23.5 De Moivre’s Theorem
AHL 1.14
- De Moivre’s theorem for powers of a complex number in polar and Euler form
- how to prove it by mathematical induction for positive integers \(n\)
- how to use it for negative integer powers
- how to find large powers of complex numbers quickly, without expanding brackets
De Moivre’s theorem
In Section 23.4 we saw that when we multiply complex numbers, the moduli multiply and the arguments add. So multiplying \(z = r\operatorname{cis}\theta\) by itself \(n\) times gives modulus \(r^n\) and argument \(n\theta\).
De Moivre’s theorem In the HL formula booklet
\[ \big[r(\cos\theta + i\sin\theta)\big]^n = r^n(\cos n\theta + i\sin n\theta) = r^n\mathrm{e}^{in\theta} = r^n\operatorname{cis} n\theta \]
- \(r\): the modulus of the complex number
- \(\theta\): the argument of the complex number
- \(n\): the power (an integer)
In Euler form the theorem is just a law of indices: \(\left(r\mathrm{e}^{i\theta}\right)^n = r^n\mathrm{e}^{in\theta}\).
Proof by induction for positive integers
The syllabus asks you to be able to prove the theorem for \(n \in \mathbb{Z}^+\). The method of proof by induction is explained in Section 6.4. It is enough to prove it for \(r = 1\), because \((r\operatorname{cis}\theta)^n = r^n(\operatorname{cis}\theta)^n\).
Statement. Let \(P(n)\) be: \((\cos\theta + i\sin\theta)^n = \cos n\theta + i\sin n\theta\).
Basis. \(n = 1\): the left side is \(\cos\theta + i\sin\theta\) and the right side is \(\cos\theta + i\sin\theta\). So \(P(1)\) is true.
Inductive step. Assume \(P(k)\) is true for some \(k \in \mathbb{Z}^+\), that is, \((\cos\theta + i\sin\theta)^k = \cos k\theta + i\sin k\theta\). We want to show \((\cos\theta + i\sin\theta)^{k+1} = \cos(k + 1)\theta + i\sin(k + 1)\theta\).
\[ \begin{aligned} &(\cos\theta + i\sin\theta)^{k+1} \\ &= (\cos\theta + i\sin\theta)^k(\cos\theta + i\sin\theta) \\ &= (\cos k\theta + i\sin k\theta)(\cos\theta + i\sin\theta) && {\color{#9F353A}\text{by the assumption}} \\ &= (\cos k\theta\cos\theta - \sin k\theta\sin\theta) && {\color{#9F353A}\text{expand}} \\ &\qquad + i(\sin k\theta\cos\theta + \cos k\theta\sin\theta) \\ &= \cos(k\theta + \theta) + i\sin(k\theta + \theta) && {\color{#9F353A}\text{compound angle formulae}} \\ &= \cos(k + 1)\theta + i\sin(k + 1)\theta . \end{aligned} \]
So \(P(k + 1)\) is true.
Conclusion. Since \(P(1)\) is true, and \(P(k)\) true implies \(P(k + 1)\) true, by the principle of mathematical induction \(P(n)\) is true for all \(n \in \mathbb{Z}^+\).
Zero and negative powers
For \(n = 0\) both sides equal \(1\). For a negative integer \(-n\) (with \(n \in \mathbb{Z}^+\)), use the positive case and divide:
\[ (r\operatorname{cis}\theta)^{-n} = \frac{1}{r^n\operatorname{cis} n\theta} = \frac{1\operatorname{cis}0}{r^n\operatorname{cis} n\theta} = r^{-n}\operatorname{cis}(-n\theta) . \]
So De Moivre’s theorem is true for all integers \(n\). The syllabus also extends it to rational powers such as \(n = \dfrac{1}{3}\). This is how we find roots in Section 23.6.
Using De Moivre’s theorem
To find \(z^n\):
- Write \(z\) in polar or Euler form (if it is not already).
- Raise the modulus to the power \(n\) and multiply the argument by \(n\).
- Bring the argument into \(-\pi < \theta \leq \pi\) by adding or subtracting multiples of \(2\pi\).
- Convert to Cartesian form if the question asks.
Example 1 Write each number in polar form and in Euler form.
(a) \(\left[2\operatorname{cis}\dfrac{\pi}{6}\right]^4\)
(b) \(\left[3\operatorname{cis}\dfrac{\pi}{4}\right]^{-2}\)
(a) \(2^4\operatorname{cis}\left(4 \times \dfrac{\pi}{6}\right) = 16\operatorname{cis}\dfrac{2\pi}{3} = 16\mathrm{e}^{i\frac{2\pi}{3}}\)
(b) \(3^{-2}\operatorname{cis}\left(-2 \times \dfrac{\pi}{4}\right) = \dfrac{1}{9}\operatorname{cis}\left(-\dfrac{\pi}{2}\right) = \dfrac{1}{9}\mathrm{e}^{-i\frac{\pi}{2}}\) (this is the number \(-\dfrac{1}{9}i\)).
Example 2 Find \((1 + \sqrt{3}\,i)^8\) in the form \(x + iy\).
\(1 + \sqrt{3}\,i = 2\operatorname{cis}\dfrac{\pi}{3}\). By De Moivre’s theorem,
\[ (1 + \sqrt{3}\,i)^8 = 2^8\operatorname{cis}\frac{8\pi}{3} = 256\operatorname{cis}\frac{2\pi}{3} \qquad\left(\frac{8\pi}{3} - 2\pi = \frac{2\pi}{3}\right) \]
\[ = 256\left(-\frac{1}{2} + \frac{\sqrt{3}}{2}i\right) = -128 + 128\sqrt{3}\,i . \]
Expanding \((1 + \sqrt{3}\,i)^8\) with the binomial theorem would take much longer.
Example 3 By first finding the modulus and argument of each part, show that
\[ z = \frac{(1 + i)^5}{(1 - \sqrt{3}\,i)^2} \]
can be written as \(z = r(\cos\theta - i\sin\theta)\), where \(r > 0\) and \(0 < \theta < \dfrac{\pi}{2}\) are to be found.
\(1 + i = \sqrt{2}\operatorname{cis}\dfrac{\pi}{4}\), so \((1 + i)^5 = (\sqrt{2})^5\operatorname{cis}\dfrac{5\pi}{4} = 4\sqrt{2}\operatorname{cis}\dfrac{5\pi}{4}\).
\(1 - \sqrt{3}\,i = 2\operatorname{cis}\left(-\dfrac{\pi}{3}\right)\), so \((1 - \sqrt{3}\,i)^2 = 4\operatorname{cis}\left(-\dfrac{2\pi}{3}\right)\).
\[ z = \frac{4\sqrt{2}}{4}\operatorname{cis}\left(\frac{5\pi}{4} + \frac{2\pi}{3}\right) = \sqrt{2}\operatorname{cis}\frac{23\pi}{12} = \sqrt{2}\operatorname{cis}\left(-\frac{\pi}{12}\right) . \]
Since \(\cos\left(-\dfrac{\pi}{12}\right) = \cos\dfrac{\pi}{12}\) and \(\sin\left(-\dfrac{\pi}{12}\right) = -\sin\dfrac{\pi}{12}\),
\[ z = \sqrt{2}\left(\cos\frac{\pi}{12} - i\sin\frac{\pi}{12}\right), \qquad r = \sqrt{2}, \;\; \theta = \frac{\pi}{12} . \]
Example 4 Let \(z = \sqrt{3} + i\). Find the smallest positive integer \(n\) for which
(a) \(z^n\) is real, \(\qquad\) (b) \(z^n\) is purely imaginary.
In each case write down the value of \(z^n\).
\(z = 2\operatorname{cis}\dfrac{\pi}{6}\), so \(z^n = 2^n\left(\cos\dfrac{n\pi}{6} + i\sin\dfrac{n\pi}{6}\right)\).
(a) Real when \(\sin\dfrac{n\pi}{6} = 0\), that is \(\dfrac{n\pi}{6}\) is a multiple of \(\pi\). The smallest is \(n = 6\): \(\;z^6 = 64\operatorname{cis}\pi = -64\).
(b) Purely imaginary when \(\cos\dfrac{n\pi}{6} = 0\), that is \(\dfrac{n\pi}{6} = \dfrac{\pi}{2}, \dfrac{3\pi}{2}, \ldots\) The smallest is \(n = 3\): \(\;z^3 = 8\operatorname{cis}\dfrac{\pi}{2} = 8i\).
Common errors
| ✗ Wrong | ✓ Right | Why |
|---|---|---|
| \((2\operatorname{cis}\theta)^4 = 2\operatorname{cis}4\theta\) | \(16\operatorname{cis}4\theta\) | The modulus is raised to the power too: \(r^n\). |
| \((2\operatorname{cis}\theta)^4 = 16\operatorname{cis}\theta^4\) | \(16\operatorname{cis}4\theta\) | The argument is multiplied by \(n\), not raised to the power \(n\). |
| \((1 + \sqrt{3}i)^8 = 256\operatorname{cis}\dfrac{8\pi}{3}\) as the final answer | \(256\operatorname{cis}\dfrac{2\pi}{3}\) | Give the principal argument. |
| \((\cos\theta - i\sin\theta)^n = \cos n\theta + i\sin n\theta\) | \(\cos n\theta - i\sin n\theta\) | First write \(\cos\theta - i\sin\theta = \operatorname{cis}(-\theta)\). |
| Induction proof with no concluding sentence | Write the full conclusion | Marks are given for the conclusion. |
Exercises
1 Find \(\left[2\operatorname{cis}\dfrac{\pi}{5}\right]^5\).
\(2^5\operatorname{cis}\pi = 32\operatorname{cis}\pi = -32\).
2 Write \(\left(\sqrt{2}\,\mathrm{e}^{i\frac{\pi}{8}}\right)^6\) in Euler form and in Cartesian form.
\((\sqrt{2})^6\mathrm{e}^{i\frac{6\pi}{8}} = 8\mathrm{e}^{i\frac{3\pi}{4}} = 8\left(-\dfrac{\sqrt{2}}{2} + \dfrac{\sqrt{2}}{2}i\right) = -4\sqrt{2} + 4\sqrt{2}\,i\).
3 Find \((1 - i)^{10}\) in the form \(x + iy\).
\(1 - i = \sqrt{2}\operatorname{cis}\left(-\dfrac{\pi}{4}\right)\), so \((1 - i)^{10} = 32\operatorname{cis}\left(-\dfrac{5\pi}{2}\right) = 32\operatorname{cis}\left(-\dfrac{\pi}{2}\right) = -32i\).
4 Find \((-\sqrt{3} + i)^7\) in the form \(x + iy\).
\(-\sqrt{3} + i = 2\operatorname{cis}\dfrac{5\pi}{6}\). Then \((-\sqrt{3} + i)^7 = 128\operatorname{cis}\dfrac{35\pi}{6}\).
\(\dfrac{35\pi}{6} - 6\pi = -\dfrac{\pi}{6}\), so the answer is \(128\left(\dfrac{\sqrt{3}}{2} - \dfrac{1}{2}i\right) = 64\sqrt{3} - 64i\).
5 Find \(\left[\operatorname{cis}\dfrac{\pi}{9}\right]^{-3}\) in the form \(x + iy\).
\(\operatorname{cis}\left(-\dfrac{3\pi}{9}\right) = \operatorname{cis}\left(-\dfrac{\pi}{3}\right) = \dfrac{1}{2} - \dfrac{\sqrt{3}}{2}i\).
6 Simplify \(\dfrac{(\cos 3\theta + i\sin 3\theta)^2}{(\cos\theta - i\sin\theta)^4}\) to the form \(\cos k\theta + i\sin k\theta\).
The top is \((\operatorname{cis}3\theta)^2 = \operatorname{cis}6\theta\). The bottom is \(\big(\operatorname{cis}(-\theta)\big)^4 = \operatorname{cis}(-4\theta)\).
\[ \frac{\operatorname{cis}6\theta}{\operatorname{cis}(-4\theta)} = \operatorname{cis}10\theta = \cos 10\theta + i\sin 10\theta . \]
7 Without a calculator, show that \(\dfrac{(\sqrt{3} + i)^4}{(1 - i)^6} = \sqrt{3} + i\).
\((\sqrt{3} + i)^4 = \left(2\operatorname{cis}\dfrac{\pi}{6}\right)^4 = 16\operatorname{cis}\dfrac{2\pi}{3}\).
\((1 - i)^6 = \left(\sqrt{2}\operatorname{cis}\left(-\dfrac{\pi}{4}\right)\right)^6 = 8\operatorname{cis}\left(-\dfrac{3\pi}{2}\right) = 8\operatorname{cis}\dfrac{\pi}{2}\).
Quotient: \(2\operatorname{cis}\left(\dfrac{2\pi}{3} - \dfrac{\pi}{2}\right) = 2\operatorname{cis}\dfrac{\pi}{6} = \sqrt{3} + i\).
8 Let \(z = 1 + i\). Find the smallest positive integer \(n\) for which
(a) \(z^n\) is real, \(\qquad\) (b) \(z^n\) is real and positive.
Give the value of \(z^n\) in each case.
\(z^n = (\sqrt{2})^n\operatorname{cis}\dfrac{n\pi}{4}\).
(a) \(\dfrac{n\pi}{4}\) must be a multiple of \(\pi\): \(n = 4\), \(\;z^4 = 4\operatorname{cis}\pi = -4\).
(b) \(\dfrac{n\pi}{4}\) must be a multiple of \(2\pi\): \(n = 8\), \(\;z^8 = 16\).
9 Let \(z = \cos\theta + i\sin\theta\).
(a) Show that \(z^n + \dfrac{1}{z^n} = 2\cos n\theta\) and \(z^n - \dfrac{1}{z^n} = 2i\sin n\theta\) for \(n \in \mathbb{Z}^+\).
(b) Hence find the value of \(z^3 + z^{-3}\) when \(\theta = \dfrac{2\pi}{9}\).
(a) By De Moivre, \(z^n = \cos n\theta + i\sin n\theta\) and \(z^{-n} = \cos(-n\theta) + i\sin(-n\theta) = \cos n\theta - i\sin n\theta\). Adding gives \(2\cos n\theta\); subtracting gives \(2i\sin n\theta\).
(b) \(z^3 + z^{-3} = 2\cos\dfrac{6\pi}{9} = 2\cos\dfrac{2\pi}{3} = -1\).
10 Let \(z = 2\operatorname{cis}\dfrac{\pi}{12}\). Find the two smallest positive integers \(n\) for which \(z^n\) is purely imaginary with a positive imaginary part, and find \(z^n\) for the smaller one.
\(\arg(z^n) = \dfrac{n\pi}{12}\) must equal \(\dfrac{\pi}{2} + 2k\pi\): \(\;n = 6 + 24k\). So \(n = 6\) and \(n = 30\).
\(z^6 = 64\operatorname{cis}\dfrac{\pi}{2} = 64i\).