Cauchy's Theorem for a disc

The purpose of this section is to prove the following basic version of Cauchy's Theorem:


Theorem (3)

Let be a disc, and let be holomorphic (i.e. complex-differentiable). Then for every closed curve ,

 


We shall reduce this, via the fundamental theorem of calculus, to the corresponding result for triangles.


Definition (3)

If , are complex numbers we denote by the curve

 


Less formally, is the line segment joining to in traversed from to .


Definition (4)

A triangle with vertices is the set of all points of the form

The is the sum of the segments , thought of as a piecewise curve. As part of the definition, we assume the points ordered so that this sum of curves traverses the boundary of in an direction.

 


Remark (5)

The fact that we have written means that the have to be real. (There is no order relation on complex numbers.)

Combinations of the form with the and summing to are called of the . The set of all convex combinations of a set of points is called the convex hull of the set of points. We have defined to be the convex hull of three points in the complex plane. We could equivalently have defined as the intersection of three closed half-spaces whose boundaries are the lines containing the segments , and . Draw some pictures.

Our definition includes infinitely thin triangles in which lie on a single line. In this case, the issue about the ordering of the points to ensure the boundary of the triangle is traversed anti-clockwise does not arise. is ignored.

 


Definition (6)

If is an open set of , we say that a triangle is contained in if .

 


Remark (7)

The triangle with vertices is not contained in the unit disc because the vertex is not in ...even though every other point of this triangle lie in .

 


The following result is often called the Cauchy--Goursat Theorem, or 'Cauchy's Theorem for triangles'.


Proposition (8)

Let be an open subset of and be holomorphic. Let be any triangle contained in . Then

 


Proof

This is an ingenious argument by subdivision, variants of which are useful in other more advanced areas of analysis. Let

Subdivide into subtriangles with vertices the original vertices of and the mid-points of its sides. The integral over is the sum of the integrals over the boundaries of these subtriangles, because the contributions from the edges which join any two of the mid-points of cancel.

If the absolute values of these integrals are , then we have by the triangle inequality

so it is not possible that of the s are . It follows that one of the s (at least) is . Denote the corresponding quarter-sized triangle by and define
(If there is more than one such quarter-sized triangle, pick one at random.)

Now we iterate. That is, by dividing into quarters in the same way as we did , we find a triangle one-sixteenth the size of the and such that

Proceeding in this way, we obtain a sequence of triangles
such that
By construction, moreover,

We claim next that is a single point , say. The intersection cannot consist of more than one point. For the distance between any two points and of is certainly bounded by the length of the boundary of . If were in the intersection of all the , then and we have a contradiction to as soon as .

To see that the intersection is non-empty, pick at random. Then by essentially the same observation as above, is a Cauchy sequence. For if , so . So as . The limit lies in the intersection because for all , so since is closed. Since this is true for any , .

We haven't yet used the complex-differentiability of , so this is what we do now, to show finally that . Given , there exists so that

(Here of course is assumed so small that .) We use this to estimate . By previous results
Hence, using
This supremum is and so all together we have
provided that is so large that (which eventually happens for any ). Combining with

 the factors  on each side cancel, leaving

us with . Since can be taken arbitrarily small (and is independent of , of course) it follows that as was required.

 


Corollary (9)

Let be holomorphic, where is a disc. Let be any (piecewise ) curve in . Then

 



Proof

Let be the centre of the disc, and for , define

in other words, we integrate along the line segment joining the centre of to the variable point . We see that knowing

 for every triangle in  is enough to prove that

in . And we have seen that in this case, the integral of around any closed curve is zero.

 


Remark (10)

There are much more general versions of Cauchy's Theorem, where is replaced by a more general domain . In this case, there needs to be a condition on the closed curve . Indeed, if , then is holomorphic in , but

(where the notation is shorthand for the unit circle traversed once in the anticlockwise direction). The difficulties are essentially topological---we shall discuss some generalizations (though not the most general version) later.

 


Remark (11)

All that was needed in the proof of Corollary~ was that we could joint every point in to a given point . This is true in more general sets than . For example in the cut plane

every point of can be joined to the point . This will be useful later[1].

 
  1. Open sets with this property are sometimes called 'starlike'.
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