Part II · Algebra Chapter 15 of 60
Complex numbers
For almost three hundred years a number whose square is −1 was thought impossible. Let’s follow how it turned from a trick into a symbol, from a symbol into a point, and from a point into a rotation, and what happened to equations after that.
Builds on: 14 · Polynomials
You will learn
- add, multiply and divide complex numbers and picture them as points of the plane
- understand multiplication as a rotation with a stretch and raise to powers with De Moivre’s formula
- find nth roots and solve quadratic equations with a negative discriminant
- explain why a polynomial of degree n has exactly n complex roots
1Why does minus times minus make plus?
The last chapter ended with a trick that wasn’t a trick. Rafael Bombelli solved the equation $x^3 = 15x + 4$ by Cardano’s formula and got $\sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$. He guessed that the cube roots were $2 \pm \sqrt{-1}$, checked it by calculating with $\sqrt{-1}$ as with an ordinary number, added them and got $4$, a genuine root. A number that “doesn’t exist” was enough to reach the answer honestly.
This chapter is the story of one number’s transformations. First it was a ghost that appeared in the middle of calculations and vanished by the answer. Then it became a symbol with rules for calculating. Then a point in the plane. And then it turned out that multiplying by it is a rotation, and that finally explained why minus times minus is plus. In chapter 2 multiplying by $-1$ turned out to be a half-turn, $180^\circ$, and we asked: is there a number that rotates by $90^\circ$ when you multiply by it? Now we’ll build it.
The ghost
The first to run into the square root of a negative number was Cardano. In the Great Art (1545) he split $10$ into two parts with product $40$, got $5 + \sqrt{-15}$ and $5 - \sqrt{-15}$, checked that the product came out right, and called the calculation as subtle as it was useless. Bombelli in his Algebra (1572) went further: he gave the new quantities names, “più di meno” and “meno di meno”, that is, “plus of minus” and “minus of minus”, and wrote out the rules for multiplying them, including the main one: “plus of minus times plus of minus makes minus”.
The name that stuck in the language was coined by Descartes. In the Geometry (1637) he wrote that the roots of an equation are not only true and false (positive and negative) but also “imaginary”, imaginaires. In 1702 Leibniz called the imaginary root a marvel of analysis, “a monstrosity of the world of ideas, almost an amphibian between being and non-being”. Everyone could calculate with them; nobody understood them.
Let’s do as we did in chapter 2 with negative numbers: demand that the new numbers obey the old laws of arithmetic, and see what comes of it. There this approach was called the principle of permanence. We add to the real numbers one new number with a single property.
The imaginary unit $i$ is a number whose square is $-1$: $i^2 = -1$. The notation $i$ was proposed by Euler in 1777.
Since $i$ can be multiplied by real numbers and added to them, expressions of the form $a + bi$ will appear. Nothing more will be needed: the sum and the product of two such expressions again have this form, as we’ll see in a moment.
A complex number is a number of the form $z = a + bi$, where $a$ and $b$ are real numbers. The number $a$ is called the real part of $z$, written $a = \operatorname{Re} z$, and $b$ the imaginary part, $b = \operatorname{Im} z$. Two complex numbers are equal when both their real parts and their imaginary parts are equal. The set of complex numbers is written $\mathbb C$.
The real numbers are the complex numbers with zero imaginary part: $3 = 3 + 0i$. Numbers of the form $bi$ are called purely imaginary. Complex numbers are added part by part: $(2 + 3i) + (4 - i) = 6 + 2i$. They’re multiplied by expanding the brackets and replacing $i^2$ with $-1$:
The powers of $i$ repeat with period four: $i^1 = i$, $i^2 = -1$, $i^3 = i^2 \cdot i = -i$, $i^4 = (i^2)^2 = 1$, and then $i$ again. So $i^{2026} = i^{4 \cdot 506 + 2} = i^2 = -1$.
Calculate $(4 + 3i)^2$.
By the square of a sum: $(4 + 3i)^2 = 16 + 2 \cdot 4 \cdot 3i + (3i)^2 = 16 + 24i + 9i^2$. Here $9i^2 = -9$, so the answer is $16 - 9 + 24i = 7 + 24i$. Note: $|4 + 3i| = 5$, and $|7 + 24i| = \sqrt{49 + 576} = 25 = 5^2$. Why the modulus of a square is the square of the modulus will become clear a little further on.
Division
To divide by a complex number, we need to get rid of $i$ in the denominator, just as in the chapter on roots we got rid of a root. The number that helps differs from the given one only in the sign of its imaginary part.
The conjugate of the number $z = a + bi$ is the number $\bar z = a - bi$.
For $z = a + bi$ the product $z\bar z$ is a non-negative real number: $z\bar z = a^2 + b^2$.
Multiply by the product formula: $(a + bi)(a - bi) = (a \cdot a - b \cdot (-b)) + (a \cdot (-b) + b \cdot a)\,i = a^2 + b^2 + 0 \cdot i$. The imaginary part cancels, and a sum of squares of real numbers is non-negative.
Complex numbers can always be added, subtracted, multiplied and divided (except by zero), and all the laws of arithmetic, commutative, associative and distributive, hold: this is checked by direct calculation with the formulas above. Number systems like this are called fields; there’s much more about them in the chapter on rings and fields.
But do such numbers exist?
We declared $i^2 = -1$ and started calculating. What if, as with division by zero, a contradiction follows from this rule sooner or later? In 1837 the Irishman William Rowan Hamilton settled the question: a complex number can be defined as an ordered pair of real numbers $(a, b)$ with the rules $(a, b) + (c, d) = (a + c, b + d)$ and $(a, b) \cdot (c, d) = (ac - bd, ad + bc)$. There’s nothing “imaginary” in this definition, only real numbers. All the laws of arithmetic for pairs are checked by direct calculation, pairs of the form $(a, 0)$ behave exactly like real numbers, and the pair $i = (0, 1)$ gives $(0, 1) \cdot (0, 1) = (-1, 0)$, that is, $-1$. Since this construction is built from real numbers and is consistent, a contradiction has nowhere to come from.
The point
For two hundred and fifty years people calculated with imaginary numbers without understanding what they were. The answer was found by three people, independently. The Norwegian Caspar Wessel, a surveyor working in Denmark, presented it to the Danish Academy in 1797; the paper came out in 1799 and went unnoticed for a hundred years. Jean-Robert Argand, an amateur mathematician in Paris from Geneva, published an anonymous pamphlet on the same thing in 1806. And in 1831 Gauss set out the same idea in detail, and everyone heard. He also gave the numbers the name “complex”, that is, composite.
The idea is this. A real number is a point on a line. A complex number has two real “coordinates”, $a$ and $b$, so its place is in the plane.
The complex plane is a plane with a rectangular coordinate system in which the number $a + bi$ is shown as the point $(a, b)$. The horizontal axis is the real axis (the ordinary numbers lie on it), the vertical one the imaginary axis.
Addition in the plane is adding arrows, as with vectors: the real parts add separately and the imaginary parts separately, so $z + w$ is the fourth vertex of the parallelogram built on the arrows $0 \to z$ and $0 \to w$. Conjugation is reflection in the real axis. The distance from zero to the point $a + bi$ is $\sqrt{a^2 + b^2}$ by Pythagoras’ theorem.
The modulus of a complex number $z = a + bi$ is the distance from zero to the point $z$: $|z| = \sqrt{a^2 + b^2}$. For real numbers it’s the ordinary absolute value from chapter 2.
The parallelogram of addition gives an inequality that we promised, in the chapter on triangles, to see in the complex plane.
For any complex $z$ and $w$, $|z + w| \le |z| + |w|$. Equality holds only when one of the numbers is zero or the arrows $0 \to z$ and $0 \to w$ point the same way.
The trick is to see in the parallelogram of addition a triangle whose sides are exactly $|z|$, $|w|$ and $|z + w|$.
The rotation
Addition in the plane is an old acquaintance, the vector operation. The real transformation happens with multiplication. Let’s start with the simplest factor, $i$. Take the point $z = 3 + 2i$ and multiply: $i(3 + 2i) = 3i + 2i^2 = -2 + 3i$. The point $(3, 2)$ has moved to the point $(-2, 3)$. Again: $i(-2 + 3i) = -3 - 2i$. Again: $2 - 3i$. And again: $3 + 2i$ once more.
Multiplying by $i$ rotates the complex plane about zero by $90^\circ$ anticlockwise: the point $(a, b)$ goes to the point $(-b, a)$.
The trick is to rotate not the point but a right triangle with that point as a vertex: wherever its legs go, the point goes too.
Now look at $i^2 = -1$ with a geometer’s eyes. Multiplying by $i$ twice is rotating by $90^\circ$ and then by $90^\circ$ again, that is, by $180^\circ$. And multiplying by $-1$ is exactly a half-turn of $180^\circ$, the same one we had in chapter 2 on the number line. So $i \cdot i = -1$: two quarter turns make a half-turn. The imaginary unit is “half a half-turn”, and it lives not on the line but beside it.
In the plane the rule of signs turns into geometry. Multiplying by $-1$ is a half-turn of $180^\circ$, multiplying by $i$ a rotation by $90^\circ$, and when you multiply, rotations add up. Two half-turns make a full turn: $(-1)(-1) = 1$. Two $90^\circ$ rotations make a half-turn: $i \cdot i = -1$. In chapter 2 we derived the rule “minus times minus” from the distributive law; here you can see what the same rule looks like in the plane.
Gauss himself thought the fog around imaginary numbers had been created by unfortunate words. In 1831 he wrote that if $+1$, $-1$ and $\sqrt{-1}$ had been called not the positive, negative and imaginary (or even impossible) units but, say, the direct, inverse and lateral units, there would never have been any such obscurity.
Length and angle
Multiplying by $i$ is a rotation. What does multiplying by an arbitrary complex number do? To answer, let’s describe a point not by two coordinates but by a length and a direction.
The argument of a non-zero complex number $z$ is the angle $\varphi = \arg z$ between the positive direction of the real axis and the ray from zero to the point $z$, measured anticlockwise. It’s defined up to a whole number of full turns: $30^\circ$ and $390^\circ$ are the same argument. Zero has no argument.
If the modulus of $z$ is $r$ and its argument is $\varphi$, then its coordinates are $r\cos\varphi$ and $r\sin\varphi$. This is the definition of sine and cosine by a point on the circle: rotating the point $(1, 0)$ about zero by the angle $\varphi$ takes it to the point $(\cos\varphi, \sin\varphi)$ (details in the chapter on trigonometry).
The polar form (or trigonometric form) of a complex number is $z = r(\cos\varphi + i\sin\varphi)$, where $r = |z|$ and $\varphi = \arg z$.
For example, $1 + i = \sqrt2\,(\cos 45^\circ + i\sin 45^\circ)$, $\ i = \cos 90^\circ + i\sin 90^\circ$, $\ -1 = \cos 180^\circ + i\sin 180^\circ$. Now the main theorem of the chapter.
When complex numbers are multiplied, their moduli multiply and their arguments add: $|zw| = |z| \cdot |w|$ and $\arg(zw) = \arg z + \arg w$ (up to a full turn). In other words, multiplying by $w \ne 0$ rotates the whole plane about zero by the angle $\arg w$ and stretches it from zero by a factor of $|w|$.
The trick is to watch not the point itself but the grid in which its “address” is read: multiplying by $w$ keeps the address and changes the grid.
In polar form the theorem takes one line.
This theorem comes with an unexpected gift. Multiply two numbers of modulus $1$ by the ordinary formula and compare with the polar form of the product.
For any angles $\alpha$ and $\beta$: $\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$ and $\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$.
By the multiplication theorem $(\cos\alpha + i\sin\alpha)(\cos\beta + i\sin\beta) = \cos(\alpha + \beta) + i\sin(\alpha + \beta)$: the moduli give $1 \cdot 1 = 1$, and the arguments add. On the other hand, by the product formula the same left side equals $(\cos\alpha\cos\beta - \sin\alpha\sin\beta) + (\cos\alpha\sin\beta + \sin\alpha\cos\beta)\,i$. Complex numbers are equal when their real parts and imaginary parts are equal; equating them gives both formulas.
Formulas that school usually derives from a drawing with two triangles have simply dropped out of the rule “angles add when you multiply”. If you multiply a number by itself, the angles add again and again.
For any natural $n$: $\bigl(r(\cos\varphi + i\sin\varphi)\bigr)^n = r^n(\cos n\varphi + i\sin n\varphi)$.
Induction on $n$. For $n = 1$ both sides coincide. Suppose it holds for $n$. Then $z^{n+1} = z^n \cdot z = r^n(\cos n\varphi + i\sin n\varphi) \cdot r(\cos\varphi + i\sin\varphi)$, and by the multiplication theorem the moduli multiply, $r^n \cdot r = r^{n+1}$, and the arguments add, $n\varphi + \varphi = (n + 1)\varphi$. That’s the formula for $n + 1$.
The formula is named after Abraham de Moivre, in whose papers of 1707 and 1722 it appears implicitly; Euler wrote it in the familiar form in 1748.
Roots of unity
How many solutions does $z^3 = 1$ have? Among the real numbers, one: $z = 1$. But if multiplication is rotation, the cube of a number of modulus $1$ is a triple rotation, and any number whose angle, tripled, gives a whole number of turns will come back to the point $1$: $0^\circ$, $120^\circ$ and $240^\circ$. Three solutions, all on the unit circle.
For any non-zero complex number $c = \rho(\cos\theta + i\sin\theta)$ and natural $n$, the equation $z^n = c$ has exactly $n$ solutions:
$$w_k = \sqrt[n]{\rho}\left(\cos\frac{\theta + 360^\circ k}{n} + i\sin\frac{\theta + 360^\circ k}{n}\right), \qquad k = 0, 1, \dots, n - 1.$$In the plane they’re the vertices of a regular $n$-gon centred at zero.
The idea: by De Moivre’s formula, raising to a power multiplies the angle by $n$, and we need to see which angles, multiplied by $n$, land on the direction of $c$.
These numbers are solutions. By De Moivre’s formula $w_k^n = \left(\sqrt[n]{\rho}\right)^n\bigl(\cos(\theta + 360^\circ k) + i\sin(\theta + 360^\circ k)\bigr)$. The first factor equals $\rho$, and the angle $\theta + 360^\circ k$ differs from $\theta$ by $k$ full turns, so the cosine and sine are the same. So $w_k^n = c$.
They’re distinct. All the $w_k$ lie on the circle of radius $\sqrt[n]{\rho}$, and their angles $\frac{\theta}{n} + \frac{360^\circ}{n}k$ for $k = 0, \dots, n - 1$ differ by less than a full turn. Neighbouring ones are $\frac{360^\circ}{n}$ apart, so the points divide the circle into $n$ equal arcs and are the vertices of a regular $n$-gon.
There are no others. All solutions of $z^n = c$ are roots of the polynomial $z^n - c$ of degree $n$. The argument from the chapter on polynomials goes through word for word for complex numbers: it used only the laws of arithmetic and the fact that a product is zero only when a factor is zero. So a polynomial of degree $n$ has at most $n$ distinct roots, and the $n$ we’ve found are all of them.
The solutions of $z^n = 1$ are called the $n$th roots of unity. They’re the numbers $\cos\frac{360^\circ k}{n} + i\sin\frac{360^\circ k}{n}$, $k = 0, 1, \dots, n - 1$.
For $n \ge 2$ the sum of all the $n$th roots of unity is zero. This can be seen in two ways. By Vieta’s formulas from the last chapter the sum of the roots of the polynomial $z^n - 1$ is the coefficient of $z^{n-1}$ with its sign reversed, and that’s zero. And geometrically: a rotation by $\frac{360^\circ}{n}$ permutes the vertices of the polygon, so it doesn’t change their sum $S$; but it multiplies it by $\cos\frac{360^\circ}{n} + i\sin\frac{360^\circ}{n} \ne 1$, so $S = 0$.
Find the root of $z^3 = -8$ with positive imaginary part.
The number $-8$ has modulus $8$ and argument $180^\circ$. The roots have modulus $\sqrt[3]{8} = 2$ and arguments $\frac{180^\circ + 360^\circ k}{3}$, that is, $60^\circ$, $180^\circ$, $300^\circ$. The one with positive imaginary part has angle $60^\circ$: $z = 2(\cos 60^\circ + i\sin 60^\circ) = 1 + i\sqrt3$. Check: $(1 + i\sqrt3)^3 = 2^3(\cos 180^\circ + i\sin 180^\circ) = -8$.
Bombelli’s riddle solved
Back to the equation $x^3 = 15x + 4$ and the formula $x = \sqrt[3]{2 + 11i} + \sqrt[3]{2 - 11i}$ (here $\sqrt{-121} = 11i$). Now we know that every number has three cube roots, and we can find them all.
The modulus of $A = 2 + 11i$ is $\sqrt{4 + 121} = \sqrt{125}$, so the modulus of each cube root is $\sqrt[3]{\sqrt{125}} = \sqrt5$. The number $2 + i$ has modulus exactly $\sqrt5$, and we’ve already checked that $(2 + i)^3 = 2 + 11i$. The other two cube roots are obtained by rotating by $120^\circ$ and $240^\circ$, that is, by multiplying by the cube roots of unity: $u_1 = (2 + i)\left(-\frac12 + \frac{\sqrt3}{2}i\right)$ and $u_2 = (2 + i)\left(-\frac12 - \frac{\sqrt3}{2}i\right)$.
Which of the three roots of $B = 2 - 11i$ goes with which? In the proof of Cardano’s formula (chapter 14) the product of the two terms had to be $-\frac p3 = 5$. If $u$ is a cube root of $A$, then its conjugate $\bar u$ is a cube root of $\bar A = B$, and $u\bar u = |u|^2 = 5$, exactly what’s needed. So the pairs are $u_k + \bar u_k = 2\operatorname{Re} u_k$: the sum of a number and its conjugate, always real.
Let’s calculate: $u_0 + \bar u_0 = 4$; $\ 2\operatorname{Re} u_1 = 2\left(-1 - \frac{\sqrt3}{2}\right) = -2 - \sqrt3$; $\ 2\operatorname{Re} u_2 = -2 + \sqrt3$. These are exactly the three roots of $x^3 = 15x + 4$ that we found in the last chapter with Horner’s method. In the irreducible case Cardano’s formula always works like this: the numbers $A$ and $B$ are conjugate, their cube roots split into conjugate pairs, and the three pairs give three real roots. And if the equation has no rational roots, then, as we saw in the last chapter, there’s no way at all to get to the real roots without passing through imaginary numbers.
Roots don’t disappear
The simplest use of the new numbers is quadratic equations. In chapter 10 Cardano split $10$ into two parts with product $40$ and arrived at the equation $x^2 - 10x + 40 = 0$ with discriminant $-60$. There was a promise there too: in this chapter the equation would get two full-fledged roots.
If $a$, $b$, $c$ are real, $a \ne 0$ and $D = b^2 - 4ac < 0$, then the equation $ax^2 + bx + c = 0$ has two complex roots $x_{1,2} = \frac{-b \pm i\sqrt{-D}}{2a}$, conjugate to each other.
Repeat the derivation of the quadratic formula from chapter 10: completing the square used only the laws of arithmetic and so holds for complex $x$ too. We get $\left(x + \frac{b}{2a}\right)^2 = \frac{D}{4a^2}$. The right side is negative, and it has exactly two square roots: $\pm\frac{i\sqrt{-D}}{2a}$. Indeed, $\left(\frac{i\sqrt{-D}}{2a}\right)^2 = \frac{i^2 \cdot (-D)}{4a^2} = \frac{D}{4a^2}$, and there are no others: if $y^2 = s^2$, then $(y - s)(y + s) = 0$, and $y = \pm s$. It remains to subtract $\frac{b}{2a}$.
For Cardano’s equation $x^2 - 10x + 40 = 0$ we get $x = \frac{10 \pm i\sqrt{60}}{2} = 5 \pm i\sqrt{15}$. Check with Vieta: the sum is $10$, the product $(5 + i\sqrt{15})(5 - i\sqrt{15}) = 25 + 15 = 40$. Cardano multiplied these numbers back in 1545 and got the same. Now we know where they live.
The picture also explains the puzzle from the last chapter, where shifting the graph of a cubic made two roots merge and disappear. They didn’t disappear: just as here, they met and left the line as a pair of conjugate numbers. This is no coincidence.
If all the coefficients of a polynomial are real and $z$ is a root, then the conjugate number $\bar z$ is a root too.
The idea: conjugation respects addition and multiplication and leaves real numbers unchanged. Let’s check the first. For $z = a + bi$ and $w = c + di$ the sum $z + w$ has imaginary part $b + d$, and $\bar z + \bar w$ has imaginary part $-b - d$; the real parts are equal, so $\overline{z + w} = \bar z + \bar w$. The product is $zw = (ac - bd) + (ad + bc)i$, and $\bar z \cdot \bar w = (a - bi)(c - di) = (ac - bd) - (ad + bc)i$, that is, $\overline{zw} = \bar z \cdot \bar w$; hence also $\overline{z^k} = \bar z^k$. Let $P(z) = a_n z^n + \dots + a_1 z + a_0 = 0$ with real $a_k$, so that $\bar a_k = a_k$. Take the conjugate of both sides: $\overline{P(z)} = a_n \bar z^n + \dots + a_1 \bar z + a_0 = P(\bar z)$, and the conjugate of zero is zero. So $P(\bar z) = 0$.
Solve the equation $x^2 + 4x + 13 = 0$ and give the root with positive imaginary part.
$D = 16 - 52 = -36$, $\sqrt{-D} = 6$. The roots are $x = \frac{-4 \pm 6i}{2} = -2 \pm 3i$. Check with Vieta: the sum is $-4$, the product $4 + 9 = 13$. Answer: $-2 + 3i$.
The trainer will help you practise arithmetic with complex numbers, division, equations and the polar form: the first level has addition, multiplication and powers of $i$, the second division, the modulus and quadratic equations, the third De Moivre’s formula and $n$th roots.
Any expression of your own with $i$ can be taken apart in the solver, and all the trainers are collected on the practice page.
All the roots are there
We added to the numbers the single root of a single equation, $x^2 = -1$. And we got the roots of all quadratic equations, of all equations $z^n = c$, and Cardano’s formula stopped running into the impossible. Will we have to add new numbers again, for equations of degree five or seven, for equations with complex coefficients? We won’t.
Every polynomial of non-zero degree with complex (in particular, real) coefficients has at least one complex root.
The first statement of this kind was made by Albert Girard in 1629; d’Alembert (1746), Euler and Lagrange tried to prove it. The proof usually regarded as the first nearly complete one was given by Gauss in his dissertation of 1799. One topological claim in it, which seemed obvious to Gauss, was rigorously justified only by Alexander Ostrowski in 1920. A complete proof needs the notion of continuity and will appear in the chapter on complex analysis, where the theorem comes out as a consequence of Liouville’s theorem. Here is an honest idea that you can see.
Let $P(z) = z^n + a_{n-1}z^{n-1} + \dots + a_0$ with $a_0 \ne 0$ (otherwise zero is a root). Move the point $z$ around the circle $|z| = r$ and watch what curve $P(z)$ draws. If $r$ is very small, all the values of $P(z)$ are close to $P(0) = a_0$: the curve is a small loop near $a_0$, and it doesn’t go round zero. If $r$ is very large, the term $z^n$ is larger in modulus than the sum of all the others, and $P(z)$ behaves almost like $z^n$. And while $z$ goes round the circle once, $z^n$ goes round zero $n$ times, by De Moivre’s formula. So does $P(z)$: the curve goes round zero $n$ times. Now increase $r$ smoothly. The number of times the curve winds round zero can change only at the moment the curve passes through zero, and it has changed: it was $0$ and became $n$. So for some $r$ the curve passed through zero: there’s a point $z$ on the circle where $P(z) = 0$.
Three things make this argument rigorous: a definition of the “number of turns” of a curve around a point, a proof that when the curve changes continuously this number changes only when the curve passes through the point, and the estimate “$z^n$ outweighs the other terms”. Each needs careful work with continuity, but the idea stays the same.
A polynomial of degree $n \ge 1$ with complex coefficients factors into $n$ factors of degree one: $P(z) = a_n(z - z_1)(z - z_2)\cdots(z - z_n)$. In other words, it has exactly $n$ complex roots, if each is counted as many times as its multiplicity.
Induction on $n$. For $n = 1$ the polynomial $a_1 z + a_0$ equals $a_1\left(z - \left(-\frac{a_0}{a_1}\right)\right)$. Suppose the statement holds for degree $n - 1$. By the fundamental theorem of algebra a polynomial $P$ of degree $n$ has a root $z_1$, and by the remainder theorem (its proof from chapter 14 goes through for complex numbers unchanged) $P(z) = (z - z_1)Q(z)$, where $Q$ has degree $n - 1$ and the same leading coefficient $a_n$. By the induction hypothesis $Q(z) = a_n(z - z_2)\cdots(z - z_n)$, and the factorization of $P$ follows.
For polynomials with real coefficients this gives one more thing. Let $z$ be a non-real root of such a polynomial $P$. By the theorem on conjugate roots $\bar z$ is a root too, and $\bar z \ne z$. By the remainder theorem $P(x) = (x - z)Q_1(x)$; substituting $\bar z$ gives $0 = (\bar z - z)Q_1(\bar z)$, so $Q_1(\bar z) = 0$ and $Q_1(x) = (x - \bar z)Q_2(x)$. So $P$ is divisible by $(x - z)(x - \bar z) = x^2 - 2\operatorname{Re} z \cdot x + |z|^2$, a quadratic with real coefficients and negative discriminant $-4(\operatorname{Im} z)^2$. The quotient $Q_2$ again has real coefficients: long division of real polynomials never leaves the real numbers, and the quotient is uniquely determined. Repeating this, we factor any polynomial with real coefficients into real factors of degree one and two; there are no other “irreducible bricks”. This also shows why a polynomial of odd degree has a real root: factors of degree two can’t make an odd total degree, so at least one factor has degree one.
Complex numbers are enough: an equation $P(z) = 0$ of any degree $n$ has exactly $n$ roots in them, counted with multiplicity. We’ll never need to extend the numbers again for the sake of polynomial roots. The existence of a root doesn’t mean the existence of a formula, and for degree five there is none (big question 3), but the roots themselves are always there.
Where next
In this chapter we made free use of the plane: rotations, lengths, similar triangles, the triangle inequality, the fact that opposite sides of a parallelogram are equal. The chapter’s main proof, that multiplication is a rotation with a stretch, rests on these facts. But why are they true? How do we know that a rotation preserves angles and that the angles of a triangle add up to $180^\circ$, and do we know it for certain or have we only seen it in pictures? More than two thousand years ago Euclid asked exactly this question and built geometry out of a few obvious rules. His Elements open the next part of the course, chapter 16.