0.6  Complex Numbers

The equation $x^2+1=0$ has no real solution, because the square of a real number is never negative. For most of school mathematics this gap is harmless. In linear algebra it is not: later in the book we meet real matrices, such as a rotation of the plane, whose most useful data cannot be found among the real numbers. This section builds the complex numbers, the number system that closes the gap, and collects the facts about them that the rest of the book uses.

0.6.1. The complex numbers as pairs of reals

The quick way to introduce complex numbers is to say “invent a symbol $i$ with $i^2=-1$ and compute as usual”. That raises a fair worry. How do we know that such a symbol does not lead to a contradiction, the way “invent a number $z$ with $0\cdot z=1$” does? The answer is to build the new numbers out of objects we already trust: pairs of real numbers.

Here is the recurring expression that suggests the definition. If we pretend $i^2=-1$ and expand, we get $$(a+bi)(c+di)=ac+adi+bci+bd{i}^2=(ac-bd)+(ad+bc)i.$$ So a “number $a+bi$” is really the pair $(a,b)$, and the product should be the pair $(ac-bd,ad+bc)$. We take this formula as the definition, so nothing is pretended.

A complex number is a pair of real numbers, multiplied as if $i^2=-1$.

In words: addition is coordinate by coordinate, as for points of the plane. Multiplication is the formula found by expanding. The last sentence is a naming convention: the real number $a$ is identified with the point $(a,0)$ on the horizontal axis, and $i$ is the point one unit up. Note that the imaginary part $b$ is a real number; it is not $bi$.

Two things need checking before this convention is safe.

• The notation $a+bi$ is honest. By the definitions, $a+bi=(a,0)+(b,0)(0,1)=(a,0)+(0,b)=(a,b)$. Two complex numbers are equal exactly when they are equal as pairs, so $a+bi=c+di$ with $a,b,c,d\in \mathbb{R}$ means $a=c$ and $b=d$. This is how we “compare real and imaginary parts”.
• Identifying $a$ with $(a,0)$ does not change real arithmetic. We have $(a,0)+(c,0)=(a+c,0)$ and $(a,0)(c,0)=(ac-0,0+0)=(ac,0)$. So adding or multiplying two real numbers inside $\mathbb{C}$ gives the same answer as in $\mathbb{R}$.

With these in place, the equation that motivated everything is now a computation: $i^2=(0,1)(0,1)=(0\cdot 0-1\cdot 1,0\cdot 1+1\cdot 0)=(-1,0)=-1$.

The degenerate case deserves a look too. A real number $a=a+0i$ is a complex number whose imaginary part is zero, and the number $0=0+0i$ is the only complex number with both parts zero. Every computation with real numbers is also a computation in $\mathbb{C}$, by the second check above.

A nonexample by minimal change. Change one sign in the multiplication rule: define $(a,b)\ast (c,d):=(ac+bd,ad+bc)$, which is what expanding gives if we pretend $j^2=+1$ instead. Addition is unchanged, and this product is still commutative and has $(1,0)$ as a unit. But $$(1,1)\ast (1,-1)=(1-1,-1+1)=(0,0),$$ so two nonzero elements multiply to zero. Division also breaks down: $(1,1)\ast (c,d)=(c+d,c+d)$ can never equal the unit $(1,0)$, because that would need $c+d=1$ and $c+d=0$ at once. So the nonzero element $(1,1)$ has no inverse. The single sign in $ac-bd$ is what makes the complex numbers a number system in which we can divide, as we are about to see.

The usual rules of arithmetic hold. Before computing freely, we should confirm that rearranging, regrouping, and expanding brackets are legal in $\mathbb{C}$. The definition does not promise this; it has to be checked.

From now on we compute in $\mathbb{C}$ exactly as in $\mathbb{R}$: expand brackets, collect terms, and replace $i^2$ by $-1$. The proposition is what makes that habit legitimate.

0.6.2. Conjugate and modulus

Division is still missing. To divide by $3+4i$ we want some number to multiply by that lands in the real numbers, where division is familiar. The computation $(a+bi)(a-bi)=a^2+b^2$ in Example (First Computations in $\mathbb{C}$) shows exactly which number does the job. Both ingredients get names.

Conjugating reflects a point in the real axis; the modulus is its distance from the origin.

In words: the conjugate keeps the real part and flips the sign of the imaginary part. The modulus is the length of the segment from $0$ to the point $(a,b)$, by Pythagoras. For a real number $a=a+0i$, we get $\bar{a}=a$ and $|a|=\sqrt{a^2}$, the usual absolute value, so the notation does not clash with its real meaning. For example, $\overline{3+4i}=3-4i$, $|3+4i|=5$, $\bar{i}=-i$, and $|i|=1$.

The following rules are used constantly, in this section and much later for complex inner products.

The first payoff is division. If $z\ne 0$, then $|z{|}^2=a^2+b^2$ is a strictly positive real number, so we may divide by it.

In practice, “multiply top and bottom by the conjugate”. For instance $$\begin{array}{rl}\frac{1}{3+4i}& =\frac{3-4i}{(3+4i)(3-4i)}=\frac{3-4i}{25},\\ \frac{5+i}{1-i}& =\frac{(5+i)(1+i)}{(1-i)(1+i)}=\frac{4+6i}{2}=2+3i,\end{array}$$ which agrees with $(2+3i)(1-i)=5+i$ from Example (First Computations in $\mathbb{C}$). Only one inverse exists; this is a special case of a general fact about fields proved in the next section.

The triangle inequality. The modulus measures distance: $|z-w|$ is the distance between the points $z$ and $w$ of the plane. The side of a triangle is never longer than the other two sides together, and this is the algebraic form of that fact.

The route “square, expand, bound the cross term” is worth remembering: it is exactly the route we will take for norms in inner product spaces. The inequality is often strict. For $z=1$ and $w=-1$, the left side is $0$ and the right side is $2$.

0.6.3. Two things that do not carry over from the reals

Complex numbers obey the rules of arithmetic, but not every habit from the real line survives.

The first loss is order. On $\mathbb{R}$ we use two facts about positivity: for every $x$, exactly one of $x>0$, $x=0$, $-x>0$ holds, and sums and products of positive numbers are positive. Suppose some relation $>$ on $\mathbb{C}$ had both properties. Since $i\ne 0$, either $i>0$ or $-i>0$, and in both cases the product rule gives $i^2=(-i{)}^2=-1>0$. The same argument applied to $1\ne 0$ gives $1=1^2=(-1{)}^2>0$. Then both $1$ and $-1$ are positive, so their sum $0$ is positive, which contradicts “exactly one”.

The second loss is the square-root rule $\sqrt{x}\sqrt{y}=\sqrt{xy}$, which on $\mathbb{R}$ is only claimed for $x,y\ge 0$.

Solving $w^2=z$ is a finite computation, done by comparing real and imaginary parts.

0.6.4. Polar form: multiplication rotates and scales

Addition in $\mathbb{C}$ is the familiar addition of arrows in the plane. Multiplication has a geometric meaning too, and it only becomes visible when we describe a point by its distance from $0$ and its angle, instead of by its coordinates.

Let $z\ne 0$ and put $r=|z|>0$. The number $z/r$ has modulus $1$ by Theorem 0.6.2 (d), so it lies on the unit circle. From trigonometry, every point $(x,y)$ with $x^2+y^2=1$ is $(\cos \theta ,\sin \theta )$ for some real $\theta$. We abbreviate $$e^{i\theta }:=\cos \theta +i\sin \theta (\theta \in \mathbb{R}),$$ and then every nonzero complex number can be written as $$z=r{e}^{i\theta },r=|z|>0,\theta \in \mathbb{R}.$$ This is the polar form of $z$, and $\theta$ is an argument of $z$. The word “an” matters: $\theta$ and $\theta +2\pi k$ give the same point for every integer $k$. Conversely, if $r{e}^{i\theta }=s{e}^{i\phi }$ with $r,s>0$, then taking moduli gives $r=s$ (as $|e^{i\theta }|=\sqrt{{\cos }^2\theta +{\sin }^2\theta }=1$). Comparing real and imaginary parts then gives $\cos \theta =\cos \phi$ and $\sin \theta =\sin \phi$, which from trigonometry means $\theta -\phi$ is an integer multiple of $2\pi$.

For now $e^{i\theta }$ is only a name for $\cos \theta +i\sin \theta$. The next result is the reason the name is a good one: these numbers multiply the way exponentials do.

In words: to multiply two complex numbers, multiply their moduli and add their arguments. Multiplying by a fixed $w=s{e}^{i\phi }$ is therefore a rotation of the plane about $0$ by the angle $\phi$, combined with a stretch by the factor $s$. The picture shows $z$, $w$, and $zw$ for $|z|=1.5$, $|w|=1.2$, and arguments ${20}^{\circ }$ and ${50}^{\circ }$.

The arrow for $zw$ has length $1.5\times 1.2=1.8$ and makes the angle ${20}^{\circ }+{50}^{\circ }={70}^{\circ }$ with the positive real axis: it is the arrow for $z$, turned by $w$’s angle and stretched by $w$’s length. The small arc marks the argument of $z$, the large arc marks the extra turn by the argument of $w$, and the dashed arc is the unit circle, where the numbers $e^{i\theta }$ live.

Two special cases are worth isolating. Multiplying by $i=e^{i\pi /2}$ is a quarter turn counterclockwise, which explains $i^2=-1$: two quarter turns send $1$ to $-1$. And $e^{i\theta }{e}^{-i\theta }=e^0=1$, so $(e^{i\theta }{)}^{-1}=e^{-i\theta }=\overline{e^{i\theta }}$.

Repeating Theorem 0.6.5 gives powers.

Powers that are painful to expand become one line in polar form.

0.6.5. Roots of unity

Polar form also turns the equation $z^n=c$ into a question about angles. The most important case is $c=1$.

Solution. Since $0^n=0\ne 1$, every solution is nonzero, so we may write $z=r{e}^{i\theta }$ with $r>0$. By Theorem 0.6.6, $z^n=r^n{e}^{in\theta }$, and $1=1\cdot e^{i0}$. By uniqueness of the polar form up to multiples of $2\pi$, $z^n=1$ holds exactly when $r^n=1$ and $n\theta =2\pi k$ for some integer $k$. Since $r>0$ is real, $r^n=1$ forces $r=1$. Hence the solutions are $$e^{2\pi ik/n},k\in \mathbb{Z}.$$ Put $\omega :=e^{2\pi i/n}$. By Theorem 0.6.6, $e^{2\pi ik/n}={\omega }^k$ for $k\ge 0$. Two integers $k,l$ give the same number exactly when $2\pi (k-l)/n$ is a multiple of $2\pi$, that is, when $n$ divides $k-l$. So there are exactly $n$ distinct solutions, $$1,\omega ,{\omega }^2,\dots ,{\omega }^{n-1},$$ the $n$-th roots of unity. They sit on the unit circle at equally spaced angles. For $n\ge 3$ they are the corners of a regular polygon with $n$ corners, one of them at $1$; for $n=2$ they are $\pm 1$.

For $n=6$, the angles are multiples of ${60}^{\circ }$, and the roots are $$1,\frac{1}{2}+\frac{\sqrt{3}}{2}i,-\frac{1}{2}+\frac{\sqrt{3}}{2}i,-1,-\frac{1}{2}-\frac{\sqrt{3}}{2}i,\frac{1}{2}-\frac{\sqrt{3}}{2}i.$$

The six roots sum to $0$: since ${\omega }^3=e^{i\pi }=-1$, we have ${\omega }^{k+3}=-{\omega }^k$, so the roots cancel in the opposite pairs $\{1,{\omega }^3\}$, $\{\omega ,{\omega }^4\}$, $\{{\omega }^2,{\omega }^5\}$. In the picture, the hexagon is symmetric about its center $0$. The sum of the $n$-th roots of unity is $0$ for every $n\ge 2$; see Exercise (C2: Roots of Unity Sum to Zero).

0.6.6. The Fundamental Theorem of Algebra

We built $\mathbb{C}$ so that one polynomial equation, $z^2+1=0$, has a solution. It turns out that nothing more ever needs to be added: every polynomial equation already has a solution in $\mathbb{C}$. Here a complex polynomial of degree $n$ means an expression $a_n{z}^n+\cdots +a_{1}z+a_0$ with $a_0,\dots ,a_n\in \mathbb{C}$ and $a_n\ne 0$; polynomials are studied properly in a later section of this chapter.

We do not prove this theorem here. Its proof needs tools we have not built, and it is proved in Chapter 6. Nothing in Chapters 0–5 depends on it.

One companion fact needs no heavy machinery. If the coefficients $a_0,\dots ,a_n$ are real and $p(z)=a_n{z}^n+\cdots +a_0$ has $p(z_0)=0$, then applying Theorem 0.6.2 (a) and (b) term by term, and $\overline{a_k}=a_k$, gives $p(\overline{z_0})=\overline{p(z_0)}=\bar{0}=0$. So the nonreal roots of a real polynomial come in conjugate pairs. For example, $z^2-2z+5$ has the roots $1+2i$ and $1-2i$: indeed $(1+2i{)}^2-2(1+2i)+5=(-3+4i)-(2+4i)+5=0$.

0.6.7. Exercises

A. Check your understanding

B. Practice

C. Going deeper