Complex exponentials & phasors

Drag Angle around the full circle and watch the two readouts, one point written two ways, always agree.

One point, two ways to write it

In 3.3 you multiplied points by rotating and scaling: multiply two complex numbers together and their lengths multiply while their angles add. I want to show you a shorthand for that rule, one you’ve probably already seen written as ”$e$ to the $j\theta$” without ever being told what it’s shorthand for.

Here’s the idea. Take the point at angle $\theta$ on the unit circle, $\cos\theta + j\sin\theta$. Multiply it by the point at angle $\phi$, and by 3.3’s rule the lengths (both 1) multiply to stay 1, and the angles add: you land on the point at angle $\theta+\phi$. That’s exactly how ordinary exponents behave, where multiplying $x^\theta$ by $x^\phi$ adds the exponents to give $x^{\theta+\phi}$. Because “the point at angle $\theta$” obeys the same adding-exponents rule, mathematicians write it using exponent notation too, with a particular number $e$ (you may have met it as the base of natural growth) raised to the power $j\theta$:

This is Euler’s formula. It earns the name “exponential” honestly, not by analogy: it obeys the one rule that defines what an exponential is, $e^{j\theta_1}\cdot e^{j\theta_2} = e^{j(\theta_1+\theta_2)}$, because that’s just 3.3’s rotate-and-scale rule in disguise. Read $e^{j\theta}$, whenever you meet it, simply as “the point on the unit circle at angle $\theta$.”

At $\theta=60°=\pi/3$ radians: $e^{j\pi/3} = \cos60° + j\sin60° = 0.5 + 0.866j$.

In the picture at the top of the page, the point’s across value is always the cosine of the angle, and its up value is always the sine: $e^{j\theta}$ isn’t a new calculation, it’s the same point you already know how to find.

A point that keeps spinning

Fix an angle and $e^{j\theta}$ sits still. But let the angle grow steadily with time instead, $\theta = \omega t$, where $\omega$ (in radians per second) sets how fast it grows, and the point starts spinning around the circle at a constant rate. A point like this, $e^{j\omega t}$, is called a phasor. Picture a lighthouse beam sweeping around at a steady rate: the beam itself is the phasor, and the patch of light it casts on a straight wall to the side is one of its shadows.

The phasor’s horizontal shadow, its across value over time, traces $\cos\omega t$. Its vertical shadow, its up value over time, traces $\sin\omega t$. These are exactly the sine and cosine waves from 3.2, now revealed as the two projections of one spinning point.

Drag Spin rate (turns per second) up and down and watch both shadow traces speed up and slow down together.

They always keep the same frequency as the spin. Notice cosine (the horizontal shadow) always leads sine (the vertical shadow) by a quarter turn, exactly the phase relationship you saw between them back in 3.2.

Two points spinning opposite ways

$e^{-j\omega t}$ is the same idea with the angle growing in the other direction: it spins clockwise at the same rate that $e^{j\omega t}$ spins counterclockwise. This second phasor turns out to be exactly what you need to get a real sinusoid, with no leftover imaginary part, out of spinning points.

Add a phasor to its opposite-spinning twin and divide by 2:

Picture two dancers spinning in mirror-image directions at the same speed: at every instant they sit the same height above or below the floor, one on each side, so their heights are always exact opposites and their sideways positions always match. Averaging the two points cancels the vertical parts completely and leaves only the shared horizontal motion, a plain cosine. Check it at $\theta=60°$: $(0.5+0.866j + 0.5-0.866j)/2 = 0.5 = \cos60°$, exactly.

This is worth sitting with, because it says every real sinusoid you’ve ever measured is secretly two counter-rotating phasors, added together and halved. The “negative frequency” $-\omega$ isn’t a strange negative quantity, it’s just the second point’s spin direction.

Switch Show the mirror pair between Shown and Hidden and watch the second, opposite-spinning point appear and disappear.

Notice, with the pair shown, the two points’ vertical coordinates are always exact opposites: whatever one is above the horizontal axis, the other is exactly as far below it.

Adding two spinning arrows

Two sinusoids at the same frequency $\omega$ are two phasors spinning together, rigidly, always the same angle apart. Because they spin together, their sum is also a plain sinusoid at that same frequency, and you can find it without touching time at all: add the two phasors as fixed arrows, the way you added complex numbers in 3.3, and only then let the combined arrow start spinning.

Picture two speakers playing the same pitch, out of step with each other: they combine into one wave at that same pitch, louder or softer depending on how their phases line up.

Take $3\cos(\omega t) + 4\sin(\omega t)$. Writing $4\sin(\omega t)$ as a cosine shifted by a quarter turn, $4\cos(\omega t - 90°)$, gives two phasors, $3\angle0°$ and $4\angle{-90°}$, which as rectangular points are $3+0j$ and $0-4j$. Adding them:

with magnitude $\sqrt{3^2+4^2}=5$ and an angle of about $-53.13°$ (the same $53.13°$ as in 3.3, but below the axis this time). So the sum is a single wave, $5\cos(\omega t - 53.13°)$.

Drag either arrow and watch all three traces below the plane: phasor 1’s wave, phasor 2’s wave, and their sum.

Notice the sum trace is always a single clean cosine at the same frequency as the other two, and its amplitude and phase read straight off the combined arrow’s own length and angle.

Two tones that almost match

Phasor addition assumed both arrows spin at exactly the same rate. Let them spin at slightly different rates instead, and the combined arrow stops keeping a fixed shape: it slowly drifts in and out of alignment with itself, so the two tones added together swell louder, then fade softer, over and over. This slow throb is called a beat.

Picture two nearly-in-tune guitar strings, plucked together: instead of one steady pitch you hear a slow “wah-wah” pulsing on top of it. Two tones at 440 Hz and 445 Hz, added together, beat at $\lvert445-440\rvert=5$ Hz: five swells and fades every second.

Drag Frequency gap (Hz) down toward zero and watch the outline curve around the wave stretch out, swelling more slowly.

Notice it swells and fades as many times per second as the frequency gap: halve the gap and you hear the throb half as often.

Spinning while shrinking or growing

Everything so far kept the phasor’s length fixed at 1, tracing a perfect circle. Let the exponent pick up a second, real part $\sigma$ alongside the spin rate $\omega$:

The $e^{j\omega t}$ piece still spins at the same steady rate as before. The new $e^{\sigma t}$ piece multiplies the radius at every instant: since it’s an ordinary (real) growth or decay factor, it doesn’t change the angle at all, only the point’s distance from the centre. The path is no longer a circle but a spiral, shrinking inward if $\sigma<0$ or growing outward if $\sigma>0$.

Picture a plucked guitar string: its actual vibration is a phasor that spins at the string’s pitch while steadily losing energy, a spinning-and-shrinking arrow whose spin rate $\omega$ sets the pitch you hear and whose decay rate $\sigma$ sets how fast the sound fades.

Drag Decay rate away from zero in either direction and watch the spiral pull inward or push outward, loop by loop.

Notice at a decay rate of exactly zero you’re back to the plain circle from the spinning-point instrument above, a pure, undamped cosine.

Where you’ll meet this

Phasors are the reason the systems of Chapters 4 and 5 turn out to be so easy to analyze: feed a spinning point into one of the well-behaved systems you’ll meet there, and what comes out is the same spinning point, just stretched and turned by one fixed complex number. Chapter 13 measures a signal’s frequencies with exactly this family of spinning points, and the spiral $e^{st}$ you just met returns as the building block of Chapters 9 and 16.

Worked example

1. Euler check. $\left(e^{j\pi/3}+e^{-j\pi/3}\right)/2 = (0.5+0.866j+0.5-0.866j)/2 = 0.5 = \cos(\pi/3)$, exactly.
2. Phasor sum. $3\cos(\omega t)+4\sin(\omega t)$: write $4\sin(\omega t)=4\cos(\omega t-90°)$, so the phasors are $3\angle0° + 4\angle{-90°} = 3-4j$. Magnitude $\sqrt{3^2+4^2}=5$, angle about $-53.13°$, giving $3\cos(\omega t)+4\sin(\omega t)=5\cos(\omega t-53.13°)$.
3. Beats. 440 Hz and 445 Hz tones added together beat at $\lvert445-440\rvert=5$ Hz.

Reference card