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<title> Cauchy-Goursat Theorem
</title>
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<p><i class="fa-solid fa-list"></i>Contents</p>
<ul>
<li><a href="#section1">Cauchy's Theorem</a></li>
<li><a href="#section2">Cauchy-Goursat Theorem</a></li>
<li><a href="#section3">Connected domains</a></li>
<li><a href="#section4">Historical notes & Proof</a></li>
<li><a href="#section5">References</a></li>
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<article>
<h1>Cauchy-Goursat Theorem</h1>
<hr>
<div id="section1">
<h2>Cauchy's Theorem</h2>
<p>
In 1825 the French mathematician Augustin-Louis Cauchy proved
one of the most important theorems in complex analysis:
</p>
<div class="theorem">
<span style="font-style: normal!important;"><strong>(Cauchy's Theorem)</strong></span>
Suppose that $f$ is analytic, with the derivative $f'$
continuous on and inside a closed simple contour $C.$
Then
\[
\int_C f(z)\, dz = 0.
\]
</div>
<figure>
<img src="../images/chp04/cauchy-theorem.svg" alt="Cauchy's Theorem" title="Cauchy's Theorem" style="width:60%;">
<figcaption>
Cauchy's Theorem implies $\displaystyle \int_C f(z)\,dz =0.$
</figcaption>
</figure>
<p>
If the function is not analytic on the whole region inside $C,$
then the integral may or may not be $0.$ For instance,
Let $C$ be the unit circle and $f(z)=1/z.$ Then $f$ is analytic
at all points except $z=0,$ and indeed the integral is
<em>not zero</em>. In fact,
\[
\int_C \frac{1}{z}\,dz = 2\pi i,
\]
as shown in example 2 from the
<a href="complex_integration.html">Complex Integration</a> section.
On the other hand, if $f(z) = 1/z^2,$ then $f$ is still analytic
at all points except $z=0,$ but now the integral <em>is</em> $0.$
This value results <em>not</em> from Cauchy's Theorem, since $f$
is not analytic everywhere inside $C,$ but rather from the fact that
$f$ has an antiderivative on $\C\setminus \{0\}.$ That is, $f$ is
the derivative of $F'(z)=-1/z.$
</p>
<p>
<em>Proof of Theorem 1.</em> The proof of this theorem is an immediate consequence
of <strong>Green's theorem</strong> in the plane, which states that, for
continuously differentiable functions $P(x,y)$ and $Q(x,y),$
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{green-theo}
\int_C P\,dx + Q\, dy & =& \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right) dA
\end{eqnarray}
</div>
<p>
In Green's theorem, $R$ represents the <em>inside</em>
of $C,$ $C$ is traversed in a counterclockwise direction, and
$P$ and $Q$ are sufficiently smooth.
</p>
<p>Setting $f=u+iv,$ we have</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C f(z)\,dz & =& \int_C \left(u+iv\right) \left(dx+idy\right)\\
& =& \int_C \left(u\,dx - v\,dy \right) + i \int_C \left(u\, dy+v \,dx\right).
\end{eqnarray*}
</div>
<p>
By applying Green's theorem to each integral, we get
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C f(z)\,dz & =& \iint_R \left(-\frac{\partial v}{\partial x} - \frac{\partial u}{\partial y}\right) dA + i \iint_R \left(\frac{\partial u}{\partial x} - \frac{\partial v}{\partial y}\right) dA \\
\end{eqnarray*}
</div>
<p>
Both integrals, on the right side, are zero by the <a href="complex_differentiation.html">Cauchy-Riemann equations</a>. $\hspace{5pt} \blacksquare$
</p>
<p>Observe that once it has been established that the value
of this integral is zero,
the orientation of $C$ becomes irrelevant.
The conclusion in Theorem 1 is also true if $C$ is taken in the
clockwise direction, since then we can use the fact that
\[
\int_C f(z)\,dz = - \int_{-C} f(z)\,dz.
\]
</p>
<div class="practice">
<p><strong>Example 1:</strong>
Consider the function $f(z)= \exp\left(z^3\right).$ If $C$ is
any simple closed contour, in either direction, then
\[
\int_C \exp\left(z^3\right)\,dz =0.
\]
In this case $f$ is the composition of two functions, both analytic everywhere.
Thus, $f$ is also analytic and its derivative
$f'(z) = 3z^2\exp\left(z^3\right)$
is continuous everywhere.
</p>
</div>
</div>
<hr>
<div id="section2">
<h2>Cauchy-Goursat Theorem</h2>
<p>
In 1900 the French mathematician
<a href="https://mathshistory.st-andrews.ac.uk/Biographies/Goursat/" target="_blank">Edouard Goursat</a>
proved that the assumption of continuity of $f'$
is not necessary to reach the conclusion of <em>Cauchy's theorem</em>.
The resulting modified version of Cauchy's theorem is known
today as the <em>Cauchy-Goursat Theorem</em>.
as we can expect,
with fewer hypotheses, the proof of this version of Cauchy's
theorem is more complicated than the one just presented.
</p>
<div class="theorem">
<span style="font-style: normal!important;"><strong>(Cauchy-Goursat Theorem)</strong></span>
If a function $f$ is analytic at all points interior to and on a simple
closed contour $C,$ then
\[
\int_C f(z) \, dz = 0.
\]
</div>
<div class="practice">
<p>
<strong>Example 2:</strong>
The function $f(z)=\exp(z)$ is entire and consequently
is analytic at all points within and on any simple closed
contour $C.$ It follows from the Cauchy-Goursat Theorem that
\[
\int_C \exp(z)\,dz = 0.
\]
Similarly, since $\sin z,$ $\cos z$ and
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
p(z) = a_nz^n+ a_{n-1}z^{n-1} + \cdots + a_1 z + a_0,
\quad (n =0,1,2,\ldots)
\end{eqnarray*}
</div>
<p>are entire, then</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C \sin z\,dz = 0,\quad \int_C \cos z\,dz = 0,\quad \int_C p(z)\,dz = 0.
\end{eqnarray*}
</div>
<p>
for any simple closed contour $C$
</p>
</div>
<div class="practice">
<p>
<strong>Example 3:</strong>
We can use Cauchy-Goursat Theorem to evaluate
\[
\int_C \frac{1}{z^2}\,dz,
\]
where $C$ is the ellipse $(x-2)^2+\dfrac{1}{4}(y-5)^2=1.$
Note that the function $f(z)=1/z^2$ is
analytic everywhere except at $z=0.$ In this case, $z=0$ is not inside
or on the simple closed elliptical contour $C.$ Hence
\[
\int_C \frac{1}{z^2}\,dz=0.
\]
</p>
<figure>
<div class="ggbElement" id="ggb-element-1"></div>
<!--<img src="../images/chp04/cauchy-theorem-example-03.svg" alt="Cauchy-Goursat Theorem example" title="Cauchy-Goursat Theorem example" style="width:55%;">-->
<figcaption>
The elliptical contour $C.$ Activate the box <code>Phase portrait</code> to show
the enhanced phase portrait of $f(z)=1/z^2,$ with level curves of the modulus.
</figcaption>
</figure>
</div>
</div>
<hr>
<div id="section3">
<h2>Simply and multiply connected domains</h2>
<p>
We say that a domain $D$ is <em>simply connected</em> if every simple
closed contour $C$ lying entirely in $D$ can be shrunk to a point
without leaving $D.$ See Figure 3. In other words, if we draw
any simple closed contour $C$ so that it lies entirely within
a simply connected domain, then $C$ encloses only points of
the domain $D.$
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-simply-connected.gif" alt="Simply connected" title="Simply connected" style="width:500px;">
<figcaption>
Simply connected domain $D.$
</figcaption>
</figure>
<p>
In other words, a simply connected domain has no "holes" in it.
The entire complex plane is an example of a simply
connected domain; the annulus defined by $1 \lt |z| \lt 2$ is
not simply connected.
</p>
<figure id="annulus">
<img src="../images/chp04/annulus.svg" alt="Annulus" title="Annulus" style="width:55%;">
<figcaption>
$1 \lt |z| \lt 2.$
</figcaption>
</figure>
<p>A domain that is not simply
connected is called a <em>multiply connected</em> domain; that is, a
multiply connected domain has "holes" in it. For example,
note in Figure 5
that if the curve $C$ enclosing the "hole" (on the left side)
is shrunk to a point, the curve would have to leave $D$ eventually.
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-not-simply-connected.gif" alt="No simply connected" title="Not simply connected" style="width:500px;">
<figcaption>
Not simply connected domain $D.$
</figcaption>
</figure>
<p>
The closed contour in the Cauchy-Goursat Theorem does not need to be
simple when the theorem is adapted to simply connected domains.
That is, the contour can actually cross itself. The following
theorem allows for this possibility.
</p>
<div class="theorem">
If a function $f$ is analytic throughout a simply connected domain $D,$
then
\[
\int_C f(z) \, dz = 0.
\]
for every closed contour $C$ lying in $D.$
</div>
<p>
The proof is easy if $C$ is a simple closed contour or if it is
a closed contour that intersects itself a finite number of times.
In the case that $C$ is simple and lies in $D,$ the function $f$ is analytic
at each point interior to and on $C.$
The Cauchy-Goursat
Theorem ensures that the conclusion of <strong>Theorem 3</strong> holds.
On the other hand, if $C$
is closed but intersects itself a finite number of times,
it consists of a finite number of simple closed contours.
This is illustrated in Figure 6, where the simple closed
contours $C_k$ $(k = 1, 2, 3, 4)$ make up $C.$ Since the value
of the integral around each $C_k$ is zero, according to the
Cauchy-Goursat theorem, it follows that
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C f(z)\,dz = \sum_{k=1}^{4}\int_{C_k}f(z)\, dz = 0.
\end{eqnarray*}
</div>
<figure>
<img src="../images/chp04/cauchy-theorem-closed-contours-sum.svg" alt=" Simply closed contours C_k" title=" Simply closed contours C_k" style="width:55%;">
<figcaption>
Simply closed contours $C_k.$
</figcaption>
</figure>
<div class="practice">
<p><strong>Example 4:</strong>
Let $C$ be any closed contour lying in the open disk $|z|\lt 2.$
Then
\[
\int_C \frac{z\sin(z)}{\left(z^2-9\right)^3}dz =0.
\]
Note that the disk is a simply connected domain and the two
singularities $z=\pm 3$ of the integrand are outside the disk.
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-example-04.svg" alt="Cauchy-Goursat Theorem example" title="Cauchy-Goursat Theorem example" style="width:75%;">
<figcaption>
The open disk $D:|z|\lt 2$ is simply connected and $C$ is any closed contour.
</figcaption>
</figure>
</div>
<div class="corollary">
A function $f$ that is analytic throughout a simply connected domain $D$
must have an antiderivative everywhere in $D.$
</div>
<p>
If $f$ is analytic in a multiply connected domain $D$ then we cannot conclude
that $\int_C f(z)\,dz=0$ for every simple closed contour $C$ in $D.$
Suppose that $D$ is multiply connected with two "holes".
Let $C,$ $C_1$ and $C_2$ be simple
closed contours such that each $C_k$ surrounds only one "hole" in the
domain and are inside $C.$ See Figure 8.
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-multiply-connected-contours-start.svg" alt="Cauchy-Goursat Theorem multiply connected" title="Cauchy-Goursat multiply connected" style="width:65%;">
<figcaption>
$D$ is multiply connected with two "holes".
</figcaption>
</figure>
<p>
Now, suppose also that
$f$ is analytic on each contour and throughout
the multiply connected domain consisting of
the points inside $C$ and exterior to each $C_k.$
Assume that $C$ is described in the counterclockwise direction and
each $C_k$ is described in the clockwise direction.
We introduce a polygonal path $L_1,$ consisting
of a finite number of line segments joined end to end,
to connect the outer contour $C$ to the inner contour $C_1.$
As shown in Figure 9, we introduce another polygonal path $L_2$ which connects
$C_1$ to $C_2;$ and finally another polygonal path
$L_3$ connecting $C_3$ to $C.$
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-multiply-connected-contours.svg" alt="Cauchy-Goursat Theorem multiply connected" title="Cauchy-Goursat multiply connected" style="width:65%;">
<figcaption>
Polygonal lines $L_k$ are introduced.
</figcaption>
</figure>
<p>
As indicated in Figure 10, two simple closed contours
$\Gamma_1$ and $\Gamma_2$ can be formed, each consisting of polygonal paths
$L_k$ or $-L_k$ and pieces of $C$ and $C_k$ and each described in
such a direction that the points enclosed by them
lie to the left. Here we can apply the
Cauchy-Goursat Theorem to $f$ on $\Gamma_1$ and $\Gamma_2,$
and the sum of the
values of the integrals over those contours is found
to be zero. Since the integrals in opposite directions
along each path $L_k$ cancel,
only the integrals along $C$
and the $C_k$ remain. Thus we obtain
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C f(z)\,dz + \int_{C_1} f(z)\,dz + \int_{C_2} f(z)\,dz =0
\end{eqnarray*}
</div>
<figure>
<img src="../images/chp04/cauchy-theorem-multiply-connected-contours-final.svg" alt="Cauchy-Goursat Theorem multiply connected" title="Cauchy-Goursat multiply connected" style="width:65%;">
<figcaption>
$\int_{L_k}f(z)\,dz = -\int_{-L_k}f(z)\,dz .$
</figcaption>
</figure>
<p>
The next theorem summarizes the general result for a
multiply connected domain with $n$ “holes.”
</p>
<div class="theorem">
Suppose that
<ol>
<li>
C is a simple closed contour, described in the counterclockwise direction;
</li>
<li>
$C_k$ $(k = 1, 2, \ldots , n)$ are simple closed contours interior
to $C,$ all described in the clockwise direction, that are
disjoint and whose interiors have no points in common.
</li>
</ol>
<p>
If $f$ is analytic on all of these contours and
throughout the multiply connected domain consisting of
the points inside $C$ and exterior to each $C_k,$ then
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C f(z)\,dz + \sum_{n=1}^{n} \int_{C_k} f(z)\,dz =0
\end{eqnarray*}
</div>
</div>
<p>
The following corollary is known as the <strong>principle of deformation of
contours</strong> since it tells us that if $C_1$ is continuously deformed
into $C_2$ always passing through points at
which $f$ is analytic, then the value of the integral
of $f$ over $C_1$ never changes.
</p>
<div class="corollary">
Let $C_1$ and $C_2$ denote positively oriented simple
closed contours, where $C_1$ is interior to $C_2.$
If $f$ is analytic in the closed region
consisting of those contours and all points between
them, then
\[
\int_{C_2} f(z) \, dz = \int_{C_1} f(z) \, dz .
\]
</div>
<figure>
<img src="../images/chp04/cauchy-theorem-simply-deformation.gif" alt="Cauchy-Goursat Theorem" title="Cauchy-Goursat" style="width:500px;">
<figcaption>
$C_1$ is continuously deformed into $C_2.$
</figcaption>
</figure>
<div class="practice">
<p>
<strong>Example 5:</strong>
We can use the previous corollary to show that
\[
\int_C \frac{1}{z} dz= 2\pi \,i
\]
for any positively oriented simple closed contour $C$
surrounding the origin.
</p>
<p>
Consider $C_0$ a positively oriented
circle with center at the origin and radius
so small that $C_0$ lies entirely inside $C.$
</p>
<p>
We know that (see example 2 from the
<a href="complex_integration.html">Complex Integration</a> section)
\[
\int_{C_0} \frac{1}{z} dz= 2\pi \,i.
\]
and since $1/z$ is analytic everywhere
except at $z = 0,$ the result follows easily.
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-simply-deformation-example.gif" alt="Cauchy-Goursat Theorem" title="Cauchy-Goursat" style="width:500px;">
<figcaption>
$\ds \int_{C_0} \frac{1}{z} dz= 2\pi \,i.$
</figcaption>
</figure>
</div>
<div class="practice">
<p>
<strong>Exercise 1:</strong>
Use the principle of deformation of contours to show that
if $z_0$ is any complex constant interior to any simple
closed contour $C,$ then for $n$ an integer we have
\[
\int_C \frac{dz}{(z-z_0)^{n}} =
\left\{
\begin{array}{ll}
2\pi i & n = 1,\\
0, & n\neq 1.
\end{array}
\right.
\]
</p>
</div>
</div>
<hr>
<div id="section4">
<h2>Historical notes & Proof</h2>
<p>
Cauchy first communicated the integral theorem to the
<em>Académie des Sciences</em> in 1814, as part of a memoir related with
other topics (improper real integrals)
[<a href="#bottazzini1984">1</a> pp. 132-133, <a href="#cauchy1841">4</a>, <a href="#smithies1997">13</a> pp. 56-57].
The first general form of the integral theorem was communicated to the
<em>Académie</em> in 1825 in a memoir entitled
<em>Mémoire sur les intégrales définies, prises entre des limites imaginaires</em>
[<a href="#bottazzini1984">1</a> pp. 151-156, <a href="#cauchy1825">5</a>, <a href="#smithies1997">13</a> pp. 89-91].
</p>
<p>
Goursat presented a proof Cauchy's Theorem in 1884 for the case of a
simple closed curve, by dividing up the interior into small squares,
showing that the integral around each square of
side $\ell$ is bounded by $\epsilon \ell ^2,$
and then adding the results. To obtain this
estimate for arbitrary $\epsilon$ he needs
the uniform continuity of $f'$
[<a href="#goursat1884">7</a>].
In his 1900 paper, Goursat finally removed
the assumption of continuity of $f'$ [<a href="#borger1921">2</a>, <a href="#goursat1900">8</a>].
The argument is as before, but he now subdivides
each square until the necessary estimate holds for each
of them, that is,
</p>
<div class="scroll-wrapper">
\[
\left|f(z) - f(z_0) - f'(z_0)(z-z_0)\right|\lt \epsilon \left|z-z_0\right|
\]
</div>
<p>
for every $z$ on the boundary of the subsquare, where $z_0$
is some fixed point in the subsquare. The important point here
is that the same $\epsilon$ will do for each subsquare.
</p>
<p>
In 1900, Eliakim H. Moore refined the proof, addressing the
treatment of the boundary curve with greater precision [<a href="#moore1900">11</a>].
He also introduced the modern approach of proof by contradiction, which
involves subdividing the domain, selecting the region where the
conclusion is most significantly contradicted, and then applying
the definition of the derivative at the resulting limit point
[<a href="#hance-olsen2008">9</a>, p. 651].
</p>
<p>
Later in 1901 Alfred Pringsheim presented a criticism of Goursat's treatment
of the boundary curve [<a href="#pringsheim1901">12</a>].
He observed that these issues vanish when the proof technique is
applied to a basic geometric shape, such as a triangle. From there,
the theorem extends to simple polygonal paths within a simply connected
domain by triangulating the interior of the path. Finally, it can be
generalized to arbitrary paths by approximating them with polygonal paths.
[<a href="#hance-olsen2008">9</a>, p. 651]. This is the method that we
will use to prove the Cauchy-Goursat Theorem.
</p>
<p>
To avoid needless repetition throughout the following discussion
we will take for granted that we are working in a simply connected
domain $D$ and that $f$ represents a complex function analytic in $D.$
</p>
<div class="lemma">
If $\Delta$ is a triangular contour lying entirely within $D,$
then $\int_{\Delta} f(z)\,dz=0.$
</div>
<div class="proof">
Let $\Delta$ be the triangular contour shown in Figure 13.
The vertices of $\Delta$ are labeled $V_1,V_2$ and $V_3.$
We form smaller triangles $\Delta_1, \Delta_2,$ $\Delta_3$ and $\Delta_4$
by joining the midpoints $M_1,M_2$ and $M_3$ of the sides of $\Delta$
by straight line segments as shown in Figure 14. Then
<div class="scroll-wrapper">
\begin{eqnarray}\label{sumcontours}
\int_{\Delta} f(z)\,dz = \int_{C_1} f(z)\,dz +\int_{C_2} f(z)\,dz +\int_{C_3} f(z)\,dz +\int_{C_4} f(z)\,dz.
\end{eqnarray}
</div>
<div class="figure-container">
<figure>
<img src="../images/chp04/cauchy-theorem-lemma-01-a.svg" alt="Triangle in D." title="Triangle in D." style="width:100%;">
<figcaption>
Triangular contour $\Delta$ within $D.$
</figcaption>
</figure>
<figure>
<img src="../images/chp04/cauchy-theorem-lemma-01-b.svg" alt="Partition." title="Partition." style="width:100%;">
<figcaption>
Triangular contours $C_k.$
</figcaption>
</figure>
</div>
<p>
Using the triangle inequality we have
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{triangle01}
\left|\int_{\Delta} f(z)\,dz\right| \leq \left|\int_{C_1} f(z)\,dz\right| + \left|\int_{C_2} f(z)\,dz\right| + \left|\int_{C_3} f(z)\,dz\right| + \left|\int_{C_4} f(z)\,dz\right|.
\end{eqnarray}
</div>
<p>
Since the four quantities on the right side of the previous inequality
are nonnegative real numbers and, as a consequence, one of
them must be greater than or equal to the other three.
Let's denote the triangular contour of the integral
with the largest modulus by the symbol $\Delta_1.$ Thus
\begin{eqnarray}\label{triangle02}
\left|\int_{\Delta} f(z)\,dz\right| \leq 4 \left|\int_{\Delta_1} f(z)\,dz\right| .
\end{eqnarray}
</p>
<p>
Now we repeat the foregoing process for the triangle $\Delta_1.$ That is,
we form triangles within $\Delta_1$ by joining the midpoints of its
sides by line segments in the same way as shown in Figure 14
and proceed to the equivalent of (\ref{sumcontours}) and (\ref{triangle01}).
The integral of $f$ along one of these new triangular contous, let's call it
$\Delta_2$ then satisfies
\[
\left|\int_{\Delta_1} f(z)\,dz\right| \leq 4 \left|\int_{\Delta_2} f(z)\,dz\right| .
\]
</p>
<p>
We combine this last inequality with (\ref{triangle02}) to obtain
\[
\left|\int_{\Delta_1} f(z)\,dz\right| \leq 4 \left|\int_{\Delta_1} f(z)\,dz\right| \leq 4^2 \left|\int_{\Delta_2} f(z)\,dz\right| .
\]
</p>
<p>
We continue in this way to obtain a sequence of "nested" triangular
contours $\Delta, \Delta_1, \Delta_2,\ldots,$ that is,
each triangle in the sequence is contained in the one immediately
preceding it. After $n$ steps we get
\begin{eqnarray}\label{bound4}
\left|\int_{\Delta_1} f(z)\,dz\right| \leq 4^n \left|\int_{\Delta_n} f(z)\,dz\right| .
\end{eqnarray}
</p>
<p>
Since the sequence triangular contours
$\Delta, \Delta_1, \Delta_2,\ldots,$ are nested, there exists
a point $z_0$ in the domain $D$ that is common to every
triangle in the sequence. Also, since $f$ is analytic, then $f'(z_0)$
exists. If we define
\begin{eqnarray}\label{lambda}
\Lambda(z) = \frac{f(z)-f(z_0)}{z-z_0} - f'(z_0),
\end{eqnarray}
then $\left|\Lambda(z)\right|$ can be made arbitrarily small whenever
$z$ is sufficiently close to $z_0.$ This is true
because $f$ is analytic in $D$ and so the limit
\begin{eqnarray*}
\lim_{z\to z_0}\frac{f(z)-f(z_0)}{z-z_0}
\end{eqnarray*}
exists and is equal to $f'(z_0).$ In other words, for every $\epsilon\gt 0,$
there exist $\delta\gt 0$ such that
</p><div class="scroll-wrapper">
\begin{eqnarray}\label{smallenough}
\left|\Lambda(z)\right|\lt \epsilon \quad \text{whenever}\quad \left|z-z_0\right|\lt \delta.
\end{eqnarray}
</div>
We can solve (\ref{lambda}) for $f(z)$
and replace this value in the integrand in (\ref{bound4}) to obtain
<p></p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{expanded}
\int_{\Delta_n}f(z)\,dz = f(z_0) \int_{\Delta_n}dz + f'(z_0)\int_{\Delta_n}(z-z_0)\,dz + \int_{\Delta_n}(z-z_0)\Lambda(z)\,dz.
\end{eqnarray}
</div>
<p>
Since
\[
\int_{\Delta_n}dz =0 \quad \text{and}\quad \int_{\Delta_n}(z-z_0)\,dz=0
\]
(Why?), the right side of (\ref{expanded}) reduces to
\begin{eqnarray}\label{reduced}
\int_{\Delta_n}f(z)\,dz = \int_{\Delta_n}(z-z_0)\Lambda(z)\,dz.
\end{eqnarray}
</p>
<p>
Now let $L$ and $L_1$ denote the lengths of the triangular contours $\Delta$ and $\Delta_1,$
respectively. Then, if we keep in mind how the triangle $\Delta_1$ was constructed,
it is a straightforward problem in similar triangles to show that
$L_1$ is related to $L$ by $L_1=\dfrac{1}{2}L.$ Likewise,
if $L_2$ is the length of $\Delta_2,$ then $L_2=\dfrac{1}{2}L_1 = \dfrac{1}{2^2}L.$
In general we have that if $L_n$ is the length of $\Delta_n,$ then
$L_n= \dfrac{1}{2^n}L.$
</p>
<p>
For any $z\in \Delta_n,$ we have that
$\abs{z-z_0}\lt L_n,$ where $\ds L_n= \frac{1}{2^n}L.$
If we choose $n$ large enough so that
\[
\abs{z-z_0}\lt \frac{1}{2^n}L\lt \delta,
\]
it then follows from (\ref{reduced}), (\ref{smallenough}) and the
<em>ML</em>-inequality
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{boundgeneral}
\abs{\int_{\Delta_n}f(z)\,dz } = \abs{ \int_{\Delta_n}(z-z_0)\Lambda(z)\,dz} \leq \frac{L}{2^n}\cdot \epsilon \cdot \frac{L}{2^n} = \frac{L^2}{4^n}\epsilon
\end{eqnarray}
</div>
<p>
Putting (\ref{bound4}) together with (\ref{boundgeneral}) gives us a bound for
the modulus of the integral on $\Delta:$
\begin{eqnarray}\label{lastbound}
\abs{\int_{\Delta}f(z)\,dz } \leq 4^n \frac{L^2}{4^n}\epsilon = L^2\epsilon.
\end{eqnarray}
Since $\epsilon \gt 0$ can be made arbitrarily small, then $\abs{\int_{\Delta}f(z)\,dz}=0.$
Therefore
\begin{eqnarray*}
\int_{\Delta}f(z)\,dz =0.
\end{eqnarray*}
</p>
</div>
<div class="lemma">
If $C$ is a closed polygonal contour lying entirely within $D,$
then $\int_C f(z)\,dz=0.$
</div>
<div class="practice">
<p>
<strong>Exercise 2:</strong>
Use <strong>Lemma 1</strong> and the fact that any closed polygonal
contour $C$ can be "triangulated" to prove <strong>Lemma 2</strong>.
</p>
<div class="figure-container">
<figure>
<img src="../images/chp04/cauchy-theorem-lemma-02-a.svg" alt="Closed polygonal contour" title="Closed polygonal contour" style="width:100%;">
<figcaption>
Closed polygonal contour $C$ within $D.$
</figcaption>
</figure>
<figure>
<img src="../images/chp04/cauchy-theorem-lemma-02-b.svg" alt="Triangulation of the polygonal contour" title="Triangulation of the polygonal contour" style="width:100%;">
<figcaption>
Triangulation of the polygonal contour $C.$
</figcaption>
</figure>
</div>
<p>
<strong>Remark:</strong>
Roughly, "triangulated" means that the closed polygon $C$ can be decomposed
into a finite number of triangles by adding lines as shown in
Figure 16.
Note that we can then proceed as in the proof of <strong>Lemma 1</strong>
and integrate twice along these added line segments but
in opposite directions. If the closed polygon $C$ has $n$ sides,
then it can be decomposed
into n triangles $C_1,C_2, \ldots, C_n$ and we would eventually arrive
at the following analogue of (\ref{sumcontours}):
\[
\int_C f(z)\,dz = \sum_{k=1}^{n} \int_{C_k}f(z)\,dz.
\]
</p>
</div>
<p>
Now the Cauchy-Goursat Theorem can easily be proved using
<strong>Lemma 2</strong> and the fact that any closed contour
$C$ can be approximated to any desired degree of accuracy
by a closed polygonal path.
</p>
<figure>
<img src="../images/chp04/cauchy-theorem-final-proof.gif" alt="Approximation by polygonal path" title="Approximation by polygonal path" style="width:550px;">
<figcaption>
The contour $C$ is approximated by a polygonal contour $P.$
</figcaption>
</figure>
<p>
<em>Proof of Cauchy-Goursat Theorem.</em>
Consider a simple closed contour $C$ and $n$ points $z_1, z_2, \ldots, z_n$
on $C$ through which a polygonal path $P$ has been constructed.
Then it can be shown that the difference
\[
\abs{\int_C f(z)\,dz - \int_P f(z)\,dz}
\]
can be made arbitrarily small as $n\to \infty.$
Thus, by <strong>Lemma 2</strong>, $\ds \int_P f(z)\,dz = 0$
for any $n.$ Therefore
$\ds \int_C f(z)\,dz=0.$ $\hspace{10pt}\blacksquare$
</p>
<p>
The Cauchy-Goursat Theorem has been established through various methods.
For instance, proofs have been constructed specifically for
rectangles or disks (see [<a href="#brown2009">3</a>, <a href="#marsden1999">10</a>]).
Beyond these approaches, many other proofs of the theorem have been developed.
Notably, John D. Dixon presented a concise and elegant proof relying only on
fundamental concepts of complex function theory in convex sets
[<a href="#dixon1971">6</a>].
Rudolf Výborný, on the other hand, offered a proof based on
differentiable homotopy [<a href="#vyborny1979">14</a>].
</p>
</div>
<hr>
<div id="section5">
<h2>References</h2>
<ol class="bbtex-list">
<li id="bottazzini1984">Bottazzini, U. (1984). <em>The higher calculus: a history of real and complex analysis from Euler to Weierstrass</em>. New York : Springer-Verlag. <a href="https://archive.org/details/highercalculushi0000bott/mode/2up" target="_blank"><i class="fas fa-external-link-alt"></i></a> </li>
<li id="borger1921">Borger, R. L. (1921). On the Cauchy-Goursat theorem, <em>Bulletin of the American Mathematical Society</em>, Vol. 27, No. 7, pp. 325-329.
<a href="https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-27/issue-7/On-the-Cauchy-Goursat-theorem/bams/1183425654.full" target="_blank"><i class="fas fa-external-link-alt"></i></a></li>
<li id="brown2009">Brown, J. W., Churchill, R. V. (2009). <em>Complex Variables and Applications.</em> 8th Edition. New
York: McGraw-Hill Higher Education.</li>
<li id="cauchy1814">Cauchy, A. (1814). <em>Mémoire sur la théorie des intégrales définies,</em>
in Œuvres complètes d'Augustin Cauchy, $\text{I}^{re}$ série, vol. I, Gauthier Villars, Paris. <a href="https://doi.org/10.1017/CBO9780511702174" target="_blank"><i class="fas fa-external-link-alt"></i></a></li>
<li id="cauchy1825">Cauchy, A. (1825). <em>Mémoire sur les intégrales définies, prises entre des limites imaginaires,</em>
in Œuvres complètes d'Augustin Cauchy, $\text{II}^{e}$ série, vol. XV, Gauthier Villars, Paris. <a href="https://archive.org/details/mmoiresurlesin00cauc/mode/2up" target="_blank"><i class="fas fa-external-link-alt"></i></a></li>
<li id="dixon1971">Dixon, J. D. (1971). A brief proof of Cauchy's integral theorem, <em>Proceedings of the American Mathematical Society</em>, Vol. 29, No. 3, pp. 625-626.
<a href="https://doi.org/10.2307/2038614" target="_blank"><i class="fas fa-external-link-alt"></i></a></li>
<li id="goursat1884">Goursat, E. (1884). Démonstration du théorème de Cauchy. <em>Acta Mathematica</em>, 4, pp. 197-200.
<a href="https://link.springer.com/article/10.1007/BF02418419" target="_blank"><i class="fas fa-external-link-alt"></i></a></li>
<li id="goursat1900">Goursat, E. (1900). Sur la définition générale des fonctions analytiques, d'après Cauchy, <em>Transactions of the American Mathematical Society</em>, Vol. 1, No. 1, pp. 14-16.
<a href="https://doi.org/10.2307/1986398" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
<li id="hance-olsen2008">Hance-Olsen, H. (2008). On Goursat's Proof of Cauchy's Integral Theorem, <em>The American Mathematical Monthly</em>, 115, pp. 648-652.
<a href="https://www.jstor.org/stable/27642560" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
<li id="marsden1999">Marsden, J. E. & Hoffman, M. J. (1999) <em>Basic Complex Analysis.</em> (3rd ed.) New York: W. H. Freeman and Co.</li>
<li id="moore1900">Moore, E. H. (1900). A Simple Proof of the Fundamental Cauchy-Goursat Theorem, <em>Transactions of the American Mathematical Society</em>, Vol. 1, No. 4, pp. 499-506.
<a href="https://doi.org/10.2307/1986368" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
<li id="pringsheim1901">Pringsheim, A. (1901). Ueber den Goursat'schen Beweis des Cauchy'schen Integralsatzes, <em>Transactions of the American Mathematical Society</em>, Vol. 2, No. 4, pp. 413-421.
<a href="https://doi.org/10.2307/1986254" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
<li id="smithies1997">Smithies, F. (1997). <em>Cauchy and the Creation of Complex Function Theory.</em> UK: Cambridge University Press.
<a href="https://archive.org/details/cauchycreationof0000smit/mode/2up" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
<li id="vynorny1979">Výborný, R. (1979). On the Use of a Differentiable Homotopy in the Proof of the Cauchy Theorem, <em>The American Mathematical Monthly</em>, Vol. 86, No. 5, pp. 380-382.
<a href="https://doi.org/10.2307/2321099" target="_blank"><i class="fas fa-external-link-alt"></i></a>
</li>
</ol>
</div>
<br>
</article>
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