Complex numbers carry this course from chapter 3 onward. The discrete-time
Fourier transform of chapter 3 is complex-valued at every frequency. The
z-transform of chapter 4 is evaluated on the complex plane. A filter's
pole-zero picture in chapter 5 plots roots of a complex polynomial. This
section builds what those chapters need: complex arithmetic, plus Euler's
formula, written in the j notation this course uses throughout. These
ideas were first built in the readiness chapter of fourier-analysis. Go
there for a deeper treatment of the algebra behind Euler's formula. What
follows here is everything this course itself reaches for.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Fourier Analysis
What point on the unit circle does \theta=\pi/2 correspond to?
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-1 is the point for \theta=\pi, a quarter turn further around than \pi/2.
1 is the point for \theta=0, the starting point before any rotation at all.
-i is the point for \theta=3\pi/2 (or equivalently -\pi/2), a rotation in the opposite direction from \pi/2.
Why
e^{i\pi/2}=\cos(\pi/2)+i\sin(\pi/2)=0+i=i.
Evaluate \cos(\pi/2) and \sin(\pi/2) separately, then combine as \cos\theta+i\sin\theta.
From Fourier Analysis
Which value equals e^{i\pi}?
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That's e^{i\pi/2}, a quarter turn — a different angle entirely.
That's e^{i\cdot 0} or e^{2\pi i}, a full turn back to the start.
That's e^{-i\pi/2} or e^{i3\pi/2}, a different angle.
Why
e^{i\pi} = \cos\pi+i\sin\pi = -1+0i = -1.
From Fourier Analysis
A 12-hour clock identifies 13 o'clock with 1 o'clock. What is the analogous identification for angles on the circle?
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Dividing by 12 has nothing to do with the actual identification a full turn produces -- 2\pi, not the number 12, is what closes the loop for angles.
\theta and -\theta are generally two different points on the circle (mirror images), not the same point.
The clock's "wraps around after a fixed amount" structure is exactly the same idea, with 2\pi playing the clock's 12-hour role.
Why
Arithmetic modulo N (a 12-hour clock, N=12) is the same wraparound idea this section builds for angles, with N evenly spaced points replaced by a full continuum on S^1 and the modulus 12 replaced by 2\pi.
Identify what number plays the clock's role of "12" for angles -- the amount after which everything repeats.
A real number lives on one line. One number is enough to say where it sits. A complex number needs two numbers instead, because it sits on a flat sheet. There are two equally complete ways to hand over those two numbers.
The first way is "how far right, how far up." Those two amounts are the a and the b in z = a + jb. The letter j is a marker. It says "the amount next to me is the up-down one." An up-down amount like b is called imaginary, which is an old name and nothing more.
The second way is "how far from zero, and which way." The distance from zero has a name: the modulus. The direction has a name too: the argument, which is an angle. Think of a spot on a map. You can give it as (east, north), or as (distance, compass heading). Both pin down the very same spot. Neither one is more correct than the other.
The conjugate of a point is its mirror image straight across the horizontal axis. Same distance from zero, opposite tilt. Flip it up-down and you have it. Now multiply a point by its own mirror image. The answer always lands back on the horizontal axis, with no up-down part left over. How far out does it land? Exactly the square of how far out the original point sat. Write that fact in the "how far right, how far up" numbers and it is the Pythagorean theorem in disguise: right-amount squared plus up-amount squared.
A complex number z=a+jb pairs a real number a (the real part) with a real number b (the imaginary part), where j is defined by j^2=-1. This course writes the imaginary unit as j rather than the i a pure-math text uses, following its primary source and the wider signal-processing field. Circuit analysis already reserves i for electric current, so "i amps at frequency i\omega" would be ambiguous on the page. Plot a on a horizontal axis and b on a vertical one, and the result is the complex plane. There z has a modulus |z|=\sqrt{a^2+b^2}, its distance from the origin. It also has an argument \theta, the angle its ray makes with the positive real axis. Together those two give the polar form z=r(\cos\theta+j\sin\theta) with r=|z|. The complex conjugate z^*=a-jb flips the sign of the imaginary part only, and satisfies zz^*=a^2+b^2=|z|^2. So multiplying a number by its own conjugate always produces a nonnegative real number. That number is its squared distance from the origin.
Start with ordinary growth, written e^{kt}. In that expression t is time, k is a fixed growth rate, and e is the number 2.718\ldots that smooth growth always runs on. Each instant, the expression stretches a length by the same fixed share of itself. It is the curve a savings account follows.
Euler's formula feeds in an up-down rate instead of a plain one. That sends the same stretching instinct sideways. The point stops growing longer. It spins in place instead. Each instant it turns through the same fixed amount of angle. Its distance from zero never changes at all.
That is why e^{j\theta} sits exactly on the unit circle, the circle of radius 1 around zero. The symbol \theta is the Greek letter theta, and here it stands for the angle turned so far. The point is not moving out and not moving in. It only turns.
A spin can always be undone by spinning the other way. So take e^{j\theta}, the point spun one way, and add e^{-j\theta}, the same point spun the other way. Their up-down parts are opposites, so those cancel to nothing. Their left-right parts are equal, so those add up to double. What is left is twice the left-right part and no up-down part at all. That is exactly what the cosine formula above says.
Euler's formula says the polar form above can be written far more compactly as z=re^{j\theta}, because e^{j\theta} = \cos\theta + j\sin\theta holds for every angle. Here is one quick way to see why. Let f(\theta) = e^{-j\theta}(\cos\theta + j\sin\theta). The product rule gives f'(\theta)=0 for every \theta, because the two exponential factors' rates of change cancel exactly. So f is a constant, and f(0)=1 fixes that constant at 1. Rearranging f(\theta)=1 recovers the formula above.
Two consequences follow that this course uses constantly. First, |e^{j\theta}|=\sqrt{\cos^2\theta+\sin^2\theta}=1 for every \theta, so e^{j\theta} always lands somewhere on the unit circle. Second, multiplying two complex numbers in polar form multiplies their moduli while adding their arguments: r_1e^{j\theta_1}\cdot r_2e^{j\theta_2}=r_1r_2\,e^{j(\theta_1+\theta_2)} That is a rotation and a scaling turned into ordinary exponent addition. Euler's formula and its conjugate, e^{-j\theta}=\cos\theta-j\sin\theta, can also be added or subtracted to isolate cosine and sine directly: \cos\theta = \frac{e^{j\theta}+e^{-j\theta}}{2}, \qquad \sin\theta = \frac{e^{j\theta}-e^{-j\theta}}{2j}.
One point, two shadows, this course's letter
Drag \theta, or press Play. The point's horizontal position IS \cos\theta's height on the right, and its vertical position IS \sin\theta's height, at the same instant.
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Multiplying rotates and scales at once
Drag z_1 or z_2. The product's angle is the SUM of the two angles, and its length is the PRODUCT of the two lengths.
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Worked example
Rectangular to polar form
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1
Write z=1+j in polar form re^{j\theta}. Compute the modulus and argument separately, then combine with Euler's formula.
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2
r=|z|=\sqrt{1^2+1^2}=\sqrt2. The ray to (1,1) makes a 45°, i.e. \pi/4, angle with the positive real axis, so \theta=\pi/4.
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3\boxed{1+j = \sqrt2\,e^{j\pi/4}}
Chapter 3 reads a signal's frequency content off numbers in exactly this form. The modulus gives a strength, and the argument gives a phase.
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Worked example
Multiplying two conjugates two ways
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1
Evaluate (1+j)(1-j) both directly and via the polar multiply-moduli rule, and confirm they agree. Multiply the rectangular forms directly first, then convert each factor to polar and apply r_1r_2\,e^{j(\theta_1+\theta_2)}.
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2
Directly: (1+j)(1-j) = 1 - j^2 = 1-(-1) = 2. In polar form, 1+j = \sqrt2\,e^{j\pi/4} (previous example) and 1-j=\sqrt2\,e^{-j\pi/4}, so their product is \sqrt2\cdot\sqrt2\,e^{j(\pi/4-\pi/4)} = 2\,e^{j0}=2.
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3\boxed{(1+j)(1-j)=2}
Both routes agree, and this is no coincidence: 1-j is the conjugate of 1+j, so this is exactly zz^*=|z|^2=(\sqrt2)^2=2 from the definition above.
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Worked example
A value that will reappear when a filter's response is measured
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1
Evaluate e^{j\pi/2} and |e^{j\pi/2}|. Euler's formula applies with \theta = \pi/2; every point on the unit circle has modulus exactly 1.
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2
e^{j\pi/2} = \cos\frac{\pi}{2} + j\sin\frac{\pi}{2} = 0 + j(1) = j, and |j| = \sqrt{0^2+1^2} = 1.
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3\boxed{e^{j\pi/2} = j, \quad |e^{j\pi/2}| = 1}
Chapter 4 evaluates expressions of exactly this shape, e^{j\omega}, at many different \omega. That shows how a system responds at each frequency. Every one of those values sits on the unit circle, for the same reason this one does.
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Worked example
Cosine as an average of exponentials
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1
Derive \cos\theta=\dfrac{e^{j\theta}+e^{-j\theta}}{2} starting from Euler's formula and its conjugate. Write both forms and add them; the sine terms cancel.
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2
e^{j\theta}=\cos\theta+j\sin\theta and e^{-j\theta}=\cos\theta- j\sin\theta. Adding: e^{j\theta}+e^{-j\theta} = 2\cos\theta.
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3\boxed{\cos\theta = \dfrac{e^{j\theta}+e^{-j\theta}}{2}}
This identity comes back in chapter 3. There it turns a real sinusoid into the complex-exponential form that chapter's frequency-domain transform is built from.
Where this goes next
The z-transform, chapters 4-5
The z-transform of chapter 4 works with a system's transfer function H(z). It evaluates that function at points z = e^{j\omega} on the unit circle. Doing so shows how the system responds at each frequency. That is the exact exponential form built above, with \omega now playing the role \theta played here. Every pole-zero picture from chapter 5 onward reads pole and zero locations against this same unit circle. A pole's distance from the origin is exactly the modulus, computed the same way as in the first worked example.
Check your understanding (counts toward your progress)
Compute |e^{j\pi/3}|.
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Why
Every point e^{j\theta} lies on the unit circle, so its modulus is always exactly 1, regardless of \theta.
Correct answer
Also accepted:
Why
What is the argument of the product e^{j\pi/6}\cdot e^{j\pi/3}?
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Multiplying moduli and adding arguments are two different operations — the arguments \pi/6 and \pi/3 are added, not multiplied.
That's only the first factor's own argument, not the sum.
This simplifies to \pi/9, not the correct sum \pi/6+\pi/3.
Why
Multiplying two points in polar form adds their arguments — \pi/6+\pi/3 = \pi/6+2\pi/6 = 3\pi/6 = \pi/2.
Both factors already have modulus 1; only the arguments need combining, and multiplication combines arguments by adding them.
Evaluate e^{j\pi} in the form a+bj (enter as a+b*j).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
e^{j\pi} = \cos\pi + j\sin\pi = -1 + j(0) = -1 — the same identity as e^{i\pi}=-1, with j in place of i.
Correct answer
Also accepted:
Why
Substitute \theta=\pi into \cos\theta+j\sin\theta and evaluate each piece separately.
Practice: Translating between $i$ and $j$ notation (3 problems · doesn’t affect your progress)
Write e^{j\pi/2} in the form a+bj (enter as a+b*j).
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
e^{j\pi/2} = \cos\frac{\pi}{2}+j\sin\frac{\pi}{2} = 0+j(1) = j — identical to the familiar e^{i\pi/2}=i, with j substituted for i.
Correct answer
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Why
Why do DSP and electrical-engineering texts write j instead of i for the imaginary unit?
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Both satisfy exactly j^2=-1 and i^2=-1 — the same equation, just a different letter.
It's a deliberate, field-wide convention, not an error.
The arithmetic is identical either way; only the symbol differs.
Why
Electrical engineering already uses i for current, so j is reserved for the imaginary unit to avoid ambiguity — a purely notational choice, not a mathematical one.
i already denotes electric current in that field
What is the modulus r of -2j written in polar form re^{j\theta}?
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Why
-2j sits on the negative imaginary axis, a distance 2 from the origin, so r=|-2j|=\sqrt{0^2+(-2)^2}=2.
Correct answer
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Why