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An analytic function integrates to zero round any closed contour in a region with no holes, proved from Green's theorem and the Cauchy-Riemann equations — and used mainly in its other form, which says a contour may be deformed onto a small circle round each point the integrand misbehaves at.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to state Cauchy's theorem with its hypotheses, reproduce its proof from Green's theorem and the Cauchy-Riemann equations, decide whether it applies to a given integrand and contour, use it to conclude that an integral vanishes, and use its deformation form to move an awkward contour onto a small circle where the integral is already known.
Contour integrals, the estimation bound, and the fact that a function with an antiderivative integrates to zero round a closed contour. Also Green's theorem from vector calculus, which converts a loop integral into a double integral over the region inside — and which is where the hypothesis about the inside comes from.
A region is simply connected when it has no holes: every loop in it can be shrunk to a point without leaving it. A disc is simply connected; an annulus is not. Cauchy-Goursat is the theorem of this lesson, named for both because Goursat removed the assumption that the partial derivatives are continuous. Deforming a contour means moving it continuously without crossing any point where the integrand fails to be analytic.
The theorem. If $f$ is analytic on and inside a simple closed contour $C$, then
$$\oint_C f(z)\,dz = 0.$$
More generally, if $f$ is analytic on a simply connected region, then $\oint_C f = 0$ for every closed contour $C$ in that region.
The proof in one paragraph. Write $f = u + iv$ and $dz = dx + i\,dy$. The contour integral splits into two real line integrals, $\oint (u\,dx - v\,dy)$ and $i\oint (v\,dx + u\,dy)$. Green's theorem turns each into a double integral over the region inside, with integrands $-(v_x + u_y)$ and $u_x - v_y$. The Cauchy-Riemann equations make both zero. So both double integrals vanish, and so does the contour integral.
Three consequences, and they are what the theorem is used for.
Path independence. On a simply connected region, $\int_C f$ depends only on the endpoints, since two paths with the same ends make a loop.
Antiderivatives exist. Fixing a base point and integrating gives an analytic $F$ with $F' = f$ — which is how an analytic function on a disc gets an antiderivative even when no formula for one can be written down.
Contours deform. If two closed contours can be slid onto one another without crossing a point where $f$ fails to be analytic, their integrals agree. This is the consequence that does the work: an integral round an awkward contour is moved onto a small circle round each bad point, where it can be computed.
Where it fails, and why that is the interesting case. $\oint_{|z| = 1} dz/z = 2\pi i \ne 0$. The integrand is analytic at every point of the contour and undefined at one point inside, and that single point is the entire difference. The theorem is not wrong there; it simply does not apply, and the number it fails to produce — $2\pi i$ — turns out to carry all the information about the missing point. Unit 4 gives that number a name: the residue.
Another way: picture
A loop in a region where $f$ is analytic can be shrunk to a point, and the integral does not change while it shrinks — so it must have been zero all along. Put a puncture inside and the loop snags on it: it can be shrunk down to a small circle round the puncture and no further. The integral is then whatever that small circle gives, and it cannot be reduced to nothing.
Another way: steps
The theorem is stated as the integral is zero and used as the contour may be moved, so it is worth seeing exactly how the second follows from the first.
Let $C_1$ and $C_2$ be closed contours, with $C_2$ inside $C_1$, and let $f$ be analytic on the region between them. That region's boundary consists of $C_1$ taken forwards and $C_2$ taken backwards — the two orientations from the winding-number lesson. Cut the region open with a pair of crosscuts and it becomes simply connected; the integrals along the two sides of each cut cancel because they are traversed in opposite directions; so the theorem applies to what is left and gives
$$\oint_{C_1} f - \oint_{C_2} f = 0.$$
So the two integrals are equal. Nothing was assumed about $f$ inside $C_2$ — it may be undefined there, and usually is.
That is the whole technique. An integral round some awkward contour is equal to the integral round a tiny circle about each bad point, and a tiny circle is where the arithmetic is easy. Every remaining computation in this course is this argument plus the one integral $\oint dz/z = 2\pi i$.
It also explains why simply connected is a hypothesis rather than a technicality. On an annulus, $1/z$ is analytic and its integral round the central circle is not zero, so the theorem is flatly false there. The hole is not a boundary case; it is the case the whole subject is about.
This is the error the whole lesson exists to prevent, and it produces confidently wrong answers rather than obviously wrong ones.
Every instinct points the wrong way. The integral being computed is written along $C$; the parametrisation is of $C$; the function is evaluated at points of $C$. It is entirely natural to check that $f$ behaves there and conclude that all is well. But the proof went through the region inside, and that is where the hypothesis lives.
$1/z$ on the unit circle is the standing counterexample and it is not exotic. Any expression with a denominator has this shape, and the question is the denominator ever zero inside the contour is the one to ask before anything else.
Two smaller errors. The converse is false: an integral can be zero round a particular contour without the integrand being analytic inside — $\oint_{|z| = 1} dz/z^{2} = 0$, and $1/z^{2}$ is certainly not analytic at the origin. And the theorem says nothing rather than something false when its hypotheses fail; the integral is then simply an open question, to be settled by deformation or, in unit 4, by a residue. Cauchy's theorem, Cauchy's integral formula and the residue theorem each ask something about the region a contour encloses, not about the contour. A function can be perfectly well behaved at every point of a circle and undefined at its centre, and then the theorem has not been applied but quoted. Before writing a contour integral down, ask where the integrand fails to be analytic, and then ask which of those points is inside.
$\oint_C (z^{3} - 4z + 7)\,dz$ round the boundary of any triangle: the integrand is a polynomial, hence entire.
Nothing inside to object to.
So the integral is $0$, without parametrising a single side.
Three parametrisations avoided.
$\oint_C \dfrac{dz}{z - 2}$ where $C$ is the square with corners $2 \pm 1 \pm i$, anticlockwise. The bad point $2$ is inside.
Locate the bad point first.
Between the square and a small circle about $2$, the integrand is analytic, so the two integrals agree.
Deform.
Round the small circle the answer is $2\pi i$. So the square gives $2\pi i$ too, and no side was ever parametrised.
Compute where it is easy.
The integrand fails to be analytic only at $z = 3$.
Find the bad point.
That point is outside the unit circle, so the integrand is analytic on and inside the contour: the integral is $0$.
Build the proof that an analytic function integrates to zero round a simple closed contour, using Green's theorem.
This task has no paper form; do it on a device.
$C$ is any closed contour whatever. What is $\oint_C z^{3}\,dz$?
Answer:
Select every condition that Cauchy's theorem actually requires.
This task has no paper form; do it on a device.
Each integral is round the unit circle. Match each integrand to what Cauchy's theorem has to say about it.
| Analytic everywhere, so the theorem applies and the integral is zero | Not analytic at a point inside, so the theorem says nothing | Analytic at no point at all, so the theorem does not apply | Its only bad point lies outside, so the theorem applies and the integral is zero | |
|---|---|---|---|---|
| $e^{z}$ | ||||
| $\dfrac{1}{z}$ | ||||
| $\bar z$ | ||||
| $\dfrac{1}{z - 2}$ |
A student applies Cauchy's theorem to $\dfrac{1}{z}$ round the circle $|z| = 6$, concludes the integral is $0$, and is told the answer is $2\pi i$. What went wrong?
$C$ is any simple closed contour winding once anticlockwise about the origin — a square, an ellipse, anything. $\oint_C \dfrac{dz}{z} = k\pi i$. What is $k$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Each integral is round the unit circle. Match each integrand to what Cauchy's theorem has to say about it.
| Analytic everywhere, so the theorem applies and the integral is zero | Not analytic at a point inside, so the theorem says nothing | Analytic at no point at all, so the theorem does not apply | Its only bad point lies outside, so the theorem applies and the integral is zero | |
|---|---|---|---|---|
| $e^{z}$ | ||||
| $\dfrac{1}{z}$ | ||||
| $\bar z$ | ||||
| $\dfrac{1}{z - 4}$ |
You can decide whether the theorem applies by looking inside the contour rather than at it, and you can deform a contour onto a circle. Say in your own words why one missing point inside is enough to break the hypothesis. Next: Cauchy's integral formula, which turns that missing point into a way of reading off values.
9. Your turn: $\oint_{|z| = 1} \dfrac{dz}{z - 3}$, anticlockwise, step 2