• Complex Analysis: #4 Cauchy`s Theorem (simplest version)

Theorem 1
Let G ⊂ ℂ be a region, and assume that the function f : G → ℂ has an antiderivative (auf deutsch: Stammfunktion) F : G → ℂ with F' = f. Let γ be a closed path in G (that is, a continuous, closed, piecewise continuously differentiable path). (Closed means that γ(t₀) = γ(t₁).) Then ∫_γ f(z)dz = 0.

Proof

Since every polynomial has an antiderivative, it follows that the path integral around a closed path for any polynomial is zero.

Of course this is all a bit too trivial. So let’s call the following theorem the simplest version of Cauchy’s integral theorem.


Theorem 2
Let Q be a (solid) triangle in the complex plane. Assume that Q ⊂ G ⊂ ℂ, and take f : G → ℂ to be an analytic function. Let γ be the closed path traveling around the three sides of Q. Then ∫_γ f(z)dz = 0.

Proof
We may assume that γ begins and ends in a corner of Q — for example the “lowest” corner γ in the complex plane. If the lower side of Q is parallel to the real number axis, then take the right-hand corner on that side. Let us now divide the sides of Q in half, connecting the half-way points with straight line segments, thus creating four equal sub-triangles, Q₁, . . . , Q₄. Let γ_j be the path traveling around the boundary of Q_j, for j = 1, . . . , 4. Again we may assume that each γ_j begins and ends in the bottom right corner of it’s triangle. So we have

Assume further that each of these paths is parameterized in the simplest way possible, so that |γ'| = 1 and |γ_j'| = 1 for all the j. Therefore L_γ is the sum of the lengths of the three sides of the triangle Q, and L_γj = L_γ/2 for each of the j.

Let’s say that γ₁ is one of the numbers between one and four such that the value of

is the greatest. Then we certainly have

The next step is to concentrate on the triangle Q_j1. As with Q, we subdivide Q_j1 into four equal sub-triangles and we take paths around their boundaries. Choose Q_j2 to be one of these sub-triangles of Q_j1 which is such that the value of

is the greatest. Here γ_j2 is the path around the boundary of Q_j2. Now we have L_γj2= L_γ/4, and

This whole process is continued indefinitely, so that we obtain a sequence of triangles, becoming smaller and smaller, converging to a point, z₀ ∈ Q say,

Q ⊃ Q_j1⊃ Q_j2 ⊃ · · · → z₀ ∈ Q. 

For each n we have L_γjn= L_γ/2ⁿ and

But we have assumed that f is analytic, in particular it is differentiable at the point z₀. Thus we can write

f(z) = f(z₀) + f '(z₀)(z − z₀) + χ(z),

for points z in G, where χ : G → ℂ is a continuous function with

So let ∈ > 0 be arbitrarily given. Then there exists some δ > 0 such that |χ(z)| < |z − z₀| for all z with 0 < |z − z₀| < δ.

Now we need only choose n so large that |z − z₀| < δ for all z ∈ Q_jn. For such z we have

On the other hand, again since the length of L_γjnis L_γ/2ⁿ, we have

Bearing in mind Theorem 1 (and remembering that f '(z₀) is simply a constant complex number), we conclude that

Since ∈ was arbitrary and L_γ remains constant, we conclude that ∫_γ f(z)dz = 0.

But now Theorem 2 can be turned around, and we obtain (almost) the converse.


Theorem 3
Assume G ⊂ ℂ is a region and f : G → ℂ is a continuous function. Assume furthermore that for any solid triangle Q contained in G we have ∫_γ f(z)dz = 0, where γ is the path around the triangle. Then f has an antiderivative in every open disc contained in G. That is, let U = {z ∈ ℂ : |z − z_*| < r} be some such disc, where z_* is a complex number (the middle point of the disc) and r > 0 is the radius of the disc. Then there exists F : U → ℂ with F'(z) = f(z) for all z ∈ U.

Proof
By replacing f with the function f_*, where f_*(z) = f(z − z_*), we obtain the situation that z_* = 0. Clearly, if the theorem is true for f_*, then it is also true for f. Therefore, without loss of generality, we may simply assume that z_* = 0.

Within U the function F is defined to be

Here, α_z is the straight line from 0 to z, that is, α_z(t) = tz. To show that F really is an antiderivative to f in U, let z₀ be some arbitrary point of U and let z be some other point of U. Let β be the straight line connecting z₀ to z. That is, β(t) = (1 − t)z₀ + tz. Being a triangle, the integral of f around the path from 0 out to z, then from z to z₀ then from z₀ back to 0 must itself be zero. That is,

Looking at the definition of the path integral, we see that

Combining this theorem with Theorems 1 and 2, we see that if D ⊂ G is a closed disc, and γ is the circle of it’s boundary, then  ∫_γ f(z)dz = 0, for any analytic function defined in the region G. In fact, if γ is any (piecewise continuously differentiable and continuous) closed path contained within this disc-like G, then  ∫_γ f(z)dz = 0. For example we can look at a rectangle [a, b] × [c, d] contained within G. Since the rectangle can be taken to be a union of two triangles, attached along one side, we see that also the path integral around the rectangle must be zero.

More generally, the following theorem will prove to be useful.


Theorem 4
Let Q = {x + iy : 0 ≤ x, y ≤ 1} be the standard unit square in ℂ. Take ζ to be the standard closed path, traveling around the boundary of Q once in a counterclockwise direction, beginning and ending at 0. Assume that a continuously differentiable mapping ϕ : Q → ℂ is given, such that ϕ(Q) ⊂ G, a region where an analytic function f : G → ℂ is defined. Let γ = ϕ ◦ ζ be the image of ζ under ϕ. Then ∫_γ f(z)dz = 0.

Proof.
Since Q is compact, ϕ(Q) is also compact. Therefore it can be covered by a finite number of γ open discs in G. But Q can now be partitioned into a finite number of sub-squares Q₁, . . ., Q_n such that ϕ(Q_j) is in each case contained in a single one of these open discs. The theorem then follows by observing that the path integral around each of these sub-square images must be zero.

 Note:-
[The inverse images of the open discs in Q form a finite open covering V₁, . . ., V_m of Q. A sequence of partitions of Q can be obtained by cutting it along horizontal and vertical lines spaced 1/n apart, for each n ∈ ℕ. Can it be that for each of these partitions, there exists a sub-square which is not contained completely in one of the open sets V_k? But that would mean that there exists a limit point q ∈ Q such that for every ∈ > 0, there are infinitely many of these sub-squares contained within a distance of from q. However q ∈ V_k, for some k, and since V_k is open, there exists an ∈-neighborhood of q contained entirely within V_k, providing us with the necessary contradiction.]

A special case is the following.


Theorem 5
Let D₁ and D₂ be closed discs in ℂ such that D₂ is contained in the interior of D₁. Let γ_j be the closed path going once, counterclockwise, around the boundary of D_j, j = 1, 2. Let G ⊂ ℂ be a region containing D₁\D₂ and also containing the boundary of D₂. Assume that f : G → ℂ is analytic. Then ∫_γ1 f(z)dz = ∫_γ2 f(z)dz.

Proof
The annulus between D₂ and D₁ can be taken to be the image of the unit square under a continuously differentiable mapping.

A convenient notation for this situation is the following. Let γ : [0, 1] → ℂ be the path γ(t) = z₀ + re^2πit . Then we simply write

With this notation, we can say that if the analytic function f is defined in a region containing the annulus {z ∈ ℂ : r ≤ |z| ≤ R}, then we must have