# 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

**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

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

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

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

```
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

```
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

**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

^{[1]}.

- ↑ Open sets with this property are sometimes called 'starlike'.