Complex Numbers

Lessons

Senior 6 · Mathematics

Complex Numbers

5 lessons

Lesson 5 of 5

Linearisation, trigonometric equations and alternating current

25 min

From powers to multiple angles, and on to circuits

Complex numbers began as a trick for solving equations, but their most common use today is practical: every electrical engineer who designs a supply network, a phone charger or a radio works with them. This lesson collects three applications of Lessons 3 and 4. Euler's formulae turn a power such as $\cos^3\theta$ into a sum of cosines of multiple angles (you will need this to integrate such powers in Unit 4); the exponential form solves every equation $a\cos x + b\sin x = c$ ; and a single complex number, the impedance, describes how a circuit resists and delays an alternating current.

Definition

Linearisation: rewriting a power or product of $\cos\theta$ and $\sin\theta$ as a sum of terms $\cos k\theta$ and $\sin k\theta$ , with no powers left.

Method: write $\cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}$ or $\sin\theta = \frac{e^{i\theta} - e^{-i\theta}}{2i}$ , expand with the binomial theorem, and pair each $e^{ik\theta}$ with $e^{-ik\theta}$ , using $e^{ik\theta} + e^{-ik\theta} = 2\cos k\theta$ and $e^{ik\theta} - e^{-ik\theta} = 2i\sin k\theta$ .

Example: $\cos^2\theta = \frac{1}{2} + \frac{1}{2}\cos 2\theta$ .

Worked example 1: linearise $\cos^3\theta$

Solution

\begin{aligned} \cos^3\theta &= \left(\frac{e^{i\theta} + e^{-i\theta}}{2}\right)^3 = \frac{e^{3i\theta} + 3e^{i\theta} + 3e^{-i\theta} + e^{-3i\theta}}{8} \\ &= \frac{2\cos 3\theta + 6\cos\theta}{8} = \frac{1}{4}\cos 3\theta + \frac{3}{4}\cos\theta . \end{aligned}

Check: this agrees with Lesson 3's $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ , rearranged. At $\theta = 0$ : $1 = \frac{1}{4} + \frac{3}{4}$ ✓.

Worked example 2: linearise $\sin^4\theta$

Solution Since $(2 i)^{4} = 16$ ,

\begin{aligned} \sin^4\theta &= \frac{\left(e^{i\theta} - e^{-i\theta}\right)^4}{16} = \frac{e^{4i\theta} - 4e^{2i\theta} + 6 - 4e^{-2i\theta} + e^{-4i\theta}}{16} \\ &= \frac{2\cos 4\theta - 8\cos 2\theta + 6}{16} = \frac{1}{8}\cos 4\theta - \frac{1}{2}\cos 2\theta + \frac{3}{8} . \end{aligned}

Check at $\theta = \frac{\pi}{2}$ : $\sin^4\frac{\pi}{2} = 1$ and $\frac{1}{8}(1) - \frac{1}{2}(-1) + \frac{3}{8} = 1$ ✓.

Equations of the form $a\cos x + b\sin x = c$

You met these in S5 (Trigonometric Formulae, Equations and Inequalities). Complex numbers show where the standard method comes from. Let $a$ and $b$ be real, not both 0, and write $a + bi = Re^{i\alpha}$ with $R = \sqrt{a^2 + b^2}$ . Then $a - bi = Re^{-i\alpha}$ , and

\begin{aligned} (a - bi)e^{ix} &= (a - bi)(\cos x + i\sin x) \\ &= (a\cos x + b\sin x) + i(a\sin x - b\cos x). \end{aligned}

The same product is $Re^{-i\alpha}e^{ix} = Re^{i(x - \alpha)}$ , whose real part is $R\cos(x - \alpha)$ . Comparing real parts:

a\cos x + b\sin x = R\cos(x - \alpha),

where $R = \sqrt{a^2 + b^2}$ and $\alpha = \arg(a + bi)$ .

So the equation $a\cos x + b\sin x = c$ has solutions if and only if $|c| \le R$ , and the expression takes every value from $- R$ to $R$ .

Worked example 3: solving $\sqrt{3}\cos x + \sin x = \sqrt{2}$

Find the general solution, and the solutions in $[0, 2\pi)$ .

Solution $a + bi = \sqrt{3} + i = 2e^{\frac{i\pi}{6}}$ , so $R = 2$ , $\alpha = \frac{\pi}{6}$ , and the equation becomes

2\cos\left(x - \frac{\pi}{6}\right) = \sqrt{2} \quad\Longrightarrow\quad \cos\left(x - \frac{\pi}{6}\right) = \frac{\sqrt{2}}{2} .

So $x - \frac{\pi}{6} = \pm\frac{\pi}{4} + 2k\pi$ , and $x = \frac{5\pi}{12} + 2k\pi$ or $x = -\frac{\pi}{12} + 2k\pi$ , $k \in \mathbb{Z}$ . In $[0, 2\pi)$ : $x = \frac{5\pi}{12}$ or $x = \frac{23\pi}{12}$ .

Check: $\sqrt{3}\cos\frac{5\pi}{12} + \sin\frac{5\pi}{12} \approx 0.448 + 0.966 = 1.414 \approx \sqrt{2}$ ✓.

$y = \sqrt{3}\cos x + \sin x = 2\cos\left(x - \frac{\pi}{6}\right)$ , plotted from computed points. It meets $y = \sqrt{2}$ twice in $[0, 2\pi)$ .

Warning

Find $\alpha$ from $a + b i$ , not from $\tan\alpha = \frac{b}{a}$ alone. For $-\cos x + \sin x$ , $\tan\alpha = -1$ suggests $-\frac{\pi}{4}$ , but $- 1 + i$ is in the second quadrant, so $\alpha = \frac{3\pi}{4}$ . Sketching $a + b i$ settles it.

Alternating current

Mains electricity is alternating current (AC): the voltage swings between positive and negative values many times a second, following $v(t) = V_m\cos(\omega t + \varphi)$ , where $V_{m}$ is the amplitude, $\omega$ the angular frequency in $\mathrm{rad\,s^{-1}}$ ( $\omega = 2\pi f$ for a frequency $f$ in hertz) and $\varphi$ the phase. Since $v(t) = \operatorname{Re}\left(V_m e^{i\varphi}e^{i\omega t}\right)$ , the whole signal is described by one complex number, the phasor $V = V_m e^{i\varphi}$ . Engineers write $j$ for $\sqrt{-1}$ , so the formulas below use $j$ .

Definition

Impedance $Z$ of a circuit element: the complex number with $V = Z I$ , where $V$ and $I$ are the phasors of the voltage and the current. It is measured in ohms ( $\Omega$ ).

Element	Impedance	Effect on the current
resistor, resistance $R$	$Z = R$	in phase with the voltage
inductor, inductance $L$	$Z = j\omega L$	lags the voltage by $90^\circ$
capacitor, capacitance $C$	$Z = \frac{1}{j\omega C} = -\frac{j}{\omega C}$	leads the voltage by $90^\circ$

Combining: in series, impedances add, $Z = Z_{1} + Z_{2}$ ; in parallel, $\frac{1}{Z} = \frac{1}{Z_1} + \frac{1}{Z_2}$ , exactly like resistances. Then $|I| = \frac{|V|}{|Z|}$ , and the current lags the voltage by $\arg Z$ .

Worked example

Worked example 4: a series circuit (illustrative values)

A resistor of $30\,\Omega$ and an inductor of reactance $\omega L = 40\,\Omega$ are in series across a supply of amplitude $100\,\mathrm{V}$ . Find the amplitude of the current and its phase relative to the voltage. What changes if a capacitor of reactance $\frac{1}{\omega C} = 80\,\Omega$ is added in series?

Solution $Z = 30 + 40 j$ , so $|Z| = \sqrt{30^2 + 40^2} = 50\,\Omega$ and $\arg Z = \tan^{-1}\frac{40}{30} \approx 0.927\,\mathrm{rad} \approx 53.1^\circ$ . The current amplitude is $\frac{100}{50} = 2\,\mathrm{A}$ , and the current lags the voltage by about $53.1^\circ$ .

With the capacitor, $Z = 30 + 40 j - 80 j = 30 - 40 j$ . Still $|Z| = 50\,\Omega$ , so the amplitude is still $2\,\mathrm{A}$ , but now $\arg Z \approx -53.1^\circ$ : the current leads the voltage by about $53.1^\circ$ .

For a resistor and an inductor in series, $Z = R + j X$ . The modulus $∣ Z ∣$ links the voltage and current amplitudes; the argument $\varphi$ is how far the current lags behind the voltage.

Tip

In circuit questions, draw the impedance as a point on an Argand diagram. Inductors push it up, capacitors push it down, and resistors push it right. The modulus and the argument can then be read like the sides and the angle of a right triangle.

Decipher Physics: Phasors and Complex Numbers in AC Electricity (Full Visual Explanation). Watch for how a rotating arrow, the phasor, stands for a sinusoidal voltage, and how the angle between the voltage and current phasors is the phase difference.

✦

Practice

Questions 6 to 8 are about circuits; give angles in degrees to one decimal place.

Use Euler's formulae to show that $\sin^3\theta = \frac{3}{4}\sin\theta - \frac{1}{4}\sin 3\theta$ .
Linearise $\cos^4\theta$ .
Write $3\cos x - 4\sin x$ in the form $R\cos(x - \alpha)$ with $R \gt 0$ and $-\pi \lt \alpha \le \pi$ . State its greatest and least values.
Solve $\cos x + \sin x = 1$ for $x \in [0, 2\pi)$ .
Show that $3\cos x + 4\sin x = 6$ has no real solution.
A resistor of $12\,\Omega$ and an inductor of reactance $5\,\Omega$ are in series across a supply of amplitude $26\,\mathrm{V}$ . Find the amplitude of the current and its phase lag.
A capacitor of $100\,\mu\mathrm{F}$ and a resistor of $20\,\Omega$ are in series on a $50\,\mathrm{Hz}$ supply. Find the capacitor's reactance, the impedance $Z$ , $∣ Z ∣$ and $\arg Z$ .
Two branches with impedances $Z_{1} = 3 + 4 j$ and $Z_{2} = 3 - 4 j$ (in ohms) are connected in parallel. Find the total impedance and comment on it.

Show answersHide answers

$\sin^3\theta = \frac{\left(e^{i\theta} - e^{-i\theta}\right)^3}{(2i)^3} = \frac{e^{3i\theta} - e^{-3i\theta} - 3\left(e^{i\theta} - e^{-i\theta}\right)}{-8i} = \frac{2i\sin 3\theta - 6i\sin\theta}{-8i} = \frac{3}{4}\sin\theta - \frac{1}{4}\sin 3\theta$ .
$\cos^4\theta = \frac{e^{4i\theta} + 4e^{2i\theta} + 6 + 4e^{-2i\theta} + e^{-4i\theta}}{16} = \frac{1}{8}\cos 4\theta + \frac{1}{2}\cos 2\theta + \frac{3}{8}$ .
$a + b i = 3 - 4 i$ : $R = 5$ , $\alpha = -\tan^{-1}\frac{4}{3} \approx -0.927$ . So $3\cos x - 4\sin x = 5\cos(x + 0.927)$ ; greatest 5, least $- 5$ .
$R = \sqrt{2}$ , $\alpha = \frac{\pi}{4}$ : $\cos\left(x - \frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$ , so $x - \frac{\pi}{4} = \pm\frac{\pi}{4} + 2k\pi$ . In $[0, 2\pi)$ : $x = 0$ or $x = \frac{\pi}{2}$ .
$R = \sqrt{9 + 16} = 5$ , so $3\cos x + 4\sin x \le 5 \lt 6$ for every $x$ .
$Z = 12 + 5 j$ , $|Z| = 13\,\Omega$ , so the current amplitude is $\frac{26}{13} = 2\,\mathrm{A}$ , lagging by $\tan^{-1}\frac{5}{12} \approx 22.6^\circ$ .
$\frac{1}{\omega C} = \frac{1}{2\pi \times 50 \times 100 \times 10^{-6}} \approx 31.8\,\Omega$ . $Z \approx 20 - 31.8j$ , $|Z| \approx 37.6\,\Omega$ , $\arg Z \approx -57.9^\circ$ (the current leads by $57.9^\circ$ ).
$Z = \frac{Z_1 Z_2}{Z_1 + Z_2} = \frac{(3 + 4j)(3 - 4j)}{6} = \frac{25}{6} \approx 4.17\,\Omega$ . It is real: the inductive and capacitive effects cancel, so the pair behaves like a pure resistor.

1From powers to multiple angles, and on to circuits

2Equations of the form acos⁡x+bsin⁡x=c a\cos x + b\sin x = c acosx+bsinx=c

3Alternating current

From powers to multiple angles, and on to circuits

Equations of the form $a\cos x + b\sin x = c$

Alternating current