Complex Exponential Representation of Oscillations

Open any wireless chip shipped in 2026 — the 5G modem in a phone, a Wi-Fi 7 radio, a Bluetooth earbud — and the heart of it is a pair of signal paths labeled I and Q: in-phase and quadrature, the real and imaginary parts of a complex baseband signal. Every symbol the radio transmits is a phasor $\tilde{A} = Ae^{i\phi}$, a point in the complex plane chosen from a constellation diagram; every received symbol is demodulated by multiplying against $e^{-i\omega t}$ and low-pass filtering — a hardware implementation of "factor out the rotation and freeze the phasor." Equation (3) is executed billions of times per second in your pocket. The same complex arithmetic runs the electric grid, where power engineers dispatch gigawatts by managing phasor angles between cities (with $j$, following Steinmetz, since $i$ was taken by current), and monitor grid stability through GPS-synchronized phasor measurement units streaming $\tilde{A}$ thirty times a second. The deepest tenants, though, are quantum. The wavefunction of a stationary state evolves as $e^{-iEt/\hbar}$ — not as a bookkeeping convenience but as physical law; interference of those complex phases is quantum mechanics. Lock-in amplifiers pull femtovolt signals out of noise a million times larger by multiplying against a reference $e^{i\omega t}$ and time-averaging — equation (10)'s constant term, weaponized. When a gravitational-wave observatory reports a chirp, the strain data was heterodyned, filtered, and matched — every step phasor arithmetic. Euler's spinning arrow turned out to be one of the load-bearing structures of the technological world.

Picture a clock hand of length $A$ spinning counterclockwise in a plane, sweeping out angle at a steady $\omega$ radians per second. Shine a light from above and watch its shadow on the horizontal axis: the shadow oscillates back and forth, compressing near the ends, sweeping fastest through the middle — exactly simple harmonic motion. That plane is the complex plane, the spinning hand is $\tilde{A}e^{i\omega t}$, and the shadow is $\operatorname{Re}(\tilde{A}e^{i\omega t}) = A\cos(\omega t + \phi)$. Every sinusoid you will ever meet is the shadow of uniform circular motion — which is why simple harmonic motion and circular motion share one $\omega$, and why the geometry of circles keeps solving the algebra of oscillations.

Freeze the arrow at $t = 0$ and you hold a single complex number $\tilde{A} = Ae^{i\phi}$: its length is the amplitude, its angle is the starting phase. This frozen snapshot is called a phasor, and it is the profound bookkeeping upgrade of this whole subject — the two real numbers you used to carry separately ($A$ and $\phi$, or equivalently the coefficients of $\cos\omega t$ and $\sin\omega t$) fuse into one algebraic object. Multiplying by $e^{i\omega t}$ sets it spinning; multiplying by any other complex number stretches it and turns it. All the algebra of oscillations becomes the geometry of rotating arrows. Why is this so much easier than trigonometry? Because $\cos$ and $\sin$ are two functions that the derivative shuffles into each other, while $e^{i\omega t}$ is a single function that the derivative merely rescales. In the language of linear algebra: the exponential is an eigenfunction of $\mathrm{d}/\mathrm{d}t$ with eigenvalue $i\omega$. Working in the eigenbasis of the operator that runs the whole problem diagonalizes everything — the same reason you diagonalize matrices, choose normal modes, or Fourier-transform a signal. And phase, that perennial source of sign errors, becomes something you can see: the angle of an arrow. When a driven oscillator's response lags its drive, the response phasor literally hangs at an angle behind the drive phasor, and sweeping the drive frequency rotates it smoothly from $0°$ through $90°$ at resonance to $180°$ — a rotation that is invisible in real-trig algebra and obvious in the complex plane.

Everything rests on one identity. It deserves a slow look before we put it to work, because it is the bridge between the exponential function — the natural language of growth and decay — and the circular functions — the natural language of oscillation. The claim is that a purely imaginary exponent converts growth into rotation:

where: $i$ — the imaginary unit, $i^2 = -1$; $\theta$ — a real angle in radians. Read it geometrically: $e^{i\theta}$ is the point on the unit circle at angle $\theta$, with the cosine as its horizontal coordinate and the sine as its vertical coordinate. Setting $\theta = \pi$ gives the famous special case $e^{i\pi} = -1$: half a turn lands you at minus one.

With Euler's formula in hand, the property that powers the whole method is one differentiation away. Ask what the derivative operator does to a complex exponential of time — the answer is the reason this page exists:

where: $\omega$ — angular frequency (rad/s); $t$ — time. The function comes back unchanged in shape, merely multiplied by $i\omega$: it is an eigenfunction of $\mathrm{d}/\mathrm{d}t$. Iterating, $\mathrm{d}^n e^{i\omega t}/\mathrm{d}t^n = (i\omega)^n e^{i\omega t}$, so an $n$-th order linear ODE acting on $e^{i\omega t}$ becomes a degree-$n$ polynomial in $i\omega$ — calculus has been demoted to algebra. Note the geometric meaning of the eigenvalue: multiplying by $i$ rotates a complex number by $90°$, so the velocity phasor leads the position phasor by a quarter turn and is $\omega$ times longer, exactly the familiar SHM relation $v_{\max} = \omega A$ with its quarter-period lead. The dictionary between the complex bookkeeping and the physical motion is the projection. A real oscillation is the real part of a spinning phasor:

where: $A = |\tilde{A}| \ge 0$ — the real amplitude; $\phi = \arg\tilde{A}$ — the phase offset; $\tilde{A}$ — the complex amplitude, or phasor. One complex number now stores what used to take two reals. In particular the initial conditions live entirely in $\tilde{A}$: the real part of $\tilde{A}$ is $x(0)$, and its imaginary part is $-\dot{x}(0)/\omega$. There is a second, more symmetric way to write the projection, and it repays attention because it is the seed of Fourier analysis. Using $\operatorname{Re}(z) = \tfrac12(z + z^{*})$ for any complex number,

where: $\tilde{A}^{*}$ — the complex conjugate of $\tilde{A}$. A real oscillation is the sum of two counter-rotating phasors: one spinning at $+\omega$, its conjugate twin at $-\omega$. The "negative frequency" that puzzles every student meeting a two-sided Fourier spectrum is nothing exotic — it is the clockwise-spinning half that any purely real signal must carry so that the imaginary parts cancel at every instant.

Now put the machine to work on the equation that owns the rest of this track. A mass $m$ on a spring of stiffness $k$ with a drag force proportional to velocity ($b\dot{x}$) obeys $m\ddot{z} + b\dot{z} + kz = 0$. Rather than guessing combinations of sines, cosines, and decaying envelopes, feed it the one function that differentiation cannot deform — a trial exponential $z = e^{st}$ with $s$ complex and unknown:

where: $s_{1,2}$ — the two roots of the characteristic equation $ms^2 + bs + k = 0$; $c_1, c_2$ — complex constants fixed by the initial position and velocity. Substituting $e^{st}$ turns each time-derivative into a factor of $s$, leaving $(ms^2 + bs + k)e^{st} = 0$; since an exponential never vanishes, the polynomial must. The quadratic formula then delivers the entire physics of damping in one line: $s = \big(-b \pm \sqrt{b^2 - 4mk}\big)/2m$. When $b^2 < 4mk$ the roots are complex, $s = -\beta \pm i\sqrt{\omega_0^2 - \beta^2}$ with $\beta = b/2m$ and $\omega_0 = \sqrt{k/m}$ — a spiral in the complex plane: rotation (oscillation) at the reduced frequency, times shrinkage (decay) at rate $\beta$. When $b^2 > 4mk$ both roots are real and negative — pure decay, no rotation, no oscillation. The full story of damping regimes is the story of where these two roots sit.

The same trial-exponential move solves the driven problem, and this is where the method pays its full fare. The physical equation has a cosine on the right-hand side; we complexify it, solve by division, and project back.

For steady-state work the two payoff formulas deserve their own display. The response phasor of the driven oscillator, and its polar decomposition, are

where: $F_0$ — drive amplitude; $\omega_0 = \sqrt{k/m}$ — natural frequency; $\beta = b/2m$ — damping rate; $\omega$ — drive frequency. One complex ratio holds the complete steady-state response at every drive frequency, and

where: $\delta$ — the phase lag of response behind drive. Amplitude is a modulus, phase is an argument: the resonance curve and the phase curve that fill the next articles of this track are just $|\tilde{A}(\omega)|$ and $\arg\tilde{A}(\omega)$ read off equation (6).

Which equation answers which question

Superposition is where the representation stops being a convenience and becomes a superpower. Add two oscillations of the same frequency with different amplitudes and phases — the kind of sum that appears in interference, in coupled oscillators, in AC circuit nodes — and trigonometry demands the sum-to-product identities. Phasors demand only that you add two arrows:

where: $\tilde{A}$ — the resultant phasor, obtained by ordinary vector addition in the complex plane. The resultant amplitude follows from the law of cosines, $A = \sqrt{A_1^2 + A_2^2 + 2A_1A_2\cos(\phi_1 - \phi_2)}$: constructive when the arrows align, destructive when they oppose, Pythagoras when they are perpendicular.

What if the two frequencies are close but not equal? Then the two phasors rotate at slightly different rates, and their relative angle drifts slowly: the arrows alternately align and oppose, so the resultant amplitude swells and shrinks. Factoring the sum of two equal-amplitude cosines makes this audible:

where: the second factor oscillates at the average frequency, while the first is a slow envelope at half the difference frequency. Your ear hears one loudness maximum per envelope half-cycle, so the perceived beat frequency is the full difference $|f_1 - f_2|$ — two guitar strings at $440$ and $442\ \mathrm{Hz}$ throb twice per second, and a tuner's job is to slow that throb to zero. One last identity carries the method's only serious danger, and it earns its own display precisely because everything up to here has encouraged you to forget the projection and compute with $z$ directly. That habit is safe for anything linear — and fatal for anything quadratic:

where: $z_1, z_2$ — any two complex signals. The product of real parts is not the real part of the product: writing each real part as a conjugate average $\tfrac12(z + z^{*})$ and multiplying out produces cross terms $z_1z_2^{*}$ that $\operatorname{Re}(z_1z_2)$ alone misses. For two phasors at the same frequency, $z_1z_2$ spins at $2\omega$ and time-averages to zero, while $z_1z_2^{*}$ is constant — which yields the single most used corollary in physics: the time-averaged product of two same-frequency oscillations is $\langle x_1x_2\rangle = \tfrac12\operatorname{Re}(\tilde{A}_1\tilde{A}_2^{*})$, and in particular $\langle x^2\rangle = \tfrac12|\tilde{A}|^2$.

1 · Two speakers, one microphone (confidence builder)

Two signals of the same frequency arrive at a microphone: one with amplitude $A_1 = 3\ \mathrm{cm}$ (in suitable units) and phase $\phi_1 = 0°$, the other with $A_2 = 4\ \mathrm{cm}$ and phase $\phi_2 = 90°$. Find the amplitude and phase of the combined oscillation.

Step 1. Encode each signal as a phasor: $\tilde{A}_1 = 3e^{i0} = 3$, and $\tilde{A}_2 = 4e^{i\pi/2} = 4i$ — one arrow along the real axis, one along the imaginary axis.

Step 2. Add the arrows: $\tilde{A} = 3 + 4i$. No identities, no angle-chasing — complex addition is the entire calculation.

Step 3. Read off the amplitude as the modulus: $A = |\tilde{A}| = \sqrt{3^2 + 4^2} = \sqrt{25} = \boxed{5\ \mathrm{cm}}$. The law-of-cosines cross term $2A_1A_2\cos(\phi_1 - \phi_2)$ vanished because the arrows are perpendicular — this is Pythagoras in phasor clothing, the 3-4-5 triangle of the figure above.

Step 4. Read off the phase as the argument: $\phi = \arctan(4/3) = \boxed{53°}$ (first quadrant, since both components are positive). The combined signal is $x(t) = 5\cos(\omega t + 53°)\ \mathrm{cm}$.

2 · A driven cart, three lines instead of two pages (exam level)

A cart of mass $m = 0.5\ \mathrm{kg}$ on a spring $k = 200\ \mathrm{N/m}$ moves through oil giving a drag coefficient $b = 4\ \mathrm{kg/s}$. It is driven by $F(t) = 10\cos(19\,t)\ \mathrm{N}$ (SI units). Find the steady-state amplitude and phase lag, and compare the amplitude with the static stretch produced by a constant 10 N force.

Step 1. Assemble the system constants: $\omega_0 = \sqrt{k/m} = \sqrt{400} = 20\ \mathrm{rad/s}$; $\beta = b/2m = 4/(2 \times 0.5) = 4\ \mathrm{s^{-1}}$; drive $\omega = 19\ \mathrm{rad/s}$, just below resonance; $F_0/m = 20\ \mathrm{N/kg}$.

Step 2. Build the complex denominator of equation (6): real part $\omega_0^2 - \omega^2 = 400 - 361 = 39\ \mathrm{s^{-2}}$; imaginary part $2\beta\omega = 2 \times 4 \times 19 = 152\ \mathrm{s^{-2}}$. Note the imaginary part dominates — we are deep in the resonance region.

Step 3. Amplitude = modulus: $|\tilde{A}| = \dfrac{20}{\sqrt{39^2 + 152^2}} = \dfrac{20}{\sqrt{1521 + 23104}} = \dfrac{20}{156.9} = \boxed{0.127\ \mathrm{m} \approx 12.7\ \mathrm{cm}}$.

Step 4. Phase lag = argument: $\delta = \arctan(152/39) = \boxed{75.6°}$ — the response trails the drive by just over a fifth of a cycle, on its way to the $90°$ lag it would show exactly at $\omega_0$.

Step 5. Compare with statics: a constant 10 N force stretches the spring by $F_0/k = 10/200 = 0.05\ \mathrm{m} = 5\ \mathrm{cm}$. Driving near resonance amplifies this to $12.7\ \mathrm{cm}$ — a factor of $2.5$, which is no accident: the quality factor of this oscillator is $Q = \omega_0/2\beta = 20/8 = 2.5$, and $Q$ is precisely the resonant amplification of the static response. The complex division knew all of resonance physics before we asked.

References. Morin, Introduction to Classical Mechanics, ch. 4 (Oscillations) · Goldstein, Poole & Safko, Classical Mechanics, 3rd ed., ch. 6 · Landau & Lifshitz, Mechanics, 3rd ed., §21–§26 · Taylor, Classical Mechanics, ch. 5 · Feynman, Lectures on Physics, vol. I, chs. 22–24 (algebra, resonance, transients).

Provenance. Euler's formula and series expansion verifiable from any analysis text. Driven steady-state amplitude/phase match Taylor Classical Mechanics ch. 5 and Goldstein ch. 6; damped characteristic roots standard. Phasor superposition reduces to the law of cosines. A verifier should recompute the complex division for $\tilde A$, confirm $\langle\cos^2\rangle=1/2$, and check the phase limits $\delta(0)=0,\ \delta(\omega_0)=90^\circ,\ \delta(\infty)=180^\circ$. Worked-example numbers on this page ($A = 5\ \mathrm{cm},\ \phi = 53°$ for the phasor sum; $|\tilde{A}| = 12.7\ \mathrm{cm},\ \delta = 75.6°,\ Q = 2.5$ for the driven cart) are illustrative arithmetic recomputed from the boxed formulas.

Builds on: Simple Harmonic Motion · Amplitude, Phase, and Angular Frequency · Small Oscillations About Equilibrium. Leads to: Damped Harmonic Oscillator · Driven Oscillations · Resonance · Transient and Steady-State Response.