Complex analysis

The last chapter ended with complex numbers simplifying Fourier series: when differentiated, the harmonic $e^{inx}$ is simply multiplied by $in$. Let’s take the next step and allow not only the value of a function but also its argument to be complex. Let $z = x + iy$ roam the whole plane, and let $f(z)$ be a point of the plane too. Such functions can be differentiated and integrated by almost the same rules as ordinary ones, but they’re far more rigid: demand a single derivative of a function, and it gets all its properties at once.

This chapter has two old debts to pay. Chapter 30 left a riddle: the function $\frac{1}{1 + x^2}$ is smooth on the whole line, yet its Taylor series $1 - x^2 + x^4 - \dots$ converges only for $|x| < 1$. Moving the centre of the series to a point $a$, we got the radius $\sqrt{1 + a^2}$, the distance to something standing one unit off the line. And in chapter 15 we promised to prove the fundamental theorem of algebra: every polynomial has a complex root. Both debts will be paid by one and the same idea, and we’ll arrive at it as we go round the exhibition.

An exhibition, because a function $f\colon \mathbb C \to \mathbb C$ takes the plane to the plane, and its graph would need four dimensions. It can’t be drawn, so we’ll hang each function on the wall as a painting and derive theorems from what the paintings show. The subject is also called the theory of functions of a complex variable.

How to hang a function on the wall

The number $f(z)$ has two characteristics: direction (the argument) and length (the modulus). We’ll show direction by colour and length by brightness. Paint each point $z$ of the plane the colour of the number $f(z)$, and the function turns into a coloured plane.

Our palette is this. Positive numbers are red, $i$ is yellow-green, negative numbers are cyan, $-i$ is violet; going round anticlockwise takes you through the rainbow: red, yellow, green, cyan, blue, magenta. Brightness changes in rings: in each ring $|f|$ doubles, from $2^k$ to $2^{k+1}$, and the ring gets lighter towards its outer edge. The legend in the corner of the picture is the colouring of the function $f(z) = z$ itself.

Now the paintings can be read. Take $z^2$. Around zero the colours go round the circle twice, anticlockwise, and the rings crowd towards the centre: there $|z^2|$ tends to zero. That’s what a zero of order two looks like. The reason is de Moivre’s formula from chapter 15: $\arg z^2 = 2\arg z$, and while $z$ goes round zero once, $z^2$ goes round it twice. Near any zero of order $n$ the function behaves like $c(z - a)^n$, and the colours go round the point $a$ exactly $n$ times.

For $1/z$ the picture is the same, but the colours go round the circle the other way, and the rings crowd in because the modulus grows to infinity: $\arg\frac1z = -\arg z$. Such a point is called a pole. Now open $\frac{1}{1 + z^2}$. On the real axis there’s nothing special: the colour is red, the brightness changes smoothly, the function is positive and at most one. But at the points $i$ and $-i$ there are two whirlpools, in which the colours spin clockwise. Remember them: the answer to the riddle from chapter 30 is hidden in them.

Around a point $a$ the colours go round a full circle three times, clockwise, and the brightness rings crowd towards $a$. What’s happening to the function at this point?

A zero of order three A pole of order three: $f(z) \approx \frac{c}{(z - a)^3}$ Three different zeros Nothing special, it’s a colouring glitch

Near $a$ the function behaves like $c(z - a)^{-3}$, and the argument of this expression is $\arg c - 3\arg(z - a)$: the point goes round $a$ anticlockwise, and the value goes round three times clockwise. The modulus meanwhile grows to infinity.

$z^2$ versus $\bar z$: rigidity

We define the derivative as in chapter 26, only now the increment $h$ is complex and can approach zero from any side: from the left, from above, along a spiral.

A complex increment. The limit must be the same however $h$ tends to zero. The change in value is a complex number too, an arrow in the plane of values. Example: for $f(z) = z^2$ the quotient equals $\frac{2z_0h + h^2}{h} = 2z_0 + h \to 2z_0$. At the point $z_0 = 1 + i$ the derivative is $2 + 2i$. Let’s check with numbers: at $h = 0.01$ the quotient is $2.01 + 2i$, at $h = 0.01i$ it’s $2 + 2.01i$. The same from both sides.

The proofs of the rules of differentiation from chapter 26 use only arithmetic and limits, and those are the same for complex numbers. So sums, products and quotients of holomorphic functions are holomorphic (quotients where the denominator isn’t zero), the chain rule works, and $(z^n)' = nz^{n-1}$. Polynomials are holomorphic on the whole plane, the fraction $\frac{1}{1 + z^2}$ everywhere except $\pm i$. And a holomorphic function is continuous: the difference $f(z_0 + h) - f(z_0)$ is $h$ times the difference quotient and tends to $0 \cdot f'(z_0) = 0$.

But the most harmless-looking function, $f(z) = \bar z = x - iy$, the reflection in the real axis, has no derivative anywhere. The difference quotient equals $\bar h / h$: for real $h$ it’s $1$, for purely imaginary $h$ it’s $-1$. As a map of the plane the reflection is perfectly smooth, but the complex derivative demands more: that near a point the function act as multiplication by a single number, that is, as a rotation with a stretch. No rotation can stand in for a reflection.

Write the function in terms of its real and imaginary parts: $f(x + iy) = u(x, y) + i\,v(x, y)$. The derivative requires that approaching the point along the $x$-axis and along the $y$-axis give the same result. This requirement is expressed by two equations in partial derivatives, the ones that Jean le Rond d’Alembert wrote down in 1752 in a work on the resistance of fluids, and that Cauchy and Riemann made the foundation of the whole theory.

The rates of change of $u$ and $v$ along the $x$-axis. Together they give the derivative: $f'(z_0) = u_x + i\,v_x$ (for short we write $u_x = \frac{\partial u}{\partial x}$, as in chapter 32). The rates of change along the $y$-axis. The equations say that these are determined by the first: a step up acts just like a step to the right, only turned through $90^\circ$. Example: $z^2 = (x^2 - y^2) + 2xy\,i$, so $u_x = 2x = v_y$ and $u_y = -2y = -v_x$. But for the fake $(x^2 - y^2) + xy\,i$, with the two lost, $u_x = 2x$ while $v_y = x$, and $u_y = -2y$ while $-v_x = -y$: the equations fail at every point except zero, and there’s no derivative there.

The trick is to approach the point $z_0$ by two roads, along the real axis and along the imaginary axis, and compare where $f$ sends a small step along each.

The derivative is the limit of the quotient $\bigl(f(z_0 + \Delta) - f(z_0)\bigr)/\Delta$ however $\Delta$ tends to zero. So the limit won’t change if we take only real increments $\Delta = h$ or only purely imaginary ones $\Delta = ih$ (in the picture, the steps $h$ and $ih$, for $f(z) = z^2$). A step along the $x$-axis. For real $h$ only $x$ changes, and the numerator is $\Delta u + i\,\Delta v$, where $\Delta u = u(x_0 + h, y_0) - u(x_0, y_0)$ and likewise $\Delta v$. Divide by $h$. Complex numbers converge when their real and imaginary parts converge, and the whole quotient tends to $f'(z_0)$. So both parts have limits, the partial derivatives, and $f'(z_0) = \p1{u_x} + i\,\p1{v_x}$. The step $h$ goes to the arrow $(u_x, v_x)\,h$. A step along the $y$-axis. For $\Delta = ih$ only $y$ changes: the numerator is $\Delta u + i\,\Delta v$, where now $\Delta u = u(x_0, y_0 + h) - u(x_0, y_0)$ and likewise $\Delta v$, and we’re dividing by $ih$. The quotient tends to $f'(z_0)$, so the numerator divided by $h$ tends to $i\,f'(z_0)$. So $u_y$ and $v_y$ exist and $\p2{u_y} + i\,\p2{v_y} = i\,f'(z_0)$. The step $ih$ goes to the arrow $(u_y, v_y)\,h$. Compare the arrows. The first is $f'(z_0)\,h$, the second $i\,f'(z_0)\,h$: it’s obtained from the first by multiplying by $i$, that is, by a $90^\circ$ rotation without stretching (chapter 15). In the picture they’re perpendicular and equal in length, wherever you move $z_0$. Expand $i\,(u_x + i\,v_x) = -v_x + i\,u_x$ and equate real and imaginary parts with $u_y + i\,v_y$: $u_y = -v_x$ and $v_y = u_x$.

The converse holds too, if the partial derivatives not only exist but are continuous.

The test gives the complex exponential at once. Put

$$e^{z} = e^{x + iy} = e^x(\cos y + i\sin y).$$

At $y = 0$ this is the ordinary exponential, at $x = 0$ it’s Euler’s formula from chapter 29, and the rule $e^{z + w} = e^z e^w$ survives, because when numbers are multiplied their moduli multiply and their arguments add. Here $u = e^x\cos y$ and $v = e^x\sin y$, and $u_x = e^x\cos y = v_y$, $u_y = -e^x\sin y = -v_x$. So $e^z$ is holomorphic on the whole plane and $(e^z)' = u_x + i\,v_x = e^z$. The sine and cosine are defined through it: $\sin z = \frac{e^{iz} - e^{-iz}}{2i}$, $\cos z = \frac{e^{iz} + e^{-iz}}{2}$. On the real axis these are the familiar functions, but on the imaginary axis they grow without limit: $\sin(iy) = i\sinh y$, and, for example, $|\sin(10i)| = \sinh 10 \approx 11{,}013$.

Which of these functions is holomorphic?

$x^2 + y^2$ $e^x\cos y + i\,e^x\sin y$ $x - iy$ $|z|$

This is $e^z$: $u_x = e^x\cos y = v_y$ and $u_y = -e^x\sin y = -v_x$ at every point, and the partial derivatives are continuous. The others break the Cauchy–Riemann equations.

$e^z$ and the Mercator map: angles don’t lie

The derivative tells us what the function looks like under a magnifier. For small $h$

$$f(z_0 + h) \approx f(z_0) + f'(z_0)\,h,$$

and multiplication by the number $f'(z_0)$ is a rotation through the angle $\arg f'(z_0)$ with a stretch by a factor of $|f'(z_0)|$ (chapter 15). So a small square near $z_0$ goes almost to a square: rotated and enlarged, but not skewed. Large figures are distorted, because the rotation and stretch differ from point to point, but small ones keep their shape.

The angle between curves is the angle between their tangents. It remains to see where $f$ sends the tangent arrows.

Let the curve $\gamma_1(t)$ pass through $z_0$ at $t = t_0$ with tangent arrow $\gamma_1'(t_0) \ne 0$, and the curve $\gamma_2$ with tangent $\gamma_2'(t_0)$. The angle $\theta$ between the curves is the angle between these arrows. In the picture the curves are straight lines and the map is $e^z$, but the argument works for any. The image of the first curve is $f(\gamma_1(t))$. By the chain rule its tangent is $f'(z_0)\,\gamma_1'(t_0)$: the proof from chapter 26 goes through word for word. The line has become a spiral, but the tangent was obtained from the old one by multiplying by $f'(z_0)$. Likewise the tangent to the image of the second curve is $f'(z_0)\,\gamma_2'(t_0)$. Both arrows have been multiplied by the same number $f'(z_0) \ne 0$, that is, rotated through the same angle $\arg f'(z_0)$ and stretched by the same factor. Such a rotation changes neither the angle between the arrows nor the direction in which it’s measured: between the images it’s $\theta$ again.

The condition $f'(z_0) \ne 0$ matters. For $z^2$ the derivative at zero is zero, and angles there are doubled: the $x$- and $y$-axes, perpendicular to each other, go to the positive and negative halves of the real axis, and an angle of $90^\circ$ has become $180^\circ$. In the widget below you can see this by moving $z_0$ to zero.

Conformal maps are older than complex numbers. In 1569 Gerardus Mercator published a world map on which the angles between directions are the same as on the globe. That matters to a navigator: a course that crosses all the meridians at the same angle is a straight line on such a map, and it can be drawn with a ruler. The price was areas: the closer to the pole, the greater the stretching, and Greenland on the map looks no smaller than Africa, although in reality it’s about $14$ times smaller. The stretching is unavoidable: a globe can’t be flattened onto a plane without distorting either angles or areas, and Mercator sacrificed areas.

The second famous application is the wing. The map $z + \frac1z$ takes the circle $|z| = 1$ to the segment $[-2, 2]$, and a circle that passes through the point $1$ and is shifted slightly left and up to a profile with a rounded nose and a sharp trailing edge. The sharp edge appears where the derivative $1 - \frac{1}{z^2}$ vanishes: at the point $z = 1$ angles are doubled, and the smooth circle folds into a point. In 1910 Nikolai Zhukovsky (Joukowsky) proposed such wing profiles. The flow of an ideal fluid round a circular cylinder can be written down explicitly, and a conformal map carries it over to the wing, because the equations of such a flow don’t change under a conformal change of coordinates. The early theory of lift grew out of these calculations.

By what factor does the map $f(z) = z^3$ stretch small figures near the point $z_0 = i$? (Find the angle it turns them through while you’re at it.)

$f'(z) = 3z^2$, and $f'(i) = 3i^2 = -3$. The stretch is $|f'(i)| = 3$, the rotation $\arg(-3) = 180^\circ$: a small square near $i$ goes to a square three times bigger, upside down, near the point $i^3 = -i$.

$1/z$: a hole in the canvas

The integral in chapter 28 was taken over an interval. In the plane there are infinitely many roads between two points, and we have to integrate along a curve. Divide the curve $\gamma$ by points $z_0, z_1, \dots, z_N$ in the order of travel, multiply the value of the function at each point by the step $z_{k+1} - z_k$, and add up. It’s the same Riemann sum, only the steps are complex numbers, arrows along the curve.

The value of the function at a partition point: a complex number. A step along the curve: a small arrow. The limit is taken as the longest step tends to zero; for a continuous $f$ on a smooth curve it exists, just as in chapter 28. If the curve is given as $z = \gamma(t)$, $a \le t \le b$, then $\int_\gamma f\,dz = \int_a^b f(\gamma(t))\,\gamma'(t)\,dt$, an ordinary integral of a complex-valued function. Example: on the segment from $0$ to $1 + i$ we have $z = (1 + i)t$, $dz = (1 + i)\,dt$, and $\int z\,dz = \int_0^1 (1 + i)^2 t\,dt = \frac{(1 + i)^2}{2} = \frac{2i}{2} = i$.

The second property shows that $\oint (z - a)^n\,dz = 0$ over any closed curve not passing through $a$, for all integers $n \ne -1$: $(z - a)^n$ has the antiderivative $\frac{(z - a)^{n+1}}{n + 1}$. For $n = -1$ the antiderivative would be the function $\ln(z - a) = \ln|z - a| + i\arg(z - a)$, but it doesn’t return to its original value: after one trip round $a$ the argument has grown by $2\pi$. Let’s work out what that amounts to.

An integer power of the distance to the centre: $n$ may be negative, in which case the function is undefined at the point $a$. A step along the circle of radius $r$, traversed once anticlockwise. The answer doesn’t depend on $r$. Examples: $\oint_{|z| = 1}\frac{dz}{z} = 2\pi i$, $\oint_{|z| = 1}\frac{dz}{z^2} = 0$, $\oint_{|z| = 5} z\,dz = 0$. On the unit circle $z = e^{ix}$, $dz = ie^{ix}\,dx$, and the equation turns into $\int_{-\pi}^{\pi}e^{i(n+1)x}\,dx = 0$ for $n + 1 \ne 0$: the very orthogonality of harmonics from chapter 33.

The trick is to inscribe a regular polygon in the circle: its vertices are roots of unity, and the sum can be computed exactly.

Divide the circle by the points $z_k = a + r\omega^k$, $k = 0, 1, \dots, N$, where $\omega = \cos\frac{2\pi}{N} + i\sin\frac{2\pi}{N}$ is an $N$th root of unity (chapter 15), and form the sum $S_N = \sum_{k=0}^{N-1}(z_k - a)^n(z_{k+1} - z_k)$. Since $z_k - a = r\omega^k$ and $z_{k+1} - z_k = r\omega^k(\omega - 1)$, each term equals $r^{n+1}(\omega - 1)\,\omega^{k(n+1)}$. In the picture $N = 12$ and $r = 1$. At $n = 0$ the terms are just the sides of the polygon, $z_{k+1} - z_k$. Placed end to end, they go round the polygon and return to the start: $S_N = z_N - z_0 = 0$. At $n = 1$ each next arrow is turned through twice the angle, and the chain goes twice round a smaller polygon and closes up again. The same happens for any $n \ne -1$: $S_N = r^{n+1}(\omega - 1)(1 + q + \dots + q^{N-1})$ with $q = \omega^{n+1}$. When $N$ is larger than $|n + 1|$, the number $q \ne 1$ while $q^N = 1$, and the sum of the progression, $\frac{q^N - 1}{q - 1}$, is zero. So the limit is zero too. At $n = -1$ the factor $\omega^{k(n+1)} = 1$: all the terms are equal to $\omega - 1$, the arrows line up in a row, and $S_N = N(\omega - 1) = N\left(\cos\frac{2\pi}{N} - 1\right) + i\,N\sin\frac{2\pi}{N}$. At $N = 12$ this is $-1.61 + 6i$, at $N = 100$ already $-0.20 + 6.28i$. By the fundamental trigonometric limit (chapter 25) $N\sin\frac{2\pi}{N} \to 2\pi$, while $N\left(1 - \cos\frac{2\pi}{N}\right) = 2N\sin^2\frac{\pi}{N} \to 0$. So $\oint\frac{dz}{z - a} = 2\pi i$, and for the other $n$ the integral is zero, whatever the radius.

The function $\frac1z$ is holomorphic everywhere except zero, and yet its integral round a circle about zero isn’t zero but $2\pi i$. The canvas on which $\frac1z$ is painted has a hole, the point $0$, and the integral feels it. If there’s no hole inside the contour, the integral of a holomorphic function is always zero. That’s the main theorem of complex analysis.

The frame decides everything

In 1825 Augustin-Louis Cauchy published a memoir on integrals “between imaginary limits”. He noticed that the integral of a holomorphic function from one point to another doesn’t depend on the road, as long as there are no singular points between the roads. In other words, such an integral round a closed contour is zero. Cauchy proved this assuming the derivative was continuous. In 1900 Édouard Goursat found a proof for which the mere existence of the derivative is enough, and that’s where we’ll start.

The trick is to drive the integral into an arbitrarily small triangle, where $f$ is almost linear, and for linear functions the integral round a closed path is zero.

Write $I = \oint_{\partial\Delta} f\,dz$ (going round anticlockwise), $p$ for the perimeter of the triangle and $d$ for its longest side. Join the midpoints of the sides. The triangle falls into four, each similar to it with ratio $\frac12$. Go round each one anticlockwise: the inner segments are traversed twice, in opposite directions, and their contributions cancel. So $I = I_1 + I_2 + I_3 + I_4$. At least one of the four triangles has $|I_k| \ge \frac{|I|}{4}$, since otherwise the modulus of the sum would be less than $|I|$. Call it $\Delta_1$ and do the same with it, then again and again. We get a chain of nested triangles $\Delta_1 \supset \Delta_2 \supset \dots$ with perimeters $\frac{p}{2^n}$ and longest sides $\frac{d}{2^n}$, for which $|I| \le 4^n\,|I(\Delta_n)|$. Nested closed triangles whose sizes tend to zero have a common point $z_0$: this is the nested intervals principle (chapter 25) applied to each coordinate. At the point $z_0$ the function is differentiable: $f(z) = f(z_0) + f'(z_0)(z - z_0) + \varepsilon(z)\,(z - z_0)$, where $\varepsilon(z) \to 0$ as $z \to z_0$. The linear part $f(z_0) + f'(z_0)(z - z_0)$ is the derivative of the polynomial $f(z_0)\,z + \frac{f'(z_0)}{2}(z - z_0)^2$, and by the second property of the integral its integral round a closed broken line is zero. What’s left is $I(\Delta_n) = \oint_{\partial\Delta_n}\varepsilon(z)(z - z_0)\,dz$. On the boundary of $\Delta_n$ the factor $|z - z_0|$ is at most $\frac{d}{2^n}$, the path has length $\frac{p}{2^n}$, and by the first property $|I(\Delta_n)| \le \max_{\Delta_n}|\varepsilon| \cdot \frac{d}{2^n} \cdot \frac{p}{2^n}$. Putting it together: $|I| \le 4^n\,|I(\Delta_n)| \le \max_{\Delta_n}|\varepsilon| \cdot dp$. As $n$ grows, the triangles shrink to $z_0$, the maximum of $|\varepsilon|$ on them tends to zero, and $dp$ doesn’t change. So $I = 0$.

From triangles let’s move on to any contours. Call a region star-shaped if it has a point $c$ from which the whole region is visible: the segment from $c$ to any point of the region lies in it. A disc, a half-plane, any convex figure are star-shaped, and so is a disc with a radius removed: from a point on the continuation of that radius everything is visible. The plane without a point isn’t star-shaped: whatever point you look from, the missing point blocks the ray behind it.

The condition on the region is essential: for $\frac1z$ in the plane without zero the theorem breaks, as we’ve already seen. Cauchy’s theorem holds for any region without holes, not only star-shaped ones, but saying precisely what “a region without holes” is takes the language of topology (chapter 57). Star-shaped regions will do for us.

The next theorem shows how tightly a holomorphic function is held in place: its values inside a contour are completely determined by its values on the contour itself, as a painting is determined by its frame.

The value at any point $a$ inside the contour. The values on the contour itself, on the “frame”. The formula needs nothing else. The same $2\pi i$ as in the going-round lemma: for $f \equiv 1$ the formula turns into it. Example: $\oint_{|z| = 1}\frac{e^z}{z}\,dz = 2\pi i \cdot e^0 = 2\pi i$. If $\gamma$ is the circle with centre $a$ and radius $r$, then $z = a + re^{it}$, $dz = i(z - a)\,dt$, and the formula says that $f(a) = \frac{1}{2\pi}\int_0^{2\pi} f(a + re^{it})\,dt$: the value at the centre is the average value on the circle.

The trick is to shrink the contour down to the point $a$ itself: there $f(z)$ is almost equal to $f(a)$, and we already know the integral of $\frac{1}{z - a}$.

The function $g(z) = \frac{f(z)}{z - a}$ is holomorphic near $K$ everywhere except the point $a$: it’s a quotient of holomorphic functions. We need its integral round $\gamma$. Cut out a small disc of radius $\varepsilon$ around $a$ and slice the remaining ring with a horizontal line through $a$. We get two halves; go round the boundary of each anticlockwise. The upper half together with its boundary lies in the region obtained by removing from a convex neighbourhood of $K$ the ray running down from $a$. This region is star-shaped: all of it is visible from a point just above $a$. The function $g$ is holomorphic in it, and by Cauchy’s theorem the integral round the boundary of the upper half is zero. For the lower half the region without the ray running up from $a$ serves in the same way. Add the two zeros. The segments on the line are traversed twice, in opposite directions, and cancel. What’s left is $\gamma$ anticlockwise and the small circle clockwise. So $\oint_\gamma g\,dz = \oint_{|z - a| = \varepsilon} g\,dz$, where on the right the circle is now traversed anticlockwise. On the small circle $g(z) = \frac{f(a)}{z - a} + \frac{f(z) - f(a)}{z - a}$. By the going-round lemma the integral of the first term is $2\pi i\,f(a)$. The second term is at most $\frac{1}{\varepsilon}\max|f(z) - f(a)|$ in absolute value, the circle has length $2\pi\varepsilon$, and its integral is at most $2\pi\max_{|z - a| = \varepsilon}|f(z) - f(a)|$. As $\varepsilon \to 0$ this tends to zero: $f$ is continuous. The left side doesn’t depend on $\varepsilon$, and the right side tends to $2\pi i\,f(a)$ as $\varepsilon \to 0$. So $\oint_\gamma\frac{f(z)}{z - a}\,dz = 2\pi i\,f(a)$ exactly.

$1/(1 + z^2)$: the answer to the riddle

Cauchy’s formula turns a holomorphic function into an integral of the simplest fraction $\frac{1}{z - a}$, and the simplest fraction expands into a geometric progression. From this it follows that every holomorphic function is the sum of a power series, and not just anywhere, but in the largest disc it fits into.

This really is the Taylor series: inside the disc of convergence a power series can be differentiated term by term as many times as you like (this is proved using uniform convergence, chapter 53), so a holomorphic function has derivatives of all orders and $a_n = \frac{f^{(n)}(c)}{n!}$, as in chapter 30.

Now the riddle is solved in two lines. The function $\frac{1}{1 + z^2}$ is holomorphic everywhere except the points $\pm i$, where the denominator vanishes, and near these points its modulus grows without limit. The largest disc centred at $0$ that doesn’t touch them has radius $1$. By the theorem the series $1 - z^2 + z^4 - \dots$ converges in this disc and nowhere outside it, including on the real line for $|x| > 1$. The real line doesn’t see the poles, but the series lives in the plane, and in the plane the region of convergence is a disc whose radius is limited by the nearest trouble in any direction. With centre at the point $a$ the nearest troubles are the same, $\pm i$, and the radius is the distance to them: $\sqrt{1 + a^2}$. That’s the number from chapter 30.

Since a holomorphic function equals its series in every disc, the complex world has no smooth but “wrong” functions like Cauchy’s example $e^{-1/x^2}$ from chapter 30. On the real axis this function is smooth, but on the imaginary axis $e^{-1/(iy)^2} = e^{1/y^2}$, which flies off to infinity as $y \to 0$: at zero it’s not even continuous as a function on the plane. Look at its picture in the first widget: something unimaginable is going on near zero.

Find the radius of convergence of the Taylor series of the function $\frac{1}{x^2 + 4}$ centred at the point $x = 1$.

The denominator vanishes at $z = \pm 2i$, and at all other points of the plane the function is holomorphic. The radius is the distance from $1$ to $2i$: $|1 - 2i| = \sqrt{1 + 4} = \sqrt5 \approx 2.236$.

Residues: an integral computed by going round

The going-round lemma says that of all the powers $(z - a)^n$, the integral round a circle feels only one, $(z - a)^{-1}$. So to integrate a function with a singular point inside the contour, it’s enough to know one coefficient, the one in front of $\frac{1}{z - a}$.

For example, $\frac{1}{1 + z^2} = \frac{1}{(z - i)(z + i)}$ has a simple pole at the point $i$ with residue $\frac{1}{i + i} = \frac{1}{2i}$, while $\frac{1}{z^2}$ has a pole of order two at zero with residue $0$: there’s no $\frac1z$ term at all.

The contribution of one trip round a pole with residue $1$: the going-round lemma. The sum of the residues at all the poles inside the contour. Poles outside take no part, and all the rest of the function gives zero by Cauchy’s theorem. Example: for $\frac{1}{1 + z^2}$ and any contour with $i$ inside but not $-i$, the integral equals $2\pi i\cdot\frac{1}{2i} = \pi$.

For poles of higher order the answer is the same if the residue is taken to be the coefficient of $\frac{1}{z - a}$: expand $g$ by Taylor, and the going-round lemma wipes out every term but that one. Cauchy developed the calculus of residues in the 1820s, and since then it’s been the main way to compute hard integrals. Here’s the first example, an integral we’ve already met.

Here the answer could have been got without complex numbers. But the same trick works where no antiderivative can be written down.

Compute $\int_{-\infty}^{\infty}\frac{dx}{1 + x^4}$. Hint: the roots of the polynomial $z^4 + 1$ are $e^{i\pi/4}$, $e^{3i\pi/4}$, $e^{5i\pi/4}$, $e^{7i\pi/4}$, and two of them lie in the upper half-plane.

Take the same half-disc. Inside it are the simple poles $a_1 = e^{i\pi/4}$ and $a_2 = e^{3i\pi/4}$. The residue is $\frac{1}{Q'(a)} = \frac{1}{4a^3} = \frac{a}{4a^4} = -\frac a4$, since $a^4 = -1$. The sum of the residues is $-\frac{e^{i\pi/4} + e^{3i\pi/4}}{4} = -\frac{i\sqrt2}{4}$, because the real parts $\pm\frac{\sqrt2}{2}$ cancel. The integral is $2\pi i\cdot\left(-\frac{i\sqrt2}{4}\right) = \frac{\pi\sqrt2}{2} = \frac{\pi}{\sqrt2} \approx 2.2214$. The contribution of the arc is at most $\frac{\pi R}{R^4 - 1}$ and tends to zero.

A blank canvas: Liouville’s theorem

On the real line the graph of $\sin x$ is locked in the strip $[-1, 1]$, but in the plane, as we’ve seen, $\sin z$ grows to infinity. That’s no accident: no entire function except a constant can stay bounded. The painting of a bounded entire function is a blank canvas of a single colour.

The theorem bears the name of Joseph Liouville, who presented it in lectures in 1847, although Cauchy was the first to publish it, in 1844. And from it, in a few lines, follows the theorem we’ve owed since chapter 15.

As shown in chapter 15, from this follow the factorisation $P(z) = a_n(z - z_1)\cdots(z - z_n)$ and exactly $n$ roots counted with multiplicity. In 1799 Gauss lacked a rigorous justification of one topological fact, and it took more than a hundred years to close the gap. In our proof continuity came in exactly once, through Weierstrass’s theorem, and Cauchy’s formula did all the rest.

How many roots are inside the loop

The fundamental theorem of algebra says there are roots but not where. In a painting, zeros show up as points around which the colours go round a circle. Let’s turn the observation around: go round a loop and count how many times the value of the function winds round zero. We saw this idea in chapter 15, in the sketch of a proof of the fundamental theorem. Now it can be made precise.

The trick is to factor the polynomial: when numbers are multiplied, their arguments add.

By the fundamental theorem of algebra $P(z) = c\,(z - z_1)(z - z_2)\cdots(z - z_n)$; a multiple root is written as many times as its multiplicity. In the picture $n = 3$. When numbers are multiplied their arguments add (chapter 15): $\arg P(z) = \arg c + \arg(z - z_1) + \dots + \arg(z - z_n)$. The number $z - z_k$ is the arrow from the root $z_k$ to the point $z$. We’ll follow how each arrow turns while $z$ goes round the loop. A root inside. A ray from an interior point of a convex region crosses its boundary exactly once. So while $z$ goes round $\gamma$, the arrow from such a root points in every direction in turn and makes exactly one turn anticlockwise: its argument grows by $2\pi$. A root outside. A convex region can be separated from an outside point by a line, so from a root outside the whole loop is seen at an angle of less than $180^\circ$. The arrow from it swings within that angle and by the end of the trip returns to its original position: its argument doesn’t change. Altogether $\arg P(z)$ grows by $2\pi$ times the number of roots inside, that is, $P(z)$ goes round zero exactly that many times. For the fraction $\frac PQ$ the arguments of the factors of the denominator are subtracted, and each pole inside reduces the number of turns by one.

If the loop is complicated and goes round some point several times, a number like “×2” appears next to the point: such a root is counted as many times as the loop goes round it. The proof is the same, only the arrow from the root makes not one turn but as many as the loop winds round it. Engineers use the argument principle to check the stability of a control system without solving its characteristic equation: this is the Nyquist criterion (1932).

A painting that’s still being finished

The rigidity of holomorphic functions goes further than Cauchy’s formula: knowing a function on a tiny segment is enough to know it everywhere.

From this comes a notion that changed number theory.

A simple example: the series $1 + z + z^2 + \dots$ converges only in the disc $|z| < 1$, where its sum is $\frac{1}{1 - z}$. But the function $\frac{1}{1 - z}$ is defined everywhere except $z = 1$: it’s the continuation of the series to the whole plane minus a point. At $z = 2$ the continuation equals $-1$, although the series $1 + 2 + 4 + \dots$ itself diverges. The value of the continuation isn’t the sum of the series but the answer to the question of what the unique holomorphic function would be that coincides with the sum where the sum exists.

The main hero of this story is the Riemann zeta function

$$\zeta(s) = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \dots = \sum_{n=1}^{\infty}\frac{1}{n^s}, \qquad n^{-s} = e^{-s\ln n}.$$

For complex $s$ the modulus of a term is $n^{-\operatorname{Re}s}$, so the series converges for $\operatorname{Re}s > 1$. At $s = 2$ this is the Basel problem from chapter 30: $\zeta(2) = \frac{\pi^2}{6}$. In 1737 Euler noticed that $\zeta(s) = \prod_p\frac{1}{1 - p^{-s}}$, a product over all primes. This is the fundamental theorem of arithmetic rewritten, and it links the zeta function to the primes (chapter 3). In 1859 Bernhard Riemann, in his only paper on number theory, continued $\zeta$ to the whole plane except the point $s = 1$ and showed that the distribution of the primes is governed by the zeros of the continuation.

The continuation has the “trivial” zeros $-2, -4, -6, \dots$ and infinitely many others in the strip $0 < \operatorname{Re}s < 1$; the first of them is $\frac12 + 14.1347\ldots\,i$. Riemann conjectured that they all lie on the line $\operatorname{Re}s = \frac12$. This is the Riemann hypothesis: it has been checked for more than ten trillion zeros, but there’s still no proof, and in 2000 the Clay Mathematics Institute put it on its list of Millennium Prize Problems, with a prize of a million dollars. That there are no zeros on the line $\operatorname{Re}s = 1$ was proved in 1896 by Jacques Hadamard and Charles-Jean de la Vallée Poussin, and this gave them the prime number theorem, the asymptotic law of the distribution of primes.

More about the zeta function and the primes is in chapter 44, and about the Riemann hypothesis among the open problems in chapter 60.

Where next

The simplest paintings in our exhibition are the maps $z \mapsto az$. Multiplication by $a = p + qi$ rotates the plane through the angle $\arg a$ and stretches it by a factor of $|a|$: one goes to $a$, the imaginary unit to $ia$, and a square grid goes to a square grid, only rotated and larger. The derivative says that under a magnifier any holomorphic function looks just like this. But what if one and $i$ were allowed to go anywhere, independently of each other? The squares of the grid would become parallelograms, straight lines would stay straight, and the origin would stay put. What transformations of the plane and of space arise this way, and how to write them down, combine them and undo them, is the subject of chapter 35, where linear algebra begins.