Cauchy Integral Formula

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.

(Cauchy's integral formula) 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}

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 \[ \int_C \frac{f(z)}{z-z_0}dz = \int_{C_r} \frac{f(z)}{z-z_0}dz. \]

Circle C_r inside the contour — Circle $C_r$ inside the contour $C.$

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,

\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}

We know that $\ds \int_{C_r}\frac{1}{z-z_0}dz = 2\pi i$ (see Exercise 1 from Cauchy-Goursat theorem section), and so (\ref{formula-01}) becomes

\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}

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

\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}

Choose the radius $r$ of the circle $C_r$ smaller thatn 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

\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*}

Thus, in view of equation (\ref{formula-02}), we have

\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*}

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 \[ f(z_0)= \frac{1}{2\pi i} \int_C \frac{f(z)}{z-z_0}dz. \]

Example 1: 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

\[ \int_C \frac{z^2-4z+4}{z+i}dz = 2\pi \, i f(-i)= 2\pi (3+ 4i) = \pi (-8+6i). \]

The contour $|z|= 2.$ Activate the box Phase portrait to show the enhanced phase portrait of $f(z)=\dfrac{z^2-4z+4}{z+i},$ with level curves of the modulus.

Exercise 1: Show that $\ds \int_C \frac{z}{z^2+9}dz = \pi i,$ where $C$ is the circle $\abs{z-2i}=4.$

An extension of the Cauchy Integral Formula

The Cauchy integral formula in Theorem 1 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}

Exercise 2: Use the formal defition of derivative to verify that $f'(z)$ exists and the expression (\ref{integral-derivative}) is in fact valid.

Hint

Note that we can write

\[ \frac{f(z+\delta z)- f(z)}{\delta z} = \frac{1}{2\pi i} \int_C \frac{f(s)}{(s-z-\delta z)(s-z)}. \]

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 $C.$

In general, we can use induction to obtain the second remarkabl formula:

\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}

which can be re-written as

\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}

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 Theory of functions of a complex variable by Alekseĭ I. Markushevich (pp. 299-301).

Example 2: We wish to evaluate \[ \int_C \frac{z+1}{z^4+2iz^3}dz , \] where $C$ is the circle $\abs{z}=1.$

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.

The contour $|z|= 1.$ Activate the box Phase portrait to show the enhanced phase portrait of $f(z)=\dfrac{z+1}{z^4+2iz^3},$ with level curves of the modulus.

By re-writting the integrand as \[ \frac{z+1}{z^4+2iz^3} = \frac{\dfrac{z+1}{z+2i}}{z^3} \] we can identiry, $z_0=0,$ $n=2$ and $f(z) = \dfrac{z+1}{z+2i}.$ Then \[ f^{(2)}(z) = \frac{2-4i}{(z+2i)^3}, \] and so $f(0) = \ds\frac{2i+1}{4i}.$ Hence, by (\ref{general-integral-der}) we find

\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*}

Exercise 3: 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 Exercise 1 in the Cauchy-Goursat Theorem section.

Some consequences of the extension

An immediate, and also remarkable, consequence of the extension of the Cauchy integral formula is the following:

If a function $f$ is analytic at a given point, then its derivatives of all orders are analytic there too.

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 $z_0.$

We can apply a similar argument to the analytic function $f'$ to conclude that $f^{(2)}(z)$ is analytic, and so on.

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 od $f'$ 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.

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.

Finally, another consequence of the Cauchy integral formula that will be essential in the next section.

(Cauchy's inequality) 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 \[ \abs{f^{(n)}(z_0)}\leq \frac{n!M_r}{r^n} \]

This inequality is called Cauchy's inequality and is an immediate consecuence of the expression (\ref{general-derivative}). From the hypothesis, \[ \abs{\frac{f(z)}{(z-z_0)^{n+1}}}= \frac{\abs{f(z)}}{r^{n+1}}\leq \frac{M_r}{r^{n+1}}. \]

Thus, from (\ref{general-derivative}) and the ML-inequality, we have

\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*}