Complex numbers

In modes, the spring’s natural rates turned out to be complex. We don’t typically think of being able to visualize complex numbers, but the spring is certainly very real. So, where does the imaginary value come from? A real exponential can only describe growth or decay growth or decay, but the spring also oscillates. We want to describe both together.

Imagine a turning arrow. Its tip’s horizontal position moves back and forth like the spring. The arrow’s length sets the size of the swing, and its angle tells us where we are in the cycle. If we shorten the arrow as it turns, we represent a decaying oscillation.

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Shrink the arrow

The horizontal position alone cannot tell us which way the tip is moving; we need the second coordinate too. Complex numbers keep both coordinates together and let multiplication rotate or scale the arrow. Let us see how that works. Call the tip’s horizontal coordinate $u$ and vertical coordinate $v$.

The imaginary unit

A complex number packages the pair into one symbol:

$$z=u+jv, \qquad j^2=-1.$$

The symbol $j$ is chosen to behave like a quarter-turn. Formally, it is the square root of $-1$; practically, it lets us combine a two-dimensional value into a single number. Two quarter-turns reverse an arrow, and multiplication by $-1$ also reverses an arrow; the rule $j^2=-1$ makes those statements identical. Indeed, multiplying $z$ by $j$ gives

$$jz=j(u+jv)=-v+ju.$$

The point $(u,v)$ has become $(-v,u)$, exactly a $90^\circ$ counterclockwise rotation.

Drag $z$ around below and $jz$ follows, always a quarter-turn ahead and always the same length. Press multiply by $j$ twice and $z$ points the opposite way - that is $j^2=-1$. Press it four times and it is back where it started.

multiply by $j$

reset

The coordinate $u$ is called the real part and $v$ is called the imaginary part. Electrical and control engineers use $j$ rather than $i$ because $i$ often denotes electric current. Nothing physical is imaginary here: $z$ is compact notation for the real pair $(u,v)$, together with an algebraic system to rotate it.

The length of the corresponding arrow is

$$|z|=\sqrt{u^2+v^2}.$$

Multiplying $z$ by a positive real number stretches this arrow without changing its direction. Thus the same notation naturally handles the two operations visible in an oscillator’s motion: rotation changes angle, while real scaling changes distance from the origin.

The unit circle

Multiplication by $j$ gives one instantaneous quarter-turn. An oscillator, however, passes continuously through every angle between those quarter-turns. Let’s build up towards fully describing this motion.

Start at the point $1$ on the positive real axis and rotate counterclockwise through an angle $\theta$. Because the point remains on the unit circle, its horizontal coordinate is $\cos\theta$ and its vertical coordinate is $\sin\theta$. The corresponding complex number is therefore

$$q(\theta)=\cos\theta+j\sin\theta.$$

The symbol $q$ is only a temporary name for the rotating point. At this stage, this is simply a geometric statement that packages two coordinates on the two-dimensional circle into one symbol as a function of angle $\theta$.

Drag the point around the circle below. Its shadow on the real axis is $\cos\theta$ and its shadow on the imaginary axis is $\sin\theta$.

$\theta$

Euler’s formula

Now I ask the reader to partake in a brief leap of faith. For a real number $x$, one exact definition of the exponential is the infinite power series

$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$$

The exclamation mark denotes a factorial: for example, $4!=4\cdot3\cdot2\cdot1$. The dots mean that the pattern continues with successively higher powers. Taking more terms gives increasingly accurate partial sums; in the limit, the series defines the exact value of $e^x$.

Every operation on the right-hand side - addition, multiplication, and division by a real number - also makes sense for complex numbers. We therefore define the complex exponential using the same recipe:

$$e^z=1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\frac{z^4}{4!}+\cdots, \qquad z\in\mathbb{C}.$$

This is a definition, not yet a claim about circles or sinusoids.

Now choose the particular complex input $z=j\theta$. The powers of $j$ repeat every four steps:

$$j^0=1,\qquad j^1=j,\qquad j^2=-1,\qquad j^3=-j,\qquad j^4=1,\ldots$$

Consequently, the even powers of $j\theta$ contribute real terms, while the odd powers contribute imaginary terms. This brings us to the result that connects the exponential to the rotating point.

Euler’s formula. For every real angle $\theta$,

$$e^{j\theta}=\cos\theta+j\sin\theta.$$

Proof. Substitute $z=j\theta$ into the definition:

$$\begin{aligned} e^{j\theta} &=1+j\theta+\frac{(j\theta)^2}{2!}+\frac{(j\theta)^3}{3!}+\frac{(j\theta)^4}{4!}+\cdots\\ &=1+j\theta-\frac{\theta^2}{2!}-j\frac{\theta^3}{3!}+\frac{\theta^4}{4!}+\cdots. \end{aligned}$$

Collecting the real terms and the imaginary terms separately gives

$$e^{j\theta}=\left(1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots\right)+j\left(\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots\right).$$

The two expressions in parentheses are precisely the power series for cosine and sine:

$$\begin{aligned} \cos\theta&=1-\frac{\theta^2}{2!}+\frac{\theta^4}{4!}-\cdots,\\ \sin\theta&=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\cdots. \end{aligned}$$

Therefore $e^{j\theta}=\cos\theta+j\sin\theta$. $\blacksquare$

The coefficients in the sine and cosine series can be found from their successive derivatives at $\theta=0$: those derivatives cycle through the same four patterns as the powers of $j$. Euler’s formula is therefore what the ordinary exponential becomes when its input points in the imaginary direction.

We show a visual for the proof by drawing each term of the Taylor series as an arrow. Each term is the previous one multiplied by $j\theta/n$, so each arrow is turned a quarter-turn from the last and, once $n>\theta$, shorter. The chain curls around and settles onto $e^{j\theta}$ on the unit circle. Things to try:

• Start with one term: the sum is just $1$. Add terms one at a time and watch the real terms (dark) and the imaginary terms (orange) alternate - exactly the split into the cosine and sine series in the proof.
• Raise $\theta$ towards $\pi$. The early terms now grow before they shrink, so the chain swings far off the circle and needs more terms before it settles.

$\theta$

terms

Comparing Euler’s formula with the geometric definition of $q(\theta)$ now gives

$$q(\theta)=e^{j\theta}.$$

The exponential notation is compact because its exponent stores the angle. At the four quarter-turns,

$$e^{j0}=1, \qquad e^{j\pi/2}=j, \qquad e^{j\pi}=-1, \qquad e^{j3\pi/2}=-j.$$

Uniform circular motion

We can now turn the angle-indexed point $q(\theta)$ into a motion through time. Let the angle increase at one radian per second, so that $\theta=t$, and write

$$z(t)=e^{jt}=\cos t+j\sin t.$$

A dot above a symbol will denote its derivative with respect to time. Differentiating gives

$$\dot z(t)=-\sin t+j\cos t=j\bigl(\cos t+j\sin t\bigr)=jz(t).$$

The exponential and the dynamic equation are thus two descriptions of the same motion.

To see the two real coordinates inside $\dot z=jz$, write $z=u+jv$. Here $u=\cos t$ and $v=\sin t$, so

$$\dot u=-v, \qquad \dot v=u.$$

At the point $(u,v)$, the velocity $(-v,u)$ is a $90^\circ$ counterclockwise rotation of the radius $(u,v)$. The velocity is therefore tangent to the circle rather than pointing toward or away from the origin.

The geometry already suggests that the radius remains fixed, which we can prove. The squared distance from the origin satisfies

$$\frac{\mathrm{d}}{\mathrm{d}t}\left(u^2+v^2\right)=2u\dot u+2v\dot v=2u(-v)+2v(u)=0.$$

Its derivative is zero, so the distance is constant. If the point begins at $z(0)=1$, it remains on the unit circle and turns counterclockwise at one radian per second.

Euler’s formula treats its argument as an angle. For a motion with angular frequency $\omega$ and initial phase $\phi$, that angle is $\theta(t)=\omega t+\phi$. Substituting this angle into Euler’s formula and choosing a radius $A_0$ gives

$$z(t)=A_0e^{j(\omega t+\phi)}=A_0\cos(\omega t+\phi)+jA_0\sin(\omega t+\phi).$$

The real part is the sinusoid we observe on a time plot; the imaginary part is its quarter-cycle-shifted companion. The radius is $A_0$, the starting angle is $\phi$, and the angle advances at $\omega$ radians per second.

Differentiating shows where the angular frequency enters the dynamics:

$$\dot z(t)=j\omega A_0e^{j(\omega t+\phi)}=j\omega z(t).$$

Compared with $\dot z=jz$, the factor $\omega$ simply changes the turning rate. For $\omega>0$, one complete turn takes

$$T=\frac{2\pi}{\omega}.$$

Complex frequency

So far, the radius remains fixed at $A_0$. To let the envelope change, multiply by a real exponential:

$$z(t)=\underbrace{A_0e^{\sigma t}}_{\text{changing radius}}\underbrace{e^{j(\omega t+\phi)}}_{\text{changing angle}}.$$

If $\sigma<0$, the radius shrinks; if $\sigma=0$, it remains constant; and if $\sigma>0$, it grows. Taking the real part gives the waveform we actually observe:

$$\operatorname{Re}\{z(t)\}=A_0e^{\sigma t}\cos(\omega t+\phi).$$

Because multiplying exponentials adds their exponents, the same motion can be written more compactly. Define

$$z(0)=A_0e^{j\phi}, \qquad s=\sigma+j\omega.$$

Then

$$z(t)=z(0)e^{st}.$$

The complex rate $s$ packages two real rates: $\sigma$ changes the radius, while $\omega$ changes the angle. The initial state determines $A_0$ and $\phi$ - how strongly the mode is excited and where it begins in its cycle. A command or disturbance can change that state; while an input continues to act, we must also account for the forced response. The mode itself determines $\sigma$ and $\omega$ - how its envelope changes and how quickly it cycles.

Finally, expand the compact dynamic equation $\dot z=sz$ back into real and imaginary components:

$$\begin{aligned} \dot u&=\sigma u-\omega v,\\ \dot v&=\omega u+\sigma v. \end{aligned}$$

This form is exactly what a control designer wants to know about a natural motion. If $\sigma<0$, the mode returns to the origin and is asymptotically stable. If $\sigma>0$, it moves away from the origin and is unstable. If $\sigma=0$, an ideal oscillation persists forever. The imaginary part $\omega$ tells us how rapidly it oscillates. Although $z(t)$ is complex, the observed waveform remains real: it is the horizontal component $\operatorname{Re}\{z(t)\}$.

In the figure below, the dark arrow is the radius $z(t)$ and the orange arrow is the velocity $\dot z=sz$. Things to try:

• Set $\sigma=0$. The velocity stays exactly perpendicular to the radius, so the point stays on the unit circle and the real part is a pure cosine.
• Drag $\sigma$ negative. The velocity now leans inward, the point spirals towards the origin, and the real part decays inside its dashed envelope $e^{\sigma t}$. Push $\sigma$ above zero and the velocity leans outward instead.
• Change $\omega$ with $\sigma$ fixed. The envelope does not move at all - only the number of turns inside it changes. Set $\omega=0$ and the point slides along the real axis without turning: a pure exponential.

$\sigma$

$\omega$