# 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(t

_{0}) = γ(t

_{1}).) 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, Q

_{1}, . . . , Q

_{4}. 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 γ

_{1}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, z

_{0}∈ Q say,

Q ⊃ Q

_{j1}⊃ Q_{j2}⊃ · · · → z_{0}∈ Q.For each n we have L

_{γjn}= L

_{γ}/2

^{n}and

But we have assumed that f is analytic, in particular it is differentiable at the point z

_{0}. Thus we can write

f(z) = f(z

_{0}) + f '(z_{0})(z − z_{0}) + 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 − z

_{0}| for all z with 0 < |z − z

_{0}| < δ.

Now we need only choose n so large that |z − z

_{0}| < δ for all z ∈ Q

_{jn}. For such z we have

On the other hand, again since the length of L

_{γjn}is L

_{γ}/2

^{n}, we have

Bearing in mind Theorem 1 (and remembering that f '(z

_{0}) 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 z

_{0}be some arbitrary point of U and let z be some other point of U. Let β be the straight line connecting z

_{0}to z. That is, β(t) = (1 − t)z

_{0}+ tz. Being a triangle, the integral of f around the path from 0 out to z, then from z to z

_{0}then from z

_{0}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 Q

_{1}, . . ., 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 V

_{1}, . . ., 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 D

_{1}and D

_{2}be closed discs in ℂ such that D

_{2}is contained in the interior of D

_{1}. 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 D

_{1}\D

_{2}and also containing the boundary of D

_{2}. Assume that f : G → ℂ is analytic. Then ∫

_{γ1}f(z)dz = ∫

_{γ2}f(z)dz.

*Proof*

The annulus between D

_{2}and D

_{1}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) = z

_{0}+ 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

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