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The integral along a contour, computed by parametrising and substituting both the point and its derivative; the bound that says it is at most the largest value times the length; and the one integral that refuses to vanish, $\oint dz/z = 2\pi i$.
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 compute a contour integral from the definition, use an antiderivative where one exists, state and apply the estimation bound, say what reversal and a constant factor do to an integral, and reproduce the computation showing that every integer power of $z$ integrates to zero round a circle except the power minus one.
Contours as functions of a real parameter, with a direction and a length. This lesson integrates along one. Everything about the definition is familiar from line integrals in the plane; what is new is that the answer is a complex number, and that one particular integral refuses to be zero.
$\displaystyle\int_C f(z)\,dz$ is the contour integral of $f$ along $C$; $\oint_C$ is written when $C$ is closed. An antiderivative of $f$ on a region is an analytic $F$ with $F' = f$ there. The estimation bound, also called the ML inequality, is $\left|\int_C f\right| \le ML$ with $M$ the largest value of $|f|$ on $C$ and $L$ its length. An integral is path independent when it depends only on the endpoints.
The definition. For a contour $z(t)$, $t \in [\alpha, \beta]$,
$$\int_C f(z)\,dz = \int_{\alpha}^{\beta} f(z(t))\,z'(t)\,dt,$$
and the right-hand side is an ordinary integral of a complex-valued function of a real variable, done by integrating the real and imaginary parts separately. The value does not depend on which parametrisation was chosen.
The rules. It is linear in $f$; it adds over contours joined end to end; and reversing the contour negates it. Where $f$ has an antiderivative $F$ on a region containing $C$,
$$\int_C f = F(\text{end}) - F(\text{start}),$$
so the integral is path independent there and every closed integral is zero.
The estimation bound. $\left|\int_C f\right| \le ML$. It is the triangle inequality applied to the defining integral, it is crude, and it is the engine of almost every proof in the rest of the course: Liouville's theorem, the vanishing of a large arc, the convergence of a Laurent series. A bound that gets small as a radius grows is usually the whole argument.
The exception. Round the circle $|z - c| = \rho$, once anticlockwise,
$$\oint (z - c)^{n}\,dz = \begin{cases} 2\pi i & n = -1 \\ 0 & n \ne -1.\end{cases}$$
Every power but one has an antiderivative on the punctured plane and so integrates to zero. The power $-1$ would need a logarithm, and the logarithm gains $2\pi i$ on a loop round the puncture — which is exactly the ambiguity of the argument from unit 1, now measured by an integral. This single line is what the residue theorem is made of.
Another way: picture
Walk round the circle and watch $\arg(z - c)$. For an integrand with an antiderivative, whatever is accumulated on the way out is given back on the way home and the loop totals nothing. For $1/(z - c)$ the antiderivative is a logarithm, whose imaginary part is the argument, and the argument does not come home: it is $2\pi$ larger than it started. The integral is measuring exactly that failure to come home.
Another way: steps
It looks like a throwaway inequality and it is the most used tool in the course, so it is worth seeing what makes it powerful.
The bound is $ML$: the largest the integrand gets, times how far you travel. Neither factor is usually sharp, and that does not matter, because the bound is almost never used to find an integral. It is used to prove an integral is zero, by showing the bound can be made smaller than any positive number.
The pattern is always the same. A contour depends on a radius $R$. Its length grows like $R$. The integrand, out there, is bounded by something that shrinks faster than $1/R$. So $ML \to 0$ while the integral itself does not change, and a fixed number smaller than every positive bound is zero.
That argument, unchanged, proves: the semicircular arc contributes nothing to a real integral evaluated by residues; the residues of a rational function with enough decay sum to zero; and — with the integral formula in place of the integral — Liouville's theorem, and through it the fundamental theorem of algebra. Four of the course's landmark results are the same estimate applied to four integrands.
The corresponding failure is worth naming too. If the integrand only decays like $1/R$, the bound tends to a constant and proves nothing at all. That is not a weakness of the method; it is the reason the hypotheses of those theorems say what they say. 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.
The estimation bound is an inequality, and reading it as an equality is the first error. $\oint_{|z| = 1} z\,dz$ is zero, while $M L = 1 \times 2\pi$: the bound is $2\pi$ and the truth is $0$. The bound never claims to be attained.
The second error is to assume a closed contour integral is always zero. It is zero when an antiderivative exists near the contour — and $1/z$ has none on a punctured disc, which is precisely why the fundamental integral of this course is $2\pi i$ rather than nothing. A learner who believes all closed integrals vanish has no residue theorem to look forward to.
The third is forgetting $z'(t)$. Writing $\int f(z(t))\,dt$ instead of $\int f(z(t))z'(t)\,dt$ produces an answer that is wrong by a factor, and it is wrong silently: nothing about the result looks strange. The habit that prevents it is to write $dz = z'(t)\,dt$ on its own line before substituting anything.
Finally, path independence is a conclusion and not an assumption. It holds where there is an antiderivative, which for an analytic function on a simply connected region is everywhere — but that is the next lesson's theorem, and the conjugate is a standing example of an integrand for which it is simply false.
$\displaystyle\int_C z^{2}\,dz$ along the segment from $0$ to $1 + i$: parametrise $z = t(1 + i)$, $t \in [0, 1]$, so $z' = 1 + i$.
Parametrise and differentiate.
The integral becomes $\int_0^1 t^{2}(1 + i)^{2}(1 + i)\,dt = (1 + i)^{3}\int_0^1 t^{2}\,dt = \tfrac13(1 + i)^{3}$.
Substitute both, then integrate in the parameter.
The antiderivative $z^{3}/3$ gives the same answer in one line, because $z^{2}$ is entire.
Two routes, one answer.
Bound $\left|\oint_{|z| = R} \dfrac{dz}{z^{2} + 1}\right|$ for large $R$. On the circle, $|z^{2} + 1| \ge R^{2} - 1$.
Bound the denominator from below.
So $M \le \dfrac{1}{R^{2} - 1}$ and $L = 2\pi R$, giving a bound of $\dfrac{2\pi R}{R^{2} - 1}$.
Largest value times length.
That tends to $0$ as $R$ grows. No value was computed, and none was needed.
The bound was the whole argument.
Substitute $w = z - 1$: the contour becomes a closed curve winding once anticlockwise about $w = 0$.
The exception is about the point the denominator vanishes at.
So the integral is $2\pi i$ — the radius and the centre made no difference, only the winding.
Let $C$ be the circle of radius $6$ about the origin, once anticlockwise, and suppose $|f(z)| \le 7$ everywhere on $C$. Fill in the table.
| Value | |
|---|---|
| The length of the contour, as a multiple of pi | |
| The estimation bound on the modulus of the integral of f, as a multiple of pi | |
| The integral of the nth power of the point round the contour | |
| The integral of one over the point round the contour, as a multiple of two pi i |
Going once anticlockwise round the unit circle, $\oint \dfrac{4}{z}\,dz = k\pi i$. What is $k$?
Answer:
Each integral is round the unit circle. Match it to its value.
| $0$ | $2\pi i$ | $-2\pi i$ | $10\pi i$ | |
|---|---|---|---|---|
| $\oint z^{5}\,dz$, anticlockwise | ||||
| $\oint \dfrac{dz}{z}$, anticlockwise | ||||
| $\oint \dfrac{dz}{z}$, clockwise | ||||
| $\oint \dfrac{5}{z}\,dz$, anticlockwise |
Put the steps of computing a contour integral directly from the definition in order.
Number the steps in order (write the number in the box):
Select every statement that is true.
This task has no paper form; do it on a device.
$C$ is the circle $|z| = 6$, once anticlockwise. $\oint_C \bar z\,dz = k\pi i$. What is $k$?
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Let $C$ be the circle of radius $2$ about the origin, once anticlockwise, and suppose $|f(z)| \le 9$ everywhere on $C$. Fill in the table.
| Value | |
|---|---|
| The length of the contour, as a multiple of pi | |
| The estimation bound on the modulus of the integral of f, as a multiple of pi | |
| The integral of the nth power of the point round the contour | |
| The integral of one over the point round the contour, as a multiple of two pi i |
You can compute a contour integral by parametrising and can bound one without computing it. Say in your own words why the power minus one behaves differently from every other. Next: Cauchy's theorem, which says when a closed contour integral has to vanish.
9. Your turn: $\displaystyle\oint_{|z| = 3} \dfrac{dz}{z - 1}$, anticlockwise, step 2