The modern version. Solve \(x^2 + 1 = 0\). Rearranging gives \(x^2 = -1\). But the square of any real number is non-negative: \(3^2 = 9\), \((-3)^2 = 9\), \(0^2 = 0\). No real number squares to \(-1\). The parabola \(y = x^2 + 1\) floats entirely above the x-axis, never touching it. In Chapter 8 this is the case of a negative discriminant, and we said "no real roots". The word real was doing a lot of work there.

The historical version. Complex numbers were not invented to solve \(x^2 + 1 = 0\). Mathematicians were perfectly happy to say that equation had no solution, just as \(x + 5 = 2\) once "had no solution" before negative numbers were accepted. What forced the issue was the cubic. In 1545 Gerolamo Cardano published a formula for solving cubics. Applied to

the formula produces

Square roots of negative numbers! Yet this equation obviously has a real solution: \(x = 4\) works, because \(4^3 = 64 = 15 \cdot 4 + 4\). Around 1572 Rafael Bombelli had what he called "a wild thought". He treated \(\sqrt{-1}\) as a symbol obeying ordinary algebra plus the rule \((\sqrt{-1})^2 = -1\), and noticed that

So \(\sqrt[3]{2 + \sqrt{-121}} = 2 + \sqrt{-1}\), and similarly \(\sqrt[3]{2 - \sqrt{-121}} = 2 - \sqrt{-1}\). Adding, the imaginary parts cancel and \(x = 4\). The "impossible" numbers were a necessary bridge between a real question and a real answer. That is the lesson of this chapter in miniature: complex numbers are a tool for getting real answers.

You have done this before. Natural numbers cannot solve \(x + 5 = 2\), so we invented negatives. Integers cannot solve \(3x = 2\), so we invented fractions. Rationals cannot solve \(x^2 = 2\), so we filled in the irrationals (Chapter 1). Each time, the new numbers were initially distrusted ("negative", "irrational", and now "imaginary" are all slightly insulting names) and each time the extension kept all the old rules working.

The difference this time is geometric. The real numbers fill a line. There is no room left on the line for a number whose square is negative. So the new numbers must live off the line — in a plane.

The form \(a + bi\) is called rectangular (or Cartesian, or algebraic) form. "Complex" here means "composite" — made of two parts — not "complicated".

The whole of complex arithmetic is: treat \(i\) as an unknown, do ordinary algebra, and replace \(i^2\) by \(-1\) whenever it appears.

Addition and subtraction combine real parts and imaginary parts separately, like collecting like terms:

Multiplication expands the brackets (the distributive law, Chapter 4):

Do not memorise that last line; just expand and simplify each time.

Worked example. Compute \((2 + 3i)(4 - i)\). Try it: Multiply (2 + 3i)(4 − i) visually.

Since \(i^2 = -1\), the powers of \(i\) cycle with period 4:

To compute \(i^n\), divide \(n\) by 4 and keep the remainder (Chapter 23): \(i^{2026} = i^{4\cdot506 + 2} = i^2 = -1\). Negative powers too: \(i^{-1} = \frac{1}{i} = \frac{i}{i^2} = -i\) (multiply top and bottom by \(i\)).

The cycle of four is a strong hint about geometry: something that returns to where it started after four steps sounds like a quarter-turn. We will confirm that shortly.

A complex number \(a + bi\) is determined by two real numbers, so it can be pictured as the point \((a, b)\) in a plane. The horizontal axis holds the real numbers (the real axis); the vertical axis holds the purely imaginary numbers (the imaginary axis). This picture is the complex plane, or Argand diagram (after Jean-Robert Argand, 1806, although Caspar Wessel published it first in 1799).

The real number line is still there, intact, as the horizontal axis. Complex numbers did not replace the reals; they surround them.

Addition now has a familiar picture: adding complex numbers is adding vectors (Chapter 26), tip to tail. Subtraction \(z - w\) is the arrow from \(w\) to \(z\), so \(|z - w|\) (defined next) is the distance between them. Try it: Add complex numbers as vectors.

A point in the plane can be described by its coordinates \((a, b)\) or by its distance from the origin and direction (Chapter 13). For complex numbers these get special names.

For \(z = 3 + 4i\): \(|z| = \sqrt{9 + 16} = 5\), and \(\tan\theta = \frac43\), so \(\theta = \arctan\frac43 \approx 0.9273\) rad \(\approx 53.13^\circ\).

Worked example. Find the modulus and argument of \(z = -1 - i\). Try it: Find arg(−1 − i) safely.

Try it: See arctan(b/a) point wrong.

The complex plane earns its keep because \(|z - w|\) is the distance between two points and adding is moving.

Addition, subtraction, and multiplication are straightforward. What is \(\frac{3 + 4i}{1 - 2i}\)? We need to put it in the form \(a + bi\), which means getting rid of \(i\) in the denominator. This is the same problem as rationalising a denominator like \(\frac{1}{1 + \sqrt2}\), and the same trick solves it.

Worked example. \(\dfrac{3 + 4i}{1 - 2i}\). Try it: Divide by the conjugate.

Check by multiplying back: \((-1 + 2i)(1 - 2i) = -1 + 2i + 2i - 4i^2 = -1 + 4i + 4 = 3 + 4i\) ✓.

Reciprocal. The same idea gives a formula for \(1/z\):

So every nonzero complex number has a reciprocal, and \(\mathbb{C}\) is a field: all four arithmetic operations work, with the usual laws.

Addition has a picture (vector addition). What does multiplication do to points in the plane? Multiplying by a positive real like 2 stretches every arrow to twice its length. Multiplying by \(-1\) flips every arrow to point the opposite way — a half-turn. What does multiplying by \(i\) do?

Take any \(z = a + bi\) and multiply by \(i\):

The point \((a, b)\) has moved to \((-b, a)\). That is exactly a rotation by \(90^\circ\) anticlockwise about the origin (try \((1, 0) \to (0, 1) \to (-1, 0) \to (0, -1) \to (1, 0)\)). Now the rule \(i^2 = -1\) has a meaning anyone can see: two quarter-turns make a half-turn. And "the square root of \(-1\)" means "the operation that, done twice, reverses direction".

Now multiply two general complex numbers. Watch the moduli and arguments in an example: \(z = 2 + i\) and \(w = 1 + i\). Try it: Watch moduli multiply, arguments add.

The moduli multiplied (\(\sqrt5\cdot\sqrt2 = \sqrt{10}\)) and the arguments added (\(26.57^\circ + 45^\circ = 71.57^\circ\)).

To see that this always happens, write each number using its modulus and argument (from the polar-form picture, \(a = r\cos\theta\), \(b = r\sin\theta\)):

Multiply:

The last step uses the angle-addition formulas from Chapter 13. So:

Multiplying by \(w\) rotates by \(\arg w\) and scales by \(|w|\). Dividing by \(w\) does the reverse: \(|z/w| = |z|/|w|\), \(\arg(z/w) = \arg z - \arg w\).

Read that derivation backwards and something remarkable appears: the angle-addition formulas, which you may once have memorised as arbitrary identities, are nothing more than the real and imaginary parts of "angles add when you multiply". If you forget them, multiply out \((\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta)\) and read them off.

This is the same trade-off as between coordinate systems in geometry: choose the representation in which your operation is simple. Converting has a cost (a square root and an arctangent), so for a single multiplication rectangular form is usually quicker; for \(z^{20}\), polar wins overwhelmingly.

Worked example. Write \(z = 1 + \sqrt3\,i\) in polar form, and convert \(4\operatorname{cis}\frac{2\pi}{3}\) to rectangular form. Try it: Convert 1 + √3 i to polar.

We found that numbers of the form \(\operatorname{cis}\theta = \cos\theta + i\sin\theta\) obey

Where have you seen a function that turns addition of inputs into multiplication of outputs? The exponential function: \(e^{a}e^{b} = e^{a+b}\) and \(e^0 = 1\) (Chapter 12, Chapter 3). This suggests that \(\operatorname{cis}\theta\) is some kind of exponential of \(\theta\). Which one?

Argument 1 — the law of exponents. As just observed, \(\operatorname{cis}\) satisfies the same law as an exponential.

Argument 2 — rates of change (a preview of Chapter 17). The function \(e^{k\theta}\) has the property that its rate of change is \(k\) times itself. What is the rate of change of \(\operatorname{cis}\theta\) as \(\theta\) increases? The point \(\operatorname{cis}\theta\) moves around the unit circle at unit speed, and its velocity is always perpendicular to its position, pointing anticlockwise — that is, velocity = position rotated by \(90^\circ\) = \(i \times\) position. So \(\operatorname{cis}\theta\) changes at a rate \(i\) times itself, exactly like \(e^{k\theta}\) with \(k = i\). (With derivatives: \(\frac{d}{d\theta}(\cos\theta + i\sin\theta) = -\sin\theta + i\cos\theta = i(\cos\theta + i\sin\theta)\).)

Argument 3 — infinite series (a preview of Taylor series, Chapter 18). The exponential function has a power-series expansion, an "infinite polynomial" in the sense of Chapter 14:

Assume this still makes sense with \(x = i\theta\). Powers of \(i\theta\) cycle through \(1, i, -1, -i\) times powers of \(\theta\):

The two brackets are precisely the power series of \(\cos\theta\) and \(\sin\theta\), as you will see in Chapter 18. All three arguments point to the same conclusion.

Strictly speaking, this is how we define the exponential of an imaginary number — but it is the only definition consistent with the law of exponents, the rate-of-change property, and the power series all at once, which is why it is universally accepted. Every complex number can now be written compactly as

The multiplication rule is now just the law of exponents.

Trust, but verify. Compare Python’s complex exponential with cosine and sine directly, and sum the power series by hand:

The tiny \(1.22 \times 10^{-16}i\) in the first output is not a flaw in Euler’s formula: math.pi is only the closest double to \(\pi\), and \(\sin\) of that number is about \(1.22 \times 10^{-16}\) (Chapter 32).

Try it: Walk to e^(iπ) = −1.

If multiplying by \(z\) rotates by \(\theta\) and scales by \(r\), then multiplying by \(z\) \(n\) times rotates by \(n\theta\) and scales by \(r^n\):

This is De Moivre’s theorem (Abraham de Moivre, 1707). With Euler’s formula it is just the law \((e^{a})^n = e^{na}\).

\((1 + i)^{10}\). \(r = \sqrt2\), \(\theta = \frac{\pi}{4}\). Try it: Spiral to (1 + i)¹⁰ = 32i.

Alternative method, as a check: \((1 + i)^2 = 1 + 2i + i^2 = 2i\), so \((1 + i)^{10} = (2i)^5 = 32\,i^5 = 32i\) ✓. Repeated squaring is a fine alternative when the base squares to something simple; polar form is the general method. Expanding \((1+i)^{10}\) by the binomial theorem would work too, but would take eleven terms.

\((1 + \sqrt3\,i)^6\). \(r = 2\), \(\theta = \frac{\pi}{3}\), so the result is \(2^6(\cos 2\pi + i\sin 2\pi) = 64\). A complex number whose sixth power is a positive real: it has rotated through exactly one full turn. Try it: See (1 + √3 i)⁶ = 64.

Expand \((\cos\theta + i\sin\theta)^3\) in two ways. By De Moivre it is \(\cos3\theta + i\sin3\theta\). By the binomial expansion \((c + is)^3 = c^3 + 3c^2(is) + 3c(is)^2 + (is)^3\):

Equate real parts, and use \(\sin^2\theta = 1 - \cos^2\theta\):

Equating imaginary parts similarly gives \(\sin3\theta = 3\sin\theta - 4\sin^3\theta\). Multiple-angle formulas need never be memorised again.

How many solutions does \(z^3 = 8\) have? Over the reals, one: \(z = 2\). But polar form lets us run De Moivre backwards. If \(z = re^{i\theta}\) then \(z^3 = r^3e^{3i\theta}\), and we need this to equal \(8 = 8e^{i\cdot0}\). Moduli: \(r^3 = 8\), so \(r = 2\). Arguments: \(3\theta\) must point in the same direction as 0 — but "the same direction" means \(3\theta = 0, 2\pi, 4\pi, \ldots\). So \(\theta = 0, \frac{2\pi}{3}, \frac{4\pi}{3}\) (after that the angles repeat). Three different cube roots: Try it: Plot the cube roots of 8.

Check one: \((-1 + \sqrt3 i)^2 = 1 - 2\sqrt3 i - 3 = -2 - 2\sqrt3 i\), then \((-2 - 2\sqrt3 i)(-1 + \sqrt3 i) = 2 - 2\sqrt3 i + 2\sqrt3 i - 6i^2 = 2 + 6 = 8\) ✓.

Try it: See the sixth roots of unity.

Why do they sum to zero? By symmetry, the regular polygon’s vertices balance around the origin. Algebraically, they form a geometric series (Chapter 14) with ratio \(\omega \ne 1\): Try it: Roots of unity sum to zero.

The last line is the sum of the sixth roots of unity: zero, up to rounding.

Worked example. Solve \(z^4 = -16\). Try it: Solve z⁴ = −16 on its circle.

Try it: A cosine at fs/4 in the FFT; DFT arrows line up; Arrows cancel off-frequency; Run this in your browser; Mirror bin k = 6; N² versus N log N.

We extended the reals to solve one equation, \(x^2 + 1 = 0\). A natural worry: will \(x^2 = i\) or \(x^5 + ix + 3 = 0\) force yet another extension? The answer, and the reason complex numbers are the end of this road, is one of the great theorems of mathematics.

The "consequently" is the factor theorem from Chapter 7: if \(z_1\) is a root, then \((z - z_1)\) is a factor, and the remaining quotient has degree \(n - 1\), which again has a root, and so on.

The proofs (Gauss gave several, starting in 1799) need tools beyond this book, but here is the intuition. For a huge circle \(|z| = R\), the polynomial behaves like its leading term \(z^n\), which winds around the origin \(n\) times as \(z\) goes once around the circle. For a tiny circle, \(p(z)\) stays close to \(p(0)\) and does not wind around the origin at all (if \(p(0) \ne 0\)). Shrinking the circle continuously from large to small, the image loop must at some point pass through the origin — and that is a root.

For a quadratic with real coefficients and negative discriminant, the quadratic formula (Chapter 8) now simply works:

Worked example. Solve \(x^2 - 4x + 13 = 0\). Try it: See the roots 2 ± 3i.

Check: \((2 + 3i)^2 - 4(2 + 3i) + 13 = (4 + 12i - 9) - 8 - 12i + 13 = 0\) ✓.

Alternative: completing the square. \(x^2 - 4x + 13 = (x - 2)^2 + 9 = 0\), so \((x - 2)^2 = -9\), \(x - 2 = \pm3i\), \(x = 2 \pm 3i\). Same answer, and it shows the geometry: the vertex of the parabola is at \((2, 9)\), above the axis; the real part of the roots is the axis of symmetry, and the imaginary part measures how far the parabola "misses" the x-axis.

Notice the roots came as \(2 + 3i\) and \(2 - 3i\), a conjugate pair. That is no accident. If a polynomial has real coefficients and \(p(z) = 0\), conjugate everything: since conjugation respects sums and products and leaves real coefficients alone, \(p(\bar z) = \overline{p(z)} = \bar 0 = 0\). So non-real roots of real polynomials come in conjugate pairs. Consequences:

Example. \(x^3 - 1 = (x - 1)(x^2 + x + 1)\). The quadratic has discriminant \(1 - 4 = -3\), so roots \(\frac{-1 \pm \sqrt3 i}{2}\). Together with 1 these are the three cube roots of unity, \(e^{0}, e^{2\pi i/3}, e^{4\pi i/3}\) — the algebra and the geometry agree.

In alternating-current circuits, voltages and currents oscillate sinusoidally, \(V(t) = V_0\cos(\omega t + \phi)\) (here \(\omega\) is the angular frequency, not a root of unity). Resistors, capacitors, and inductors each respond differently: a resistor’s current is in step with the voltage; an inductor’s current lags by a quarter cycle; a capacitor’s leads by a quarter cycle. Tracking these phase shifts with trigonometric identities is miserable.

Engineers instead represent each oscillation by a complex number, its phasor \(V_0e^{i\phi}\) (the oscillation is the real part of \(V_0e^{i\phi}e^{i\omega t}\)). A "quarter-cycle shift" is multiplication by \(\pm i\) — the quarter-turn again. Each component gets a complex impedance (generalised resistance):

and Ohm’s law \(V = IZ\) works with complex numbers. Components in series add their impedances; in parallel, the reciprocals add — exactly as for plain resistors.

Worked example. A 30 Ω resistor in series with an inductor of reactance \(\omega L = 40\) Ω is driven by a voltage of amplitude 100 V. Then \(Z = 30 + 40i\) Ω, and Try it: Solve the AC circuit with phasors.

The current’s amplitude is \(|I| = \sqrt{1.44 + 2.56} = 2\) A (equivalently \(100/|Z| = 100/50\)), and its phase is \(\arg I = -\arctan\frac{1.6}{1.2} \approx -53.1^\circ\): it lags the voltage by \(53.1^\circ\). One complex division replaced a page of trigonometry.

Any reasonable periodic signal is a sum of sines and cosines (Fourier series), and Euler’s formula packs each cosine–sine pair into one complex exponential: \(\cos\omega t = \frac{e^{i\omega t} + e^{-i\omega t}}{2}\). Filtering a signal (removing hiss, isolating a radio station, equalising audio) becomes multiplying each frequency component by a complex number that scales its amplitude and shifts its phase. The DFT and FFT from the sidebar above are the computational form.

In quantum mechanics, complex numbers are not a convenience but part of the foundations. The state of a system is described by complex amplitudes; the probability of an outcome is the squared modulus \(|\psi|^2\) of its amplitude. Because amplitudes have phases, two paths to the same outcome can cancel (destructive interference) or reinforce, which is what the double-slit experiment shows. Schrödinger’s equation, \(i\hbar\frac{\partial\psi}{\partial t} = \hat H\psi\), has \(i\) right at the front. Quantum computing algorithms (including Shor’s factoring algorithm, which uses a quantum Fourier transform built from roots of unity) are manipulations of these complex amplitudes.

Pick a complex number \(c\). Start with \(z_0 = 0\) and repeatedly apply \(z_{n+1} = z_n^2 + c\) (a recurrence, Chapter 14). Does the sequence stay bounded, or does it run off to infinity? The Mandelbrot set is the set of all \(c\) for which it stays bounded.

Try a few by hand: Try it: Follow the orbit of c = i.

A useful fact makes computation possible: once \(|z_n| > 2\), the sequence is guaranteed to escape to infinity. (Roughly: when \(|c| \le 2\) and \(|z| > 2\), squaring makes \(|z|\) grow faster than adding \(c\) can pull it back; and if \(|c| > 2\) the very first step already exceeds 2.) So we iterate until \(|z| > 2\) or until we give up after some maximum number of steps.

Try it: Zoom past double precision.