Cauchy’s integral formula
The values of an analytic function on a closed curve completely determine its value everywhere inside. There is nothing like it in real calculus, and from that rigidity come Liouville’s theorem and the fundamental theorem of algebra.
01The value inside, read from the boundary
The previous theorem says when an integral is zero. This one does something much stronger: it says that the values of an analytic function on a closed curve completely determine what the function is worth everywhere in the interior.
There is nothing equivalent in real calculus. A real function can take any value in the interior of an interval without the endpoints knowing about it. An analytic one cannot: it is so tied to itself that the boundary defines it.
The real part of \( f \) is a harmonic function, that is, a potential. That the boundary determines the interior is exactly the Dirichlet problem: once the potential is fixed on the electrodes, the potential is determined throughout the region. Cauchy’s formula is the complex version of that fact, and from it come the Poisson formulas used to solve it.
02Where it comes from
The proof is a good example of how Cauchy’s theorem is used. The function \( f(z)/(z - z_0) \) is analytic everywhere except at \( z_0 \), so the contour \( C \) can be deformed into a tiny circle of radius \( \varepsilon \) around \( z_0 \) without changing the value. On that circle, with \( z = z_0 + \varepsilon e^{j\theta} \):
The \( \varepsilon \) cancels, and since the result does not depend on \( \varepsilon \), it can be taken as small as desired. There continuity does the rest: \( f(z_0 + \varepsilon e^{j\theta}) \) tends to \( f(z_0) \), which comes out of the integral and leaves \( 2\pi j\,f(z_0) \).
Incidentally, that intermediate equality holds for any radius and is a result in its own right:
Taking real parts, this says that the potential at the center of a disk is the average of the potential on the boundary. It is the property used by any relaxation method to solve Laplace numerically: the neighbors are averaged and you iterate.
03The average that does not depend on the radius
On the left, the circle; on the right, the curve traced by the values of \( f \) on it. The yellow point is the average of that curve, and it has to fall right on \( f(z_0) \). Change the radius: as long as the function is analytic inside, the average does not move.
04And the derivatives, as a bonus
Differentiating the formula with respect to \( z_0 \) —which is allowed, and the formal proof is in any book— yields a whole family of formulas:
Read forwards, it lets you compute the nth derivative with an integral. Read backwards, it is what turns many contour integrals into a simple derivative:
And read as a statement, it says something much deeper: if \( f \) is differentiable once on a neighborhood, the formula guarantees that all the derivatives exist, because the right-hand side is defined for any \( n \). That is the proof of what had been announced since topic 3: in the complex plane, differentiable once implies differentiable infinitely many times.
05What follows
| Result | What it says | Where it is used |
|---|---|---|
| Cauchy’s inequality | \( |f^{(n)}(z_0)| \le \dfrac{n!\,M}{r^{n}} \) | Bounding derivatives without computing them |
| Liouville’s theorem | Entire and bounded \( \Rightarrow \) constant | The fundamental theorem of algebra follows from it |
| Maximum modulus | \( |f| \) attains its maximum on the boundary, never inside | The peak of \( |H| \) over a region is on the edge |
| Mean value | \( f(z_0) \) is the average over any circle | Relaxation methods for solving Laplace |
Liouville’s deserves a line of its own because of how unexpected it is: an entire function that never exceeds a fixed value has to be constant. Applied to \( 1/p(z) \) with \( p \) a polynomial with no roots, it leads to an absurdity, and hence every polynomial of degree \( \ge 1 \) has at least one complex root. The fundamental theorem of algebra comes out as a corollary of a contour integral.
06In the lab
For \( f(z) = z^2 \) and \( z_0 = 1 \), evaluate the function at eight equally spaced points of the circle of radius 0.5, average the eight complex values and check that you get \( 1 \), which is \( f(1) \). Repeat with radius 1 and verify that the average did not change.
Compute \( \oint \dfrac{e^{z}}{(z-1)^3}\,dz \) over \( |z| = 2 \) using the formula for derivatives with \( n = 2 \). Verify that it gives \( \pi j\,e \).
Set up a 20 × 20 grid in a spreadsheet, fix the potential at the edges and replace each interior cell with the average of its four neighbors, iterating until it settles. It is the mean value property turned into a computational method.
07Common mistakes
- Using the formula with singularities of \( f \) inside. \( f \) has to be analytic on the whole interior; the only special point allowed is \( z_0 \).
- Forgetting the \( n! \) in the derivative formula. It is the mistake that costs the most on an exam.
- Mixing up the exponent. For the nth derivative the denominator is raised to \( n+1 \).
- Believing that the average depends on the radius. It does not, and that is precisely the result.
- Applying it to non-analytic functions. With \( |z|^2 \) the average over the circle is \( |z_0|^2 + r^2 \): it does not match the value at the center and, on top of that, it depends on the radius.
08Self-assessment
Write Cauchy’s formula.
\( f(z_0) = \dfrac{1}{2\pi j}\displaystyle\oint_C \dfrac{f(z)}{z-z_0}\,dz \), with \( f \) analytic inside and on \( C \).
What is \( \oint \dfrac{e^z}{z}\,dz \) over \( |z| = 1 \)?
\( 2\pi j\,e^{0} = 2\pi j \): it is the formula with \( f(z) = e^z \) and \( z_0 = 0 \).
And \( \oint \dfrac{\operatorname{sin} z}{(z - \pi/2)^2}\,dz \)?
With \( n = 1 \): \( 2\pi j \cdot \cos(\pi/2) = 0 \). The derivative of the sine at \( \pi/2 \) vanishes.
What does the mean value property say?
That \( f(z_0) \) is the average of the values of \( f \) over any circle centered there, regardless of the radius.
Why is an analytic function infinitely differentiable?
Because the derivative formula gives an expression valid for every \( n \) starting from a single integral. Being differentiable once on a neighborhood is enough.
Where is the maximum of \( |f| \) in a region?
On the boundary, never in the interior, unless \( f \) is constant. This is the maximum modulus principle.
09Further reading
- Ruel V. Churchill and James W. Brown. Complex Variables and Applications. 7th ed. (Spanish edition, “Variable compleja y aplicaciones”), McGraw-Hill, 2004. The formula, the derivative formula, Liouville and maximum modulus, in that order and with the proofs.
- A. David Wunsch. Complex Variables with Applications. 2nd ed. (Spanish edition, “Variable compleja con aplicaciones”), Addison-Wesley Iberoamericana, 1997. Poisson’s formula and the Dirichlet problem solved with this tool.
- Murray R. Spiegel. Complex Variables. Schaum’s Outline Series (Spanish edition, “Variable compleja”), McGraw-Hill, 2011. Contour integrals solved with the derivative formula, which is half of the exam exercises.