De Moivre's theorem, $n$th roots, and the connection between complex characteristic roots and real-valued solutions
For any integer $n$:
Equivalently, $\bigl(e^{i\theta}\bigr)^n = e^{in\theta}$. This follows directly from the laws of exponents applied to Euler's formula.
De Moivre's theorem has two main applications: computing high powers of complex numbers (covered in 7.2), and finding roots.
Compute $\bigl(\frac{\sqrt{3}}{2} + \frac{1}{2}i\bigr)^{12}$.
Recognize: $\frac{\sqrt{3}}{2} + \frac{1}{2}i = \cos\frac{\pi}{6} + i\sin\frac{\pi}{6}$ (modulus = 1).
By De Moivre:
De Moivre's theorem, run in reverse, lets us find all $n$ distinct $n$th roots of any complex number.
The $n$ distinct $n$th roots of $z = re^{i\theta}$ are:
These $n$ roots are equally spaced around a circle of radius $r^{1/n}$, separated by an angle of $\frac{2\pi}{n}$.
The $n$th roots of any complex number form the vertices of a regular $n$-gon inscribed in a circle. The cube roots of 1 form an equilateral triangle; the 4th roots of 1 form a square; the 5th roots of 1 form a regular pentagon.
Find all three cube roots of $z = -8$.
Write $-8 = 8 e^{i\pi}$ (modulus 8, argument $\pi$).
$k = 0$: $z_0 = 2e^{i\pi/3} = 2\bigl(\frac{1}{2} + \frac{\sqrt{3}}{2}i\bigr) = 1 + \sqrt{3}\,i$
$k = 1$: $z_1 = 2e^{i\pi} = -2$ (the obvious real root)
$k = 2$: $z_2 = 2e^{i5\pi/3} = 2\bigl(\frac{1}{2} - \frac{\sqrt{3}}{2}i\bigr) = 1 - \sqrt{3}\,i$
Verify: $(-2)^3 = -8$ and $(1+\sqrt{3}\,i)^3 = -8$ (check it!).
Find all solutions to $z^4 = 1$.
$1 = 1 \cdot e^{i \cdot 0}$, so:
These form a square in the complex plane (vertices on the unit circle at $0°, 90°, 180°, 270°$).
This is the culmination of Chapter 7 and its direct connection to Chapter 2. When we solve a second-order linear ODE with constant coefficients, the characteristic equation may have complex roots. Here is the complete procedure:
1 Write the characteristic equation. For $ay'' + by' + cy = 0$, form $ar^2 + br + c = 0$.
2 Solve for $r$. Use the quadratic formula: $r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.
3 Identify the case.
| Discriminant | Root Type | General Solution |
|---|---|---|
| $b^2 - 4ac > 0$ | Two distinct real roots $r_1, r_2$ | $y = c_1 e^{r_1 t} + c_2 e^{r_2 t}$ |
| $b^2 - 4ac = 0$ | Repeated real root $r$ | $y = (c_1 + c_2 t)e^{rt}$ |
| $b^2 - 4ac < 0$ | Complex conjugate pair $\alpha \pm \beta i$ | $y = e^{\alpha t}(c_1 \cos\beta t + c_2 \sin\beta t)$ |
4 For complex roots $r = \alpha \pm \beta i$: Extract $\alpha = \frac{-b}{2a}$ and $\beta = \frac{\sqrt{4ac - b^2}}{2a}$, then write the real-valued general solution.
The two complex solutions $e^{(\alpha+\beta i)t}$ and $e^{(\alpha-\beta i)t}$ are complex conjugates of each other. By taking their sum and difference (divided by 2 and $2i$ respectively), we get two independent real solutions: $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$. This is exactly what Euler's formula enables.
Solve $y'' + 4y = 0$, $y(0) = 3$, $y'(0) = -2$.
Step 1: Characteristic equation: $r^2 + 4 = 0$.
Step 2: $r = \pm 2i$. So $\alpha = 0$, $\beta = 2$.
Step 3: General solution:
Step 4: Apply initial conditions:
$y(0) = c_1 = 3$
$y'(t) = -2c_1 \sin 2t + 2c_2 \cos 2t$, so $y'(0) = 2c_2 = -2$, giving $c_2 = -1$.
Pure oscillation (no growth or decay since $\alpha = 0$).
Solve $y'' + 2y' + 5y = 0$, $y(0) = 1$, $y'(0) = 0$.
Step 1: Characteristic equation: $r^2 + 2r + 5 = 0$.
Step 2: $r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i$
So $\alpha = -1$, $\beta = 2$.
Step 3: General solution:
Step 4: $y(0) = c_1 = 1$.
$y'(t) = e^{-t}[(-c_1 + 2c_2)\cos 2t + (-c_2 - 2c_1)\sin 2t]$
$y'(0) = -c_1 + 2c_2 = -1 + 2c_2 = 0$, so $c_2 = \frac{1}{2}$.
The factor $e^{-t}$ causes the amplitude to decay to zero as $t \to \infty$ (damped oscillation).
Find the general solution of $y''' - y'' + y' - y = 0$.
Step 1: $r^3 - r^2 + r - 1 = 0$. Factor by grouping: $r^2(r-1) + 1(r-1) = (r^2+1)(r-1) = 0$.
Step 2: Roots: $r = 1$ and $r = \pm i$.
Step 3: From $r = 1$: $e^t$. From $r = \pm i$ ($\alpha = 0$, $\beta = 1$): $\cos t$ and $\sin t$.
The form $y = e^{\alpha t}(c_1 \cos\beta t + c_2 \sin\beta t)$ has a clear physical meaning that depends on the sign of $\alpha$:
| Condition | Behavior | Physical Example |
|---|---|---|
| $\alpha < 0$ | Decaying oscillation | Underdamped spring-mass, RLC circuit with resistance |
| $\alpha = 0$ | Sustained oscillation (no damping) | Ideal LC circuit, frictionless spring |
| $\alpha > 0$ | Growing oscillation (unstable) | Positive feedback in amplifier, flutter |
The angular frequency is $\beta$ rad/s, so the natural frequency is $f = \frac{\beta}{2\pi}$ Hz and the period is $T = \frac{2\pi}{\beta}$ seconds.
When $\alpha < 0$, the time constant is $\tau = \frac{1}{|\alpha|}$ — the time for the amplitude to decay by a factor of $e \approx 2.718$.
For the series RLC circuit $L\ddot{q} + R\dot{q} + \frac{1}{C}q = 0$, the characteristic roots are $r = -\frac{R}{2L} \pm i\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$ when $R^2 < \frac{4L}{C}$. Here $\alpha = -\frac{R}{2L}$ (always negative, since $R > 0$) and $\beta = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$. The current oscillates at the damped frequency $\beta$ while decaying exponentially.
These problems are similar in style and difficulty to past exam questions. Click each problem to reveal the step-by-step solution.
Step 1: Write and solve the characteristic equation
$$r^2 + 6r + 13 = 0 \;\Longrightarrow\; r = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2} = \frac{-6 \pm 4i}{2}$$ $$r = -3 \pm 2i \qquad\text{so}\qquad \alpha = -3,\; \beta = 2$$Step 2: Convert the complex roots to the real general solution
Roots $\alpha \pm \beta i$ always give the pair $e^{\alpha t}\cos\beta t$ and $e^{\alpha t}\sin\beta t$:
$$y(t) = e^{-3t}\left(c_1\cos 2t + c_2\sin 2t\right)$$Step 3: Apply $y(0) = 2$
$$y(0) = e^{0}(c_1 \cdot 1 + c_2 \cdot 0) = c_1 = 2$$Step 4: Differentiate with the product rule
$$y'(t) = e^{-3t}\left[(-3c_1 + 2c_2)\cos 2t + (-3c_2 - 2c_1)\sin 2t\right]$$At $t = 0$ only the cosine term survives:
$$y'(0) = -3c_1 + 2c_2 = -6 + 2c_2 = -1 \;\Longrightarrow\; c_2 = \frac{5}{2}$$Step 5: Write the particular solution
$$\boxed{y(t) = e^{-3t}\left(2\cos 2t + \tfrac{5}{2}\sin 2t\right)}$$Step 1: Put $-16$ in polar form
$$-16 = 16e^{i\pi} \qquad (r = 16,\; \theta = \pi)$$Step 2: Apply the $n$th-root formula with $n = 4$
$$w_k = 16^{1/4}\exp\!\left(i\,\frac{\pi + 2\pi k}{4}\right) = 2\exp\!\left(i\,\frac{\pi + 2\pi k}{4}\right), \qquad k = 0, 1, 2, 3$$All four roots have modulus $2$ and are spaced $2\pi/4 = 90^\circ$ apart.
Step 3: List the angles and convert
$$k=0:\; 2e^{i\pi/4} = \sqrt{2} + i\sqrt{2} \qquad k=1:\; 2e^{i3\pi/4} = -\sqrt{2} + i\sqrt{2}$$ $$k=2:\; 2e^{i5\pi/4} = -\sqrt{2} - i\sqrt{2} \qquad k=3:\; 2e^{i7\pi/4} = \sqrt{2} - i\sqrt{2}$$Taking $k = 4$ would give the angle $9\pi/4 = \pi/4 + 2\pi$, which is the same number as $k = 0$ — there are exactly four distinct roots.
Step 4: Recognise these as characteristic roots
The equation $y^{(4)} + 16y = 0$ has characteristic equation $r^4 + 16 = 0$, i.e. $r^4 = -16$. So the roots we just found are the characteristic roots, arriving as two conjugate pairs:
$$r = \sqrt{2} \pm i\sqrt{2} \qquad\text{and}\qquad r = -\sqrt{2} \pm i\sqrt{2}$$Step 5: Translate each pair into real solutions
$$\boxed{y(t) = e^{\sqrt{2}\,t}\!\left(c_1\cos\sqrt{2}\,t + c_2\sin\sqrt{2}\,t\right) + e^{-\sqrt{2}\,t}\!\left(c_3\cos\sqrt{2}\,t + c_4\sin\sqrt{2}\,t\right)}$$A $1$ kg mass on a spring with stiffness $k = 20$ N/m and damping $c = 4$ N·s/m is struck so that it starts at equilibrium with velocity $8$ m/s.
Step 1: Characteristic equation
$$r^2 + 4r + 20 = 0 \;\Longrightarrow\; r = \frac{-4 \pm \sqrt{16 - 80}}{2} = \frac{-4 \pm 8i}{2} = -2 \pm 4i$$The discriminant is negative, which is precisely the underdamped case $c^2 < 4mk$: here $16 < 80$.
Step 2: General real solution
$$x(t) = e^{-2t}\left(c_1\cos 4t + c_2\sin 4t\right)$$Step 3: Apply $x(0) = 0$
$$x(0) = c_1 = 0 \;\Longrightarrow\; x(t) = c_2 e^{-2t}\sin 4t$$Step 4: Apply $x'(0) = 8$
$$x'(t) = c_2 e^{-2t}\left(-2\sin 4t + 4\cos 4t\right), \qquad x'(0) = 4c_2 = 8 \;\Longrightarrow\; c_2 = 2$$Step 5: Particular solution
$$\boxed{x(t) = 2e^{-2t}\sin 4t}$$Step 6: Read the physics off the root $-2 + 4i$
Find all five 5th roots of $32$.
$32 = 32 e^{i \cdot 0}$, so $z_k = 32^{1/5} e^{i \cdot 2\pi k/5} = 2 e^{i \cdot 2\pi k/5}$ for $k=0,1,2,3,4$.
In rectangular form: $z_0 = 2$, $z_1 \approx 0.618 + 1.902i$, $z_2 \approx -1.618 + 1.176i$, $z_3 \approx -1.618 - 1.176i$, $z_4 \approx 0.618 - 1.902i$.
These form a regular pentagon inscribed in a circle of radius 2.
Solve $y'' + 6y' + 13y = 0$, $y(0) = 2$, $y'(0) = -1$.
Characteristic equation: $r^2 + 6r + 13 = 0$.
$r = \frac{-6 \pm \sqrt{36 - 52}}{2} = \frac{-6 \pm \sqrt{-16}}{2} = -3 \pm 2i$
General solution: $y = e^{-3t}(c_1\cos 2t + c_2\sin 2t)$.
$y(0) = c_1 = 2$.
$y'(t) = e^{-3t}[(-3c_1+2c_2)\cos 2t + (-3c_2-2c_1)\sin 2t]$.
$y'(0) = -3c_1 + 2c_2 = -6 + 2c_2 = -1$, so $c_2 = \frac{5}{2}$.
For the equation $y'' + \omega^2 y = 0$, show that the general solution can be written in the amplitude-phase form $y = A\cos(\omega t - \phi)$ and express $A$ and $\phi$ in terms of initial conditions $y(0) = y_0$ and $y'(0) = v_0$.
General solution: $y = c_1\cos\omega t + c_2\sin\omega t$ with $c_1 = y_0$ and $c_2 = v_0/\omega$.
Using the identity $c_1\cos\omega t + c_2\sin\omega t = A\cos(\omega t - \phi)$:
$A$ is the amplitude and $\phi$ is the phase shift. This form is preferred in engineering because it explicitly separates amplitude from timing.
An RLC circuit has $L = 0.5$ H, $R = 4$ $\Omega$, $C = 0.04$ F. Write the general solution for the charge $q(t)$.
ODE: $0.5\ddot{q} + 4\dot{q} + 25q = 0$, or equivalently $\ddot{q} + 8\dot{q} + 50q = 0$.
$r = \frac{-8 \pm \sqrt{64 - 200}}{2} = \frac{-8 \pm \sqrt{-136}}{2} = -4 \pm i\sqrt{34}$
Since $\sqrt{34} \approx 5.83$:
This is underdamped: oscillation at $\approx 5.83$ rad/s with exponential decay at rate 4. The damped frequency is $\frac{\sqrt{34}}{2\pi} \approx 0.93$ Hz.
Complex numbers are not an abstract curiosity — they are the natural language for oscillatory phenomena. Every time a differential equation produces complex roots, those roots encode both the frequency and the growth/decay rate of the physical solution. Euler's formula $e^{i\theta} = \cos\theta + i\sin\theta$ is the translator between the compact complex exponential form and the real trigonometric form you can measure in the lab.
| Concept | Formula / Rule |
|---|---|
| Complex arithmetic | FOIL with $i^2 = -1$; divide by multiplying by $\bar{z}/\bar{z}$ |
| Polar form | $z = re^{i\theta}$; multiply moduli, add arguments |
| Euler's formula | $e^{i\theta} = \cos\theta + i\sin\theta$ |
| $n$th roots | $n$ equally-spaced points on circle of radius $r^{1/n}$ |
| Complex DE roots $\alpha \pm \beta i$ | $y = e^{\alpha t}(c_1\cos\beta t + c_2\sin\beta t)$ |
Six questions covering the whole chapter. Your answers are recorded so the AI tutor can tailor its recommendations — answer honestly rather than guessing.