# 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 ${\displaystyle D\subset \mathbb {C} }$ be a disc, and let ${\displaystyle f:D\to \mathbb {C} }$ be holomorphic (i.e. complex-differentiable). Then for every closed curve ${\displaystyle \gamma :[a,b]\to D}$,

${\displaystyle \int _{\gamma }f(z)\,{\mbox{d}}z=0.}$

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

Definition (3)

If ${\displaystyle z_{0}}$, ${\displaystyle z_{1}}$ are complex numbers we denote by ${\displaystyle [z_{0},z_{1}]}$ the curve

${\displaystyle \gamma (t)=(1-t)z_{0}+tz_{1},0\leqslant t\leqslant 1.}$

Less formally, ${\displaystyle [z_{0},z_{1}]}$ is the line segment joining ${\displaystyle z_{0}}$ to ${\displaystyle z_{1}}$ in ${\displaystyle \mathbb {C} }$ traversed from ${\displaystyle z_{0}}$ to ${\displaystyle z_{1}}$.

Definition (4)

A triangle ${\displaystyle \Delta \subset \mathbb {C} }$ with vertices ${\displaystyle z_{1},z_{2},z_{3}}$ is the set of all points of the form

${\displaystyle \Delta =\{z=t_{1}z_{1}+t_{2}z_{2}+t_{3}z_{3}:t_{j}\geqslant 0,t_{1}+t_{2}+t_{3}=1\}}$
The ${\displaystyle \partial \Delta }$ is the sum of the segments ${\displaystyle [z_{1},z_{2}]+[z_{2},z_{3}]+[z_{3},z_{1}]}$, thought of as a piecewise ${\displaystyle C^{1}}$ curve. As part of the definition, we assume the points ${\displaystyle z_{1},z_{2},z_{3}}$ ordered so that this sum of curves traverses the boundary of ${\displaystyle \Delta }$ in an direction.

Remark (5)

The fact that we have written ${\displaystyle t_{j}\geqslant 0}$ means that the ${\displaystyle t_{j}}$ have to be real. (There is no order relation on complex numbers.)

Combinations of the form ${\displaystyle \sum t_{j}z_{j}}$ with the ${\displaystyle t_{j}\geqslant 0}$ and summing to ${\displaystyle 1}$ are called of the ${\displaystyle z_{j}}$. The set of all convex combinations of a set of points is called the convex hull of the set of points. We have defined ${\displaystyle \Delta }$ to be the convex hull of three points in the complex plane. We could equivalently have defined ${\displaystyle \Delta }$ as the intersection of three closed half-spaces whose boundaries are the lines containing the segments ${\displaystyle [z_{1},z_{2}]}$, ${\displaystyle [z_{2},z_{3}]}$ and ${\displaystyle [z_{3},z_{1}]}$. Draw some pictures.

Our definition includes infinitely thin triangles in which ${\displaystyle z_{1},z_{2},z_{3}}$ 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 ${\displaystyle \Omega }$ is an open set of ${\displaystyle \mathbb {C} }$, we say that a triangle ${\displaystyle \Delta }$ is contained in ${\displaystyle \Omega }$ if ${\displaystyle \Delta \subset \Omega }$.

Remark (7)

The triangle with vertices ${\displaystyle 0,1,i/2}$ is not contained in the unit disc ${\displaystyle D=\{z:|z|<1\}}$ because the vertex ${\displaystyle 1}$ is not in ${\displaystyle D}$...even though every other point of this triangle lie in ${\displaystyle D}$.

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

Proposition (8)

Let ${\displaystyle \Omega }$ be an open subset of ${\displaystyle \mathbb {C} }$ and ${\displaystyle f:\Omega \to \mathbb {C} }$ be holomorphic. Let ${\displaystyle \Delta }$ be any triangle contained in ${\displaystyle \Omega }$. Then

${\displaystyle \int _{\partial \Delta }f(z)\,{\mbox{d}}z=0.}$

Proof

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

${\displaystyle I=\left|\int _{\partial \Delta }f(z)\,{\mbox{d}}z\right|}$
Subdivide ${\displaystyle \Delta }$ into ${\displaystyle 4}$ subtriangles with vertices the original vertices of ${\displaystyle \Delta }$ and the mid-points of its sides. The integral over ${\displaystyle \partial \Delta }$ is the sum of the integrals over the boundaries of these ${\displaystyle 4}$ subtriangles, because the contributions from the edges which join any two of the mid-points of ${\displaystyle \Delta }$ cancel.

If the absolute values of these ${\displaystyle 4}$ integrals are ${\displaystyle J_{1},J_{2},J_{3},J_{4}}$, then we have by the triangle inequality

${\displaystyle I\leqslant J_{1}+J_{2}+J_{3}+J_{4}}$
so it is not possible that of the ${\displaystyle J}$s are ${\displaystyle . It follows that one of the ${\displaystyle J}$s (at least) is ${\displaystyle \geqslant I/4}$. Denote the corresponding quarter-sized triangle by ${\displaystyle \Delta _{1}}$ and define
${\displaystyle I_{1}=\left|\int _{\partial \Delta _{1}}f(z)\,{\mbox{d}}z\right|}$
(If there is more than one such quarter-sized triangle, pick one at random.)

Now we iterate. That is, by dividing ${\displaystyle \Delta _{1}}$ into quarters in the same way as we did ${\displaystyle \Delta }$, we find a triangle ${\displaystyle \Delta _{2}}$ one-sixteenth the size of the ${\displaystyle \Delta }$ and such that

${\displaystyle I_{2}=\left|\int _{\partial \Delta _{2}}f(z)\,{\mbox{d}}z\right|\geqslant I_{1}/4.}$
Proceeding in this way, we obtain a sequence of triangles
${\displaystyle \Delta \supset \Delta _{1}\supset \Delta _{2}\supset \ldots }$
such that
${\displaystyle I_{n}=\left|\int _{\partial \Delta _{n}}f(z)\,{\mbox{d}}z\right|\geqslant 4^{-n}I.}$
By construction, moreover,
${\displaystyle \operatorname {length} (\partial \Delta _{n})=2^{-n}L,\;\;L=\operatorname {length} (\partial \Delta ).}$

We claim next that ${\displaystyle \bigcap \Delta _{n}}$ is a single point ${\displaystyle z_{0}}$, say. The intersection cannot consist of more than one point. For the distance between any two points ${\displaystyle z}$ and ${\displaystyle w}$ of ${\displaystyle \Delta _{n}}$ is certainly bounded by the length of the boundary of ${\displaystyle \Delta _{n}}$. If ${\displaystyle z\neq w}$ were in the intersection of all the ${\displaystyle \Delta _{n}}$, then ${\displaystyle |z-w|>0}$ and we have a contradiction to ${\displaystyle z,w\in \Delta _{n}}$ as soon as ${\displaystyle 2^{-n}L<|z-w|}$.

To see that the intersection is non-empty, pick ${\displaystyle z_{n}\in \Delta _{n}}$ at random. Then by essentially the same observation as above, ${\displaystyle (z_{n})}$ is a Cauchy sequence. For if ${\displaystyle n, ${\displaystyle z_{n},z_{m}\in \Delta _{n}}$ so ${\displaystyle |z_{n}-z_{m}|\leqslant 2^{-n}L}$. So ${\displaystyle |z_{n}-z_{m}|\to 0}$ as ${\displaystyle n,m\to \infty }$. The limit ${\displaystyle z_{0}}$ lies in the intersection because ${\displaystyle z_{m}\in \Delta _{n}}$ for all ${\displaystyle m\geqslant n}$, so ${\displaystyle z_{0}=\lim z_{m}\in \Delta _{n}}$ since ${\displaystyle \Delta _{n}}$ is closed. Since this is true for any ${\displaystyle n}$, ${\displaystyle z_{0}\in \bigcap \Delta _{n}}$.

We haven't yet used the complex-differentiability of ${\displaystyle f}$, so this is what we do now, to show finally that ${\displaystyle I=0}$. Given ${\displaystyle \varepsilon >0}$, there exists ${\displaystyle \delta >0}$ so that

${\displaystyle |f(z)-f(z_{0})-(z-z_{0})f'(z_{0})|<\varepsilon |z-z_{0}|{\mbox{ for all }}|z-z_{0}|<\delta .}$
(Here of course ${\displaystyle \delta }$ is assumed so small that ${\displaystyle |z-z_{0}|<\delta \Rightarrow z\in \Omega }$.) We use this to estimate ${\displaystyle I_{n}}$. By previous results
${\displaystyle \int _{\partial \Delta _{n}}\left(f(z)-f(z_{0})-(z-z_{0})f'(z_{0})\right)\,{\mbox{d}}z=\int _{\partial \Delta _{n}}f(z)\,{\mbox{d}}z}$
Hence, using
{\displaystyle {\begin{aligned}I_{n}&=\left|\int _{\partial \Delta _{n}}f(z)\,{\mbox{d}}z\right|\\&=\left|\int _{\partial \Delta _{n}}\left(f(z)-f(z_{0})-(z-z_{0})f'(z_{0})\right)\,{\mbox{d}}z\right|\\&\leqslant \operatorname {length} (\partial \Delta _{n})\varepsilon \sup _{z\in \partial \Delta _{n}}|z-z_{0}|.\end{aligned}}}
This supremum is ${\displaystyle \leqslant 2^{-n}L}$ and so all together we have
${\displaystyle I_{n}\leqslant 4^{-n}L\varepsilon }$
provided that ${\displaystyle n}$ is so large that ${\displaystyle 2^{-n}L<\delta }$ (which eventually happens for any ${\displaystyle \delta }$). Combining with

 the factors ${\displaystyle 4^{-n}}$ on each side cancel, leaving


us with ${\displaystyle I\leqslant L\varepsilon }$. Since ${\displaystyle \varepsilon }$ can be taken arbitrarily small (and ${\displaystyle I}$ is independent of ${\displaystyle \varepsilon }$, of course) it follows that ${\displaystyle I=0}$ as was required.

Corollary (9)

Let ${\displaystyle f:D\to \mathbb {C} }$ be holomorphic, where ${\displaystyle D}$ is a disc. Let ${\displaystyle \gamma :[t_{0},t_{1}]\to D}$ be any (piecewise ${\displaystyle C^{1}}$) curve in ${\displaystyle D}$. Then

${\displaystyle \int _{\gamma }f(z)\,{\mbox{d}}z=0.}$

Proof

Let ${\displaystyle z_{0}}$ be the centre of the disc, and for ${\displaystyle z\in D}$, define

${\displaystyle F(z)=\int _{z_{0}}^{z}f(w)\,{\mbox{d}}w;}$
in other words, we integrate ${\displaystyle f}$ along the line segment joining the centre of ${\displaystyle D}$ to the variable point ${\displaystyle z}$. We see that knowing

 for every triangle in ${\displaystyle D}$ is enough to prove that


${\displaystyle F'(z)=f(z)}$ in ${\displaystyle D}$. And we have seen that in this case, the integral of ${\displaystyle f}$ around any closed curve ${\displaystyle \gamma }$ is zero.

Remark (10)

There are much more general versions of Cauchy's Theorem, where ${\displaystyle D}$ is replaced by a more general domain ${\displaystyle \Omega }$. In this case, there needs to be a condition on the closed curve ${\displaystyle \gamma }$. Indeed, if ${\displaystyle \Omega =\mathbb {C} \setminus \{0\}}$, then ${\displaystyle f(z)=1/z}$ is holomorphic in ${\displaystyle \Omega }$, but

${\displaystyle \int _{|z|=1}{\frac {1}{z}}\,{\mbox{d}}z=2\pi i}$
(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 ${\displaystyle D}$ to a given point ${\displaystyle z_{0}}$. This is true in more general sets than ${\displaystyle D}$. For example in the cut plane

${\displaystyle \Omega =\mathbb {C} \setminus \{{\mbox{Re}}(z)\geqslant 0,{\mbox{Im}}(z)=0\},}$
every point ${\displaystyle z}$ of ${\displaystyle \Omega }$ can be joined to the point ${\displaystyle -1}$. This will be useful later[1].

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