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The value of an analytic function at an interior point, and every one of its derivatives, read off an integral round the boundary — which proves that analytic functions are infinitely differentiable and gives the estimate that the rest of the unit is built on.
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 evaluate a contour integral by Cauchy's integral formula and by its derivative version, check that the numerator is analytic inside before quoting either, say why the formula proves that an analytic function has derivatives of every order, and derive the Cauchy estimate on a derivative from the estimation bound.
Cauchy's theorem, the deformation of contours it licenses, and the one integral that refuses to vanish. This lesson puts a function on top of that integral and discovers that the value it returns is the function's value at the point the denominator vanishes at.
The integral formula expresses $f(z_0)$ as an integral of $f$ round a contour enclosing $z_0$. The derivative formula does the same for $f^{(n)}(z_0)$ with the denominator raised to the power $n + 1$. The Cauchy estimate is the bound on $|f^{(n)}(z_0)|$ obtained by applying the estimation bound to that integral. A function is entire when it is analytic on the whole plane.
The formula. If $f$ is analytic on and inside a simple closed contour $C$, taken anticlockwise, and $z_0$ is inside $C$, then
$$f(z_0) = \frac{1}{2\pi i}\oint_C \frac{f(z)}{z - z_0}\,dz.$$
Where it comes from. Deform $C$ onto a tiny circle of radius $\rho$ about $z_0$, which is legitimate because the integrand is analytic between them. On that tiny circle $f(z)$ is nearly the constant $f(z_0)$, by continuity, and $\oint dz/(z - z_0) = 2\pi i$. Letting $\rho$ shrink turns nearly into exactly. So the formula is the exceptional integral of the last two lessons with a function riding on it.
Differentiating under the integral gives the rest:
$$f^{(n)}(z_0) = \frac{n!}{2\pi i}\oint_C \frac{f(z)}{(z - z_0)^{n+1}}\,dz.$$
Read that backwards and it is astonishing. The right-hand side makes sense as soon as $f$ is analytic and continuous on $C$ — so every analytic function has derivatives of every order, and each of them is an integral of the function itself. Differentiating once on an open set buys differentiating for ever, which is flatly false over the real numbers.
The estimate. Applying the estimation bound to that integral over the circle of radius $R$, with $|f| \le M$ on it:
$$\left|f^{(n)}(z_0)\right| \le \frac{n!\,M}{R^{n}}.$$
The factors of $2\pi$ and one power of $R$ cancel between the integrand and the length. This inequality is small and it proves Liouville's theorem, the fundamental theorem of algebra and the convergence of Taylor series, which is most of what is left of the course.
And the mean value property. Taking $C$ to be the circle of radius $R$ about $z_0$ and parametrising, the formula becomes $f(z_0) = \frac{1}{2\pi}\int_0^{2\pi} f(z_0 + Re^{it})\,dt$: the value at the centre is the average round the circle. Taking real parts gives the mean value property for harmonic functions that lesson 9 quoted and could not prove.
Another way: picture
An analytic function on a disc is like a drum skin held by its rim: fix what it does on the boundary circle and every point inside is decided. There is no freedom in the interior at all. That is why one integral round the edge can produce the value at a point, and it is the same rigidity that makes a bounded entire function constant.
Another way: steps
Three consequences are worth stating plainly, because each sounds impossible and each follows from the one displayed line.
Analytic once means analytic for ever. Over the reals, $f(x) = x^{2}\sin(1/x)$ is differentiable everywhere and its derivative is not continuous. Nothing of that kind exists here. A complex derivative on an open set forces derivatives of every order, all of them analytic, because each is an integral of the original.
The boundary determines the interior. Two functions analytic on a disc and agreeing on the boundary circle agree everywhere inside, since the formula computes every interior value from the boundary values alone. This is what makes a boundary value problem well posed, and it is why unit 5 can solve a physical problem by moving its boundary somewhere convenient.
A derivative can be bounded without being computed. The Cauchy estimate takes a bound on the function and returns a bound on every derivative. Nothing in real analysis does this: a real function can be tiny and have an enormous derivative, because it can oscillate. An analytic function cannot oscillate without paying for it on the boundary.
That third point is the one to carry into the next lesson. Liouville's theorem is nothing but the estimate with $n = 1$ and the radius allowed to grow. Complex differentiability is not the real kind with a letter added. The difference quotient has to settle down to the same number along every one of the infinitely many directions a point can be approached from in a plane, and that single demand is strong enough to force a function to have derivatives of every order, to equal its own Taylor series, and to be determined on a whole region by its values on a curve. Nothing in real calculus behaves like that, so a fact carried over from it without checking is a guess.
The formula is quoted as the integral is $2\pi i$ times the numerator at the point, and applied to integrands whose numerator has its own singularity inside the contour. It does not apply there.
Take $\displaystyle\oint_{|z| = 3} \dfrac{1}{z(z - 1)}\,dz$ and read it as having numerator $1/z$ and point $1$. That numerator is not analytic inside the contour — it fails at the origin — so the formula says nothing. The integral has to be handled another way, and unit 4's residue theorem is the way.
The second error is the factorial. With denominator $(z - z_0)^{n+1}$ the formula carries $n!$, so a cubed denominator brings a $2!$ and a fourth power a $3!$. A simple denominator has $n = 0$ and $0! = 1$, which is why the basic version looks as if it has no factorial in it.
The third is the direction. Everything above assumes the contour winds once anticlockwise. A clockwise contour gives the negative, and a contour winding twice gives double. The general statement carries the winding number as a factor, and for the contours in this course that factor is $1$ — which is exactly why it is easy to forget it is there.
$\displaystyle\oint_{|z| = 2} \dfrac{z^{3} + 1}{z - 1}\,dz$: the numerator is entire and the point $1$ is inside.
Check the hypotheses first.
The formula gives $2\pi i$ times the numerator at $1$, which is $2$: the integral is $4\pi i$.
One evaluation, no parametrisation.
$\displaystyle\oint_{|z| = 1} \dfrac{\cos z}{z^{3}}\,dz$: the denominator is the cube, so $n + 1 = 3$ and $n = 2$.
Match the power to the order of the derivative.
The formula gives $\dfrac{2\pi i}{2!}f''(0)$ with $f = \cos$, and $f''(0) = -1$.
The factorial belongs to the derivative version.
So the integral is $-\pi i$.
A second derivative, from one integral.
The numerator $z$ is entire and the point $2$ is inside the circle of radius $3$.
Locate the point, check the numerator.
So the integral is $2\pi i$ times the numerator at $2$, which is $4\pi i$.
$C$ is a simple closed contour winding once anticlockwise about $z = 3$, and $f(z) = z^{2} + 8$. Fill in the table.
| Value | |
|---|---|
| The value of the numerator at the enclosed point | |
| The integral with a simple denominator, as a multiple of two pi i | |
| The same integral, as a multiple of pi i | |
| The derivative of the numerator at the enclosed point | |
| The integral with a squared denominator, as a multiple of two pi i |
$C$ encloses $z = 4$ once anticlockwise, and nothing else of interest. $\oint_C \dfrac{z^{2} + 6}{z - 4}\,dz = k\pi i$. What is $k$?
Answer:
Put the steps of evaluating a contour integral by Cauchy's integral formula in order.
Number the steps in order (write the number in the box):
Each integral is round the circle $|z| = 5$, once anticlockwise, which encloses both $0$ and $1$. Match each to its value.
| $2\pi i$ | $0$ | $4\pi i$ | $6\pi i$ | |
|---|---|---|---|---|
| $\displaystyle\oint \dfrac{z^{2}}{z - 1}\,dz$ | ||||
| $\displaystyle\oint \dfrac{z^{2}}{z}\,dz$ | ||||
| $\displaystyle\oint \dfrac{z^{2}}{(z - 1)^{2}}\,dz$ | ||||
| $\displaystyle\oint \dfrac{z^{3}}{(z - 1)^{2}}\,dz$ |
$\displaystyle\oint_{|z| = 3} \dfrac{7z^{2}}{(z - 1)^{2}}\,dz = k\pi i$, anticlockwise. What is $k$?
Answer:
$f$ is analytic on and inside the circle $|z| = 8$, and $|f| \le 7$ on that circle. What bound does the Cauchy estimate give for $|f'(0)|$? Give a fraction.
Answer:
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
$C$ is a simple closed contour winding once anticlockwise about $z = 4$, and $f(z) = z^{2} + 4$. Fill in the table.
| Value | |
|---|---|
| The value of the numerator at the enclosed point | |
| The integral with a simple denominator, as a multiple of two pi i | |
| The same integral, as a multiple of pi i | |
| The derivative of the numerator at the enclosed point | |
| The integral with a squared denominator, as a multiple of two pi i |
You can read a value or a derivative off a contour integral, and you can bound a derivative without computing it. Say in your own words why the boundary values decide the interior ones. Next: Liouville's theorem, which is the estimate with the radius allowed to grow.
9. Your turn: $\displaystyle\oint_{|z| = 3} \dfrac{z}{z - 2}\,dz$, anticlockwise, step 2