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<!DOCTYPE html>
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<title> Cauchy Integral Formula
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<section data-type="chapter" id="cauchy-integral-formula">
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<article>
<h1>Cauchy Integral Formula</h1>
<hr>
<p>
If $f$ is analytic in a simply connected domain $D$ and
$z_0\in D$, the quotient $f(z)/(z - z_0)$
is undefined at $z_0$, making it non-analytic in $D.$
Consequently, the Cauchy-Goursat theorem does not allow
us to conclude that the integral
\[
\int_C \frac{f(z)}{z-z_0}dz,
\]
around a simple closed contour $C$ containing $z_0$, is zero.
However, as we shall see, the value of this integral is $2\pi i f(z_0)$. This result is
the first of two remarkable formulas.
</p>
<div class="theorem">
<span style="font-style: normal!important;font-weight: bold!important;">(Cauchy's integral formula)</span>
Let $f$ be analytic everywhere inside and on a simple closed
contour $C,$ taken in the positive sense. If $z_0$ is any point
interior to $C,$ then
\begin{eqnarray}\label{cauchy-int-form}
f(z_0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0}dz
\end{eqnarray}
</div>
<div class="proof">
Let $C_r$ be a positively oriented circle $\abs{z-z_0}=r,$
where $r\gt 0$ is small enough that $C_r$ is inside $C.$ See Figure 1.
Now, since the quotient $f(z)/(z - z_0)$ is analytic between
and on the contours $C$ and $C_r,$ then by the principle of deformation
of contours we have
\[
\int_C \frac{f(z)}{z-z_0}dz = \int_{C_r} \frac{f(z)}{z-z_0}dz.
\]
<figure>
<img src="../images/chp04/cauchy-integral-formula-01.svg" alt="Circle C_r inside the contour" title="Circle C_r inside the contour" style="width:60%">
<figcaption>
Circle $C_r$ inside the contour $C.$
</figcaption>
</figure>
<p>
We wish to show that the value of the integral on
the right is $2\pi i f(z_0).$ To do this we add and subtract
the constant $f(z_0)$ in the numerator of the integrand,
</p>
<div class="scroll-wrapper">
\begin{eqnarray}
\int_{C_r} \frac{f(z)}{z-z_0}dz &=& \int_{C_r} \frac{f(z_0) - f(z_0)+ f(z)}{z-z_0}dz \nonumber\\
& = & f(z_0)\int_{C_r} \frac{1}{z-z_0}dz + \int_{C_r} \frac{f(z)- f(z_0)}{z-z_0}dz \label{formula-01}.
\end{eqnarray}
</div>
<p>
We know that $\ds \int_{C_r}\frac{1}{z-z_0}dz = 2\pi i$ (see <strong>Exercise 1</strong> from
<a href="cauchy_goursat_theorem.html">Cauchy-Goursat theorem</a> section), and so
(\ref{formula-01}) becomes
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{formula-02}
\int_{C_r} \frac{f(z)}{z-z_0}dz & = & 2\pi i f(z_0) + \int_{C_r} \frac{f(z)- f(z_0)}{z-z_0}dz .
\end{eqnarray}
</div>
<p>
Now the fact that $f$ is analytic, and therefore continuous, at $z_0$
ensures that for every $\epsilon\gt 0,$
there is $\delta\gt 0$ such that
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{formula-03}
\abs{f(z)-f(z_0)}\lt \epsilon \quad \text{whenever} \quad \abs{z-z_0}\lt \delta.
\end{eqnarray}
</div>
<p>
Choose the radius $r$ of the circle $C_r$ smaller than the number
$\delta$ in the second of these inequalities. Since
$\abs{z-z_0}=r\lt \delta$ when $z $ is on $C_r,$ it follows that
the first of inequalities in (\ref{formula-03}) holds
when $z$ has this condition. Then, by the ML-inequality,
the absolute value of the integral on the right side of
the equality in (\ref{formula-02}) satisfies
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}\label{formula-04}
\abs{ \int_{C_r} \frac{f(z)- f(z_0)}{z-z_0}dz }\lt \frac{\epsilon}{r} 2 \pi r = 2\pi \epsilon.
\end{eqnarray*}
</div>
<p>
Thus, in view of equation (\ref{formula-02}), we have
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}\label{formula-05}
\abs{ \int_{C_r} \frac{f(z)}{z-z_0}dz - 2\pi i f(z_0) }\lt 2\pi \epsilon.
\end{eqnarray*}
</div>
<p>
Since the left-hand side of this inequality is a nonnegative constant
that is less than an arbitrarily small positive number, it must
be equal to zero. Therefore
\[
f(z_0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0}dz.
\]
</p>
</div>
<div class="practice">
<p>
<strong>Example 1:</strong>
Let $C$ be the circle $|z|= 2.$
To compute the integral
\[
\int_C \frac{z^2-4z+4}{z+i}dz
\]
notice that $f(z) = z^2-4z+4$
is analytic at all points within and on the contour $C.$
Since the point $z_0=- i $ is interior to $C$ (see Figure 2),
by the Cauchy integral formula, we obtain
</p>
<div class="scroll-wrapper">
\[
\int_C \frac{z^2-4z+4}{z+i}dz = 2\pi \, i f(-i)= 2\pi (3+ 4i) = \pi (-8+6i).
\]
</div>
<figure>
<div class="ggbElement" id="ggb-element-1"></div>
<figcaption>
The contour $|z|= 2.$ Activate the box <code>Phase portrait</code> to show
the enhanced phase portrait of $f(z)=\dfrac{z^2-4z+4}{z+i},$ with level curves of the modulus.
</figcaption>
</figure>
</div>
<div class="practice">
<p>
<strong>Exercise 1:</strong>
Show that $\ds \int_C \frac{z}{z^2+9}dz = \pi i,$
where $C$ is the circle $\abs{z-2i}=4.$
</p>
</div>
<hr>
<h2>An extension of the Cauchy Integral Formula</h2>
<p>
The Cauchy integral formula in <strong>Theorem 1</strong> can
be extended to provide an integral representation for
derivatives of $f$ at $z_0.$ To obtain such extension, we consider
a function $f$ that is analytic everywhere inside and on a simple closed
contour $C,$ positively oriented. Then we write the Cauchy
Integral Formula as
\[
f(z)= \frac{1}{2\pi i} \int_C \frac{f(s)}{s-z}ds,
\]
where $z$ is inside $C$ and $s$ denotes points on $C.$
If we compute the derivative with respect to $z,$ we find
\begin{eqnarray}\label{integral-derivative}
f'(z)= \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z)^2}ds.
\end{eqnarray}
For the second derivative we have
\begin{eqnarray}\label{integral-derivative-02}
f''(z) = f^{(2)}(z)= \frac{1}{\pi i} \int_C \frac{f(s)}{(s-z)^3}ds.
\end{eqnarray}
</p>
<div class="practice">
<p>
<strong>Exercise 2:</strong>
Use the formal definition of derivative to verify that
$f'(z)$ exists and the expression (\ref{integral-derivative})
is in fact valid.
</p>
<details>
<summary>
Hint
</summary>
<p>
Note that we can write
</p>
<div class="scroll-wrapper">
\[
\frac{f(z+\delta z)- f(z)}{\delta z} = \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z-\delta z)(s-z)}.
\]
</div>
<p>You can also use the fact that $f$ is continuous on the contour $C,$
which guarantees that there exists $M\gt 0 $ such that
$f(z)\leq M$ for all points on $C.$
</p>
</details>
</div>
<p>
In general, we can use induction to obtain the second remarkable formula:
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{general-derivative}
f^{(n)}(z)= \frac{n!}{2\pi i} \int_C \frac{f(s)}{(s-z)^{n+1}}ds \quad (n= 0,1,2,\ldots),
\end{eqnarray}
</div>
<p>
which can be re-written as
</p>
<div class="scroll-wrapper">
\begin{eqnarray}\label{general-integral-der}
\int_C \frac{f(s)}{(s-z)^{n+1}}ds = \frac{2\pi i}{n!} f^{(n)}(z) \quad (n= 0,1,2,\ldots).
\end{eqnarray}
</div>
<p>
Verifying (\ref{general-derivative}) is more complicated than the cases
$n=1$ and $n=2.$ For more details, I recommend you to consult the book
<a href="https://archive.org/details/theoryoffunction0001mark/mode/2up" target="_blank"><em>Theory of functions
of a complex variable</em></a>
by Alekseĭ I. Markushevich (pp. 299-301).
</p>
<div class="practice">
<p>
<strong>Example 2:</strong>
We wish to evaluate
\[
\int_C \frac{z+1}{z^4+2iz^3}dz ,
\]
where $C$ is the circle $\abs{z}=1.$
</p>
<p>
Note that the integrand is not analytic at $z=0$ and $z=-2i.$
However, only $z=0$ lies inside the closed contour, as shown in Figure 3.
</p>
<figure>
<div class="ggbElement" id="ggb-element-2"></div>
<figcaption>
The contour $|z|= 1.$ Activate the box <code>Phase portrait</code> to show
the enhanced phase portrait of $f(z)=\dfrac{z+1}{z^4+2iz^3},$ with level curves of the modulus.
</figcaption>
</figure>
<p>
By re-writting the integrand as
\[
\frac{z+1}{z^4+2iz^3} = \frac{\dfrac{z+1}{z+2i}}{z^3}
\]
we can identify $z_0=0,$ $n=2$ and $f(z) = \dfrac{z+1}{z+2i}.$
Then
\[
f^{(2)}(z) = \frac{2-4i}{(z+2i)^3},
\]
and so $f^{(2)}(0) = \ds\frac{i+2}{4}.$ Hence, by (\ref{general-integral-der}) we find
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\int_C \frac{z+1}{z^4+2iz^3} dz = \frac{2\pi i}{2!} f^{(2)}(0) = -\frac{\pi}{4} + \frac{\pi}{2}i.
\end{eqnarray*}
</div>
</div>
<div class="practice">
<p>
<strong>Exercise 3:</strong>
Let $z_0$ be any point interior to a positively oriented simple
closed contour $C.$ If $f (z) = 1,$ expression (\ref{general-integral-der}), show that
\[
\int_C \frac{dz}{z-z_0} = 2\pi i
\]
and
\[
\int_C \frac{dz}{(z-z_0)^{n+1}} =0 \quad (n=1, 2, \ldots).
\]
Compare with <strong>Exercise 1</strong> in the <a href="cauchy_goursat_theorem.html">Cauchy-Goursat
Theorem</a> section.
</p>
</div>
<hr>
<h2>Some consequences of the extension</h2>
<p>
An immediate, and also remarkable, consequence of the extension of
the Cauchy integral formula is the following:
</p>
<div class="theorem">
If a function $f$ is analytic at a given point,
then its derivatives of all orders are analytic there too.
</div>
<div class="proof">
Assume that $f$ is analytic at $z_0.$ Then there must be
a neighborhood $\abs{z-z_0}\lt \epsilon$ of $z_0$ throughout
$f$ is analytic. This means that there is a positively oriented
circle centered at $z_0$ with radius $\epsilon /2,$ such that
$f$ is also analytic inside and on $C_0.$ From (\ref{general-derivative}) we
know that
\[
f^{(2)}(z) = \frac{1}{2\pi i} \int_{C_0} \frac{f(s)}{(s-z)^3}ds
\]
at each point $z$ inside $C_0,$ and the existence of $f^{(2)}(z)$
throughout the neighborhood $|z - z_0| \lt \epsilon /2$
means that $f'$ is analytic at $z_0.$
<p>
We can apply a similar argument to the analytic function
$f'$ to conclude that $f^{(2)}(z)$ is analytic,
and so on.
</p>
</div>
<p>
As a consequence, when a function $ f (z) = u(x, y) + iv(x, y)$
is analytic at a point $z = (x, y),$ the differentiability of
$f'$ ensures the continuity of $f'$ there. Then, since
\[
f'(z) = u_x + iv_x = v_y - i u_y,
\]
the first-order partial derivatives of $u$ and $v$ are
continuous at that point. Furthermore, since
\begin{eqnarray*}
f''(z)=& u_{xx} + iv_{xx} &= v_{yx} - i u_{yx},\\
& \vdots &
\end{eqnarray*}
we can also conclude that the real functions $u$ and $v$
have continuous partial derivatives of all orders at a
point of analyticity.
</p>
<div class="corollary">
If $ f (z) = u(x, y) + iv(x, y)$ is analytic at a
point $z = (x, y),$ then the functions $u$ and $v$ have
continuous partial derivatives of all orders at that point.
</div>
<p>
</p>
<p>
Finally, another consequence of the Cauchy integral formula that
will be essential in the next section.
</p>
<div class="theorem" id="cauchy-inequality">
<span style="font-style: normal!important;font-weight: bold!important;">(Cauchy's inequality)</span>
Suppose that $f$ is analytic inside and on a positively
oriented circle $C_r$ centered at $z_0$ and with radius $r.$
If $\abs{f(z)}\leq M_r$ for all points $z$ on $C_r,$ then
\[
\abs{f^{(n)}(z_0)}\leq \frac{n!M_r}{r^n}
\]
</div>
<div class="proof">
This inequality is called <em>Cauchy's inequality</em> and is
an immediate consecuence of the expression (\ref{general-derivative}).
From the hypothesis,
\[
\abs{\frac{f(z)}{(z-z_0)^{n+1}}}= \frac{\abs{f(z)}}{r^{n+1}}\leq \frac{M_r}{r^{n+1}}.
\]
<p>
Thus, from (\ref{general-derivative}) and the ML-inequality, we have
</p>
<div class="scroll-wrapper">
\begin{eqnarray*}
\abs{f^{(n)}(z_0)} = \frac{n!}{2\pi} \abs{\int_C\frac{f(z)}{(z-z_0)^{n+1}}dz}\leq \frac{n!}{2\pi}
\frac{M_r}{r^{n+1}}2 \pi r = \frac{n!M_r}{r^n}.
\end{eqnarray*}
</div>
</div>
<br>
</article>
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