Cauchy=s Integral Theorem

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Augustin Louis Cauchy (1789 Aug 21 B 1857 May 23) gave us a method of evaluating certain integrals that we cannot solve by other methods. He did so by leading us into the realm of the complex numbers and showing us that by taking our calculations on little forays into the complex plane we can solve our recalcitrant integrals with a little finesse.

We describe the complex plane as the two-dimensional set of points that puts all of the real numbers on the x-axis, puts all of the imaginary numbers (the square roots of the negative real numbers) on the y-axis, and thus makes each and every point in the set represent a complex number,

(Eq=n 1)

in which . For each such number we also define its complex conjugate,

(Eq=n 2)

in which I have used the physicists= notation (the asterisk on z: mathematicians indicate the complex conjugate by drawing a bar over the z). We can also express complex numbers in polar form. As we represent a number in the complex plane with the coordinates of a point, so we can also represent it with a straight line extending from the origin of the coordinate grid to that point. Then we have

(Eq=n 3)

in which we have the amplitude

(Eq=n 4)

and the unit phasor

(Eq=n 5)

in which

(Eq=n 6)

As with real numbers, we can have functions of complex numbers, algebraic recipes that map one set of complex numbers onto another set of complex numbers or that map the values of some parameter onto a set of complex numbers. I will assume that the reader has a full familiarity with the basics of complex arithmetic and thus proceed now to the calculus of complex numbers.

Complex Differentiation

In order to maintain a connection of consistency with real-valued analysis, we must define a complex derivative (the derivative of a complex-valued function with respect to a complex variable) as

(Eq=n 7)

in which we have made the substitution f(z)=u(x,y)+iv(x,y). In order for that derivative to exist, to have a fixed value, its value cannot depend upon the path we follow in approaching the limit. If we conceive all possible values of dz in this case as representing the points comprising a minuscule circle in the complex plane, one centered on the point at which we want to calculate the derivative, then the value of the derivative cannot depend upon where in that circle we begin the limiting move to the central point. If we set dy=0 on that circle, then we have

(Eq=n 8)

If we instead set dx=0 on that minuscule circle, then we have

(Eq=n 9)

Because we can begin at any point on the dz circle, we must have the actual derivative as a linear combination of those two solutions;

(Eq=n 10)

The value of the derivative cannot depend upon our choice of values for α and β, which criterion our function satisfies if and only if

(Eq=n 11)

and

(Eq=n 12)

Equations 11 and 12 express the Cauchy-Riemann conditions. So long as a complex function is continuous (of course) on some region of the complex plane and satisfies those equations, it will be differentiable on that region.

Having thus established the derivative of a complex function with respect to a complex variable, we now conceive the idea of differentiating a complex function with respect to the complex conjugate of a complex variable. Adapting Equation 7, we have

(Eq=n 13)

Following the steps we took above brings us to

(Eq=n 14)

and

(Eq=n 15)

Using Equations 11 and 12 to make the substitutions in Equation 15 gives us

(Eq=n 16)

according to Equation 14. That equation can stand true to mathematics if and only if

(Eq=n 17)

So now we know that we can differentiate the complex function f(z) on the complex plane if and only if it has no algebraic dependence on z*, the complex conjugate of z.

For the first of two examples, let=s differentiate the function f(z)=z3. The binomial theorem allows us to express that function as

(Eq=n 18)

so we have

(Eq=n 19)

and

(Eq=n 20)

By Equation 8 we have

(Eq=n 21)

By Equation 9 we have

(Eq=n 22)

We expect that equality between Equations 21 and 22 because the functions in Equations 19 and 20 satisfy the Cauchy-Riemann equations.

For our second example, let=s differentiate

(Eq=n 23)

That function gives us

(Eq=n 24)

and

(Eq=n 25)

Those functions do not satisfy the Cauchy-Riemann equations and we can easily see the reason why if we make the substitutions x=(z+z*)/2 and y=(z-z*)/2i into Equation 23. We get

(Eq=n 26)

which depends explicitly on z*, so, even though it is continuous on the complex plane, we cannot differentiate it.

Now that we know how to differentiate complex functions on the complex plane with respect to complex variables, we can turn our attention to integrating functions on the complex plane.

Complex Integration

As with real-number integrals, we define the complex-number integral as the appropriate limit of a Riemannian sum,

(Eq=n 27)

We can break that integral down into its real and imaginary parts for convenience and thereby get

(Eq=n 28)

which we recognize as the sum of two line integrals.

Between the points A and B in the complex plane we have an endless array of smooth curves along which we can carry out our integration. As with complex differentiation, we want complex integration to yield results that do not depend upon which of those paths we follow. If we integrate a function for which the integral has that path independence, then we can combine two different paths with endpoints A and B into a closed loop and integrate the function around the loop;

(Eq=n 29)

Thus we integrate the function from A to B along one path and then from B to A along the other path. Reversing the limits on the integration simply multiplies the unreversed integral by minus one, so we have

(Eq=n 30)

which expresses the Cauchy-Goursat theorem. That integral seems too trivial to warrant granting the status of a theorem, but sometimes the most profound ideas come expressed in the simplest statements.

Now we want to apply Stokes= theorem to Equation 30. If we have a two-component function F=(fx,fy) on the x-y plane, then we have

(Eq=n 31)

in which C represents the boundary of the surface patch S. If we rewrite Equation 30 by way of Equation 28, then Stokes= theorem gives us

(Eq=n 32)

We can only guarantee that the integral equals zero by requiring that the two integrands equal zero. But that requirement simply gives us Equations 11 and 12, the Cauchy-Riemann equations. So now we know that we can only properly integrate a complex function on some region of the complex plane if we can differentiate that function on the given region.

If we have some continuous and differentiable function on a given region of the complex plane, Cauchy tells us that we can determine its value at some point in that region by carrying out a suitable integration over an arbitrary loop in that region that encloses the chosen point. To that end we contrive a discontinuity in our function at the point in question, a rather easy thing to do. We designate our chosen point as z0 and make the replacement

(AEq=n@ 33)

At the point z=z0 the integrand in that new integral blows up in our faces (i.e. it goes toward infinity, which makes it incalculable), so we have no way in which we can integrate that function directly. Indeed, that substitution would seem to render Equation 30 no longer valid, because we have violated one of the assumptions upon which we founded it, the continuity of the function in the integrand. But, of course, M=sieur Cauchy has a little trick that he wants us to play.

We know that we could integrate our contrived function over the closed curve C if the region S the curve encloses does not contain the point z0, so we cut a minuscule patch that includes z0 out of the region S. To achieve that cut we must define a minuscule circle K (from the Greek Kyklos) centered on z0 and draw a line L from one point on that circle to a point on the curve C. Beginning at the point where L meets C, integrate all the way around C in the counterclockwise direction (which makes the integral positive), then integrate along L to K, integrate all the way around K in the clockwise direction (which makes the integral negative), and integrate back along L to the point of beginning. Because we have thus excluded the patch on which z0 resides, our integral now conforms to Equation 30. We integrated along L twice, in opposite directions, so those contributions to the line integral cancel each other, which leaves us with

(Eq=n 34)

To evaluate the second integral in that equation we explicitly describe the circle K as the set of points for which

(Eq=n 35)

in which r represents some small real-valued number and 0# θ# 2π. We can substitute that into the integral and then let r go to zero to obtain a limit, so we rewrite Equation 34 as

(Eq=n 36)

If we divide that equation by 2πi, we get the Cauchy integral formula,

(Eq=n 37)

Because the integral in Equation 37 is purely contrived, we can put the point z0 anywhere in the complex plane that suits our needs. Thus we can treat it as a variable distinct from z; that is, z0=Z. That fact means that we can differentiate our function with respect to that variable;

(Eq=n 38)

We can repeat that process indefinitely to get, after n iterations,

(Eq=n 39)

Thus we get differentiation through integration (yes, I know, it sounds like some sappy political slogan).

The Residue Theorem

We have already seen a hint of the residue theorem in our parlaying of Cauchy=s integral theorem into Cauchy=s integral formula. When we cut from our region of integration a minuscule patch containing a singularity, that cut left a residue that aided us in solving the integral. In that case we had a residue of f(z0). But if we have a sufficiently complicated function, we might have more than one residue to consider and they may have different mathematical forms. Those possibilities fall within the residue theorem, which tells us that:

If a function f(z) is analytic on some region S of the complex plane except at a finite number of isolated singular points (zj), then over a positively-oriented (counterclockwise), simple closed contour C that encloses those singularities we must have

(Eq=n 40)

standing true to mathematics.

That looks simple enough, but before we can begin to calculate residues we need to know what kind of singularities we have to deal with.

On some region of the complex plane we have a complex function with a singularity at the point zj (and, as the index implies, we can have more than one such point). Then for some positive, real-valued number R we have an annular region R>|z-zj|>0 on which we can represent f(z) by the Laurent series,

(Eq=n 41)

The part of that series with coefficients bn, the principle part of f(z) at zj, gives us the means to distinguish three types of isolated singularities:

1. If bn=0 for all n\$ 1, then the singularity at zj is a removable point of f(z). In this case we need only set f(zj)=a0 to make the function analytic at zj.

2. If bn=0 for all n>m and bm 0, then zj is a pole of order m. If m=1, we have a simple pole.

3. If the coefficients bn 0 comprise an infinite set, then zj is an essential singularity. An obvious example is

(Eq=n 42)

For solving actual integrals we want to focus our attention on singularities of type 2 and plan on finessing the other kinds. If we have a pole of order m at zj, then we calculate its residue at that point from

(Eq=n 43)

With that equation we can now solve integrals that we cannot solve by other means, as we shall see in the following essays.

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