frac{\pi}{2} - \theta\right)\).

Worked example The induction proof for positive n

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

Base case, \(n = 1\): the left side is \(\cos\theta + i\sin\theta\) and the right side is \(\cos\theta + i\sin\theta\). True.
Assume the result is true for \(n = k\): \((\cos\theta + i\sin\theta)^{k} = \cos k\theta + i\sin k\theta\).
Then \[(\cos\theta + i\sin\theta)^{k+1} = (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta)\]
\[= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta)\]
By the compound angle formulae, \(= \cos(k + 1)\theta + i\sin(k + 1)\theta\), which is the result for \(n = k + 1\).
Conclusion: the result is true for \(n = 1\), and if it is true for \(n = k\) then it is true for \(n = k + 1\). So by mathematical induction it is true for every positive integer \(n\).

Answer: \((\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta\) for all \(n \in \mathbb{Z}^{+}\)

Worked example Extending to negative integers

Given the result for positive integers, show that \((\cos\theta + i\sin\theta)^{n} = \cos n\theta + i\sin n\theta\) when \(n\) is a negative integer.

Let \(n = -m\), where \(m\) is a positive integer.
\[\frac{1}{\cos\theta + i\sin\theta} = \frac{\cos\theta - i\sin\theta}{\cos^{2}\theta + \sin^{2}\theta} = \cos\theta - i\sin\theta\]
Since cosine is even and sine is odd, \(\cos\theta - i\sin\theta = \cos(-\theta) + i\sin(-\theta)\).
So \((\cos\theta + i\sin\theta)^{-m} = \big(\cos(-\theta) + i\sin(-\theta)\big)^{m} = \cos(-m\theta) + i\sin(-m\theta)\), by the positive-integer result with angle \(-\theta\).
That is \(\cos n\theta + i\sin n\theta\). (For \(n = 0\) both sides equal 1.)

Answer: De Moivre's theorem holds for every integer \(n\)

Your turn

Simplify \((\cos 5\theta + i\sin 5\theta)(\cos 2\theta - i\sin 2\theta)^{3}\) to the form \(\cos k\theta + i\sin k\theta\).

Examiner trap

The power acts on the ANGLE, not on each term: \((\cos\theta + i\sin\theta)^{3}\) is \(\cos 3\theta + i\sin 3\theta\), not \(\cos^{3}\theta + i\sin^{3}\theta\). In the induction, an answer without the closing sentence (true for \(n = 1\); true for \(k\) implies true for \(k + 1\); so true for all positive integers) loses the final mark.

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

Simplify \(\left(\cos\frac{\pi}{5} + i\sin\frac{\pi}{5}\right)^{10}\).

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\(|z| = 2\) and \(\arg z = \dfrac{\pi}{6}\). Find the modulus and argument of \(z^{3}\).

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Which expression is equal to \(\dfrac{1}{\cos\theta + i\sin\theta}\)?

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Multiplying every complex number by \(w = 3\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right)\) has which effect on the Argand diagram?

Section 03

Exact powers in a few lines.

To raise \(a + bi\) to a large or negative power, change to modulus-argument form, apply de Moivre's theorem, reduce the argument to the principal range, and convert back. Find the argument from a sketch of the point: the quadrant decides the angle, not a calculator's \(\tan^{-1}\).

Two results for the next lesson. If \(z = \cos\theta + i\sin\theta\) (so \(|z| = 1\)), then \(\dfrac{1}{z^{n}} = z^{-n} = \cos n\theta - i\sin n\theta\). Adding and subtracting:

\[z^{n} + \frac{1}{z^{n}} = 2\cos n\theta, \qquad z^{n} - \frac{1}{z^{n}} = 2i\sin n\theta.\]

These turn powers of \(\cos\theta\) and \(\sin\theta\) into multiple angles, and the reverse, which is the work of the next lesson.

Worked example A positive power

Find \((1 + i\sqrt{3})^{8}\) in the form \(a + bi\), where \(a\) and \(b\) are exact.

\(|1 + i\sqrt{3}| = \sqrt{1 + 3} = 2\); the point is in the first quadrant with \(\tan\theta = \sqrt{3}\), so \(\arg = \dfrac{\pi}{3}\).
\[(1 + i\sqrt{3})^{8} = 2^{8}\left(\cos\tfrac{8\pi}{3} + i\sin\tfrac{8\pi}{3}\right)\]
\(\dfrac{8\pi}{3} - 2\pi = \dfrac{2\pi}{3}\), so this is \(256\left(\cos\frac{2\pi}{3} + i\sin\frac{2\pi}{3}\right)\).
\(= 256\left(-\dfrac{1}{2} + \dfrac{\sqrt{3}}{2}i\right)\).

Answer: \(-128 + 128\sqrt{3}\,i\)

Worked example A negative power

Find the exact value of \((\sqrt{3} - i)^{-6}\).

\(|\sqrt{3} - i| = \sqrt{3 + 1} = 2\); the point is in the fourth quadrant with \(\tan\alpha = \dfrac{1}{\sqrt{3}}\), so \(\arg = -\dfrac{\pi}{6}\).
\[(\sqrt{3} - i)^{-6} = 2^{-6}\left(\cos\pi + i\sin\pi\right)\]
since \(-6 \times \left(-\dfrac{\pi}{6}\right) = \pi\).
\(\cos\pi + i\sin\pi = -1\).

Answer: \(-\dfrac{1}{64}\)

Your turn

Find \((1 - i)^{10}\) in the form \(a + bi\).

Examiner trap

The argument of \(\sqrt{3} - i\) is \(-\dfrac{\pi}{6}\), not \(\dfrac{\pi}{6}\) (wrong quadrant) and not \(-\dfrac{\pi}{3}\) (\(\tan\) inverted). And raise the MODULUS to the power too: \((1 + i\sqrt{3})^{8}\) has modulus \(2^{8} = 256\), not 2 or 16.

Stage 2 · Exam

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Find \((1 + i)^{6}\).

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Find the exact value of \((-1 + i\sqrt{3})^{-3}\).

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Simplify \(\dfrac{(\cos 3\theta + i\sin 3\theta)^{2}}{(\cos\theta - i\sin\theta)^{4}}\).

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Given that \(z = \cos\theta + i\sin\theta\), which expression is equal to \(z^{3} + \dfrac{1}{z^{3}}\)?

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Powers turned into rotations, and the theorem proved.

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Six original Cambridge-style questions

Products and quotients in modulus-argument form, the induction proof of de Moivre's theorem and its extension to negative powers, exact powers of \(\sqrt{3} + i\), the first real power of \(1 + i\sqrt{3}\), and the spiral of powers of \(1 + i\).