Lecture 7: Cauchy's Integral Formula

The Cauchy’s Theorem combined with the Method of Contour Deformation allows us to evaluate integrals on contours that enclose singular points, i.e. points at which the function fail to be analytic, of the integrands. Cauchy’s Integral Formula pushes this one step further to handle more complicated integrands.

Cauchy’s Integral Formula

Cauchy’s Integral Formula is given in the following theorem.

Theorem 1 (Cauchy’s Integral Formula). Let $f(z)$ be analytic interior to and on a simple closed contour $C$ . Then at any interior point $z$ $\label{EQ:CIF} f(z) = \dfrac{1}{2\pi i} \int_C \dfrac{f(\zeta)}{\zeta - z} d\zeta.$

Proof. Let us use the Method of Contour Deformation to deform the contour to include a small circle with radius $\delta$ centered at $z$ , and we denote the circle by $C_\delta$ . Then Cauchy’s Theorem indicates that $\int_C \dfrac{f(\zeta)}{\zeta - z} d\zeta=\int_{C_\delta} \dfrac{f(\zeta)}{\zeta - z} d\zeta.$ To evaluate this integral, we have $\begin{gathered} \int_{C_\delta} \dfrac{f(\zeta)}{\zeta - z} d\zeta=\int_{C_\delta} \dfrac{f(\zeta)-f(z)+f(z)}{\zeta - z} d\zeta=\int_{C_\delta} \left(\dfrac{f(z)}{\zeta - z}+\dfrac{f(\zeta)-f(z)}{\zeta - z}\right) d\zeta \\ = 2\pi i f(z) + \int_{C_\delta} \dfrac{f(\zeta)-f(z)}{\zeta - z} d\zeta \end{gathered}$ Use the fact that $f$ is continuous, we have that $\forall \varepsilon>0$ , $\exists \delta>0$ such that $|\zeta-z|\le\delta$ implies $|f(\zeta)-f(z)|\le\varepsilon$ . Therefore, $\forall \varepsilon$ , $\exists \delta$ such that $|\int_{C_\delta} \dfrac{f(\zeta)-f(z)}{\zeta - z} d\zeta|= |\int_{C_\delta} \dfrac{f(\zeta)-f(z)}{\zeta - z} \zeta'(\theta) d\theta| \le \int_0^{2\pi} \dfrac{\varepsilon}{\delta} \delta d\theta =2\pi \varepsilon.$ The result then follows by taking $\varepsilon\to 0$ . ◻

When proving Cauchy’s Integral Formula, we used the following well-known result about integrals.

Theorem 2. Let $C$ be a contour and $f(z)$ be continuous on $C$ . Then $|\int_C f(z) dz| \le M |C|,$ where $M$ is an upper bound for $|f|$ and $|C|$ denote the length of $C$ .

Proof. Let $C$ be parameterized by $t$ with $a\le t\le b$ . The result then comes from the following calculation: $\left|\int_C f(z) dz\right| =\left |\int_a^b f(z(t)) z'(t) dt\right| \le \int_a^b|f(z(t))||z'(t)|dt\le M |\int_a^b |z'(t)|dt =M |C|.$ The last step comes from the definition of the length of $C$ . ◻

Applications of Cauchy’s Integral Formula

The Cauchy’s Integral Formula provides another way to evaluate certain contour integrals where the integrands have the form: $\dfrac{g(z)}{z-z_0}$ , with $g(z)$ analytic and $z_0$ enclosed in the contour. More precisely, let $C$ be a simple closed contour that enclose $z_0$ , and $f(z)$ an analytic function in the region enclosed by $C$ . Then $\int_C \dfrac{f(z)}{z-z_0} dz = f(z_0) \cdot 2\pi i.$

Example 3. Let $C$ be a simple closed contour that encloses $z=0$ , then $\int_C \dfrac{1}{z} dz = 1\cdot 2 \pi i = 2 \pi i.$

$\int_C \dfrac{\sin z}{z} dz = \sin 0 \cdot 2 \pi i = 0.$

$\int_C \dfrac{\cos z}{z} dz = \cos 0 \cdot 2 \pi i = 2\pi i.$

$\int_C \dfrac{e^z}{z} dz = e^0 \cdot 2 \pi i = 2\pi i.$

Example 4. Let $C$ be the unit circle centered at the origin. Then $\int_C \dfrac{1}{z-1/2} dz = 1 \cdot 2 \pi i = 2\pi i.$ $\int_C \dfrac{1}{(z-1/2)(z+1/2)} dz = \int_C \left[\dfrac{1}{z-1/2} - \dfrac{1}{z+1/2}\right] dz = 1 \cdot 2 \pi i - 1 \cdot 2 \pi i =0.$

$\begin{gathered} \int_C \dfrac{\sin(\pi z)}{(z-1/2)(z+1/2)} dz = \int_C \sin(\pi z) \left[\dfrac{1}{z-1/2} - \dfrac{1}{z+1/2}\right] dz \qquad = \sin(\pi/2) \cdot 2 \pi i - \sin(-\pi/2) \cdot 2 \pi i =4 \pi i. \end{gathered}$

Example 5. Let $C$ be the circle of radius $2\pi$ centered at the origin. $\int_C \dfrac{e^{iz}}{z(z-\pi)} dz =\int_C \dfrac{e^{iz}}{\pi} \left[\dfrac{1}{z-\pi}-\dfrac{1}{z} \right] dz \qquad =\left(\dfrac{e^{i\pi}}{\pi}-\dfrac{e^{i0}}{\pi} \right)\cdot 2 \pi i=-4 i.$

Cauchy’s Derivative Formula

Cauchy’s Integral Formula allows us to establish an important property about analytic functions: if $f(z)$ is analytic, then $f^{(k)}(z)$ exists for all $k\ge 1$ and there is a formula to find all of them.

Theorem 6 (Cauchy’s Derivative Formula). Let $f(z)$ be analytic in the interior to and on a simple closed contour $C$ . Then $f^{(k)}$ ( $k\ge 1$ ) exists at all $z$ in the interior of $C$ . Moreover, $\label{EQ:Cauchy Derivative Formula} f^{(k)}(z) =\dfrac{k!}{2\pi i}\int_C \dfrac{f(\zeta)}{(\zeta-z)^{k+1}} d\zeta.$

Proof. (i) For any $z$ inside $C$ , the Cauchy Integral Formula holds. Therefore $\begin{gathered} \dfrac{f(z+h)-f(z)}{h} =\dfrac{1}{2\pi i}\dfrac{1}{h} \int_C f(\zeta)\left(\dfrac{1}{\zeta-(z+h)}-\dfrac{1}{\zeta-z}\right) d\zeta\\ =\dfrac{1}{2\pi i} \int_C f(\zeta) \dfrac{1}{(\zeta-z-h)(\zeta-z)} d\zeta = \dfrac{1}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^2}d\zeta+ \dfrac{h}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^2(\zeta-z-h)}d\zeta.\end{gathered}$ Let $\min|\zeta-z|=2\delta>0$ . Then when $|h|<\delta$ for all $\zeta$ , we have that $|\zeta-(z+h)|\ge |\zeta-z|-|h|>\delta$ by the Triangle Inequality. Moreover, we know that $|f(\zeta)|<M$ for some $M$ , $\forall \zeta\in C$ . Therefore the last term can be bounded by $|\dfrac{h}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^2(\zeta-z-h)}d\zeta| \le \dfrac{|h|}{2\pi}\dfrac{M}{(2\delta)^2 \delta} |C|,$ where $|C|$ denote the length of the contour $C$ .

(ii) We can now take $h\to 0$ to find that $f'(z)=\dfrac{1}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^2}d\zeta.$

(iii) Assume that the formula [EQ:Cauchy Derivative Formula] holds for $k$ , we then have follow the same calculation as in (i) and (ii): $\begin{gathered} \dfrac{f^{(k)}(z+h)-f^{(k)}(z)}{h} =\dfrac{k!}{2\pi i}\dfrac{1}{h} \int_C f(\zeta)\left(\dfrac{1}{(\zeta-(z+h))^{k+1}}-\dfrac{1}{(\zeta-z)^{k+1}}\right) d\zeta\\ =\dfrac{k!}{2\pi i} \int_C f(\zeta) \dfrac{(k+1)(\zeta-z)^k h}{[(\zeta-z-h)(\zeta-z)]^{k+1}} d\zeta +H.O.T. \\ = \dfrac{(k+1)!}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^{k+2}}d\zeta+ \dfrac{(k+1)!h}{2\pi i} \int_C \dfrac{f(\zeta)}{(\zeta-z)^{k+2}(\zeta-z-h)}d\zeta + H.O.T,\end{gathered}$ where $H.O.T$ denotes terms with $h^m$ for $m\ge 2$ . In the same manner the second term and H.O.T. vanishes when $h\to 0$ . Therefore, we should have [EQ:Cauchy Derivative Formula] hold for $k+1$ .

(iv) By mathematical induction, we can see that [EQ:Cauchy Derivative Formula] holds for any $k\ge 1$ .

(v) Analyticity of each derivative of $f(z)$ is easily seen by the fact that the derivatives are well-defined. ◻

The above result says that if $f(z)=u(x,y)+iv(x,y)$ is analytic in some neighborhood, its real and imaginary parts can be differentiated infinite number of times with respect to $x$ and $y$ .

Application of Cauchy’s Derivative Formula

In fact, the Cauchy derivative formula [EQ:Cauchy Derivative Formula] can be very useful in evaluating integrals as well. Let us re-write the formula a little bit to get $\int_C \dfrac{f(z)}{(z-z_0)^{k+1}} d\zeta = f^{(k)}(z_0) \cdot \dfrac{2\pi i}{k!}.$

Example 7. Let $C$ be the unit circle centered at the origin. To evaluate $\int_C \dfrac{1}{z^2} dz$ , we observe that $\int_C \dfrac{1}{z^2} dz =\int_C \dfrac{f(z)}{(z-z_0)^{k+1}} dz$ where $f(z)=1$ and $z_0=0$ . By Cauchy’s Derivative Formula, we have that $\int_C \dfrac{1}{z^2} dz =\int_C \dfrac{f(z)}{(z-z_0)^{1+1}} dz = f'(z_0) \dfrac{2\pi i}{1!} = 0.$ In fact, the same approach shows that $\int_C \dfrac{1}{z^m} dz=0$ $\forall m\ge 2$ as we have proved before.

Example 8. Let $C$ be the circle of radius $4$ centered at the origin.

(a) Using the fact that $(e^z)' = e^z$ , we can compute $\int_C \dfrac{e^z}{(z-\pi)^2} dz =e^\pi \cdot\dfrac{2\pi i}{1!} = 2\pi e^\pi i.$

(b) Using the fact that $(\sin z)''' = \cos z$ , we can compute $\int_C \dfrac{\sin z}{(z-\pi)^4} dz =\cos \pi \cdot\dfrac{2\pi i}{3!} = - \cdot\dfrac{\pi i}{3}$

Cauchy’s Derivative Formula expresses derivatives of holomorphic functions in terms of the functions themselves. This often allows us to bound the derivatives of those functions using bounds on the function values. Here is a general result.

Theorem 9 (The Cauchy Estimates). Let $f(z)$ be analytic in an open set $\Omega$ that contains the disk $\mathcal B_r(z_0)$ ( $r>0$ ). Assume that $|f(z)|\le M$ $\forall z\in\overline{\mathcal B_r(z_0)}$ . Then $|f^{(k)}(z_0)| \le \dfrac{k! M}{r^{k}},\ \ k=1, 2, \cdots\,.$

Proof. By Cauchy’s Deriative Formula, we have $|f^{(k)}(z_0)| =|\dfrac{k!}{2\pi i}\int_{\partial \mathcal B_r(z_0)} \dfrac{f(\zeta)}{(\zeta-z_0)^{k+1}} d\zeta| \le \dfrac{k!}{2\pi} \int_{\partial \mathcal B_r(z_0)} |\dfrac{f(\zeta)}{(\zeta-z_0)^{k+1}} ||d\zeta|.$ We then use the parameterization $\zeta-z_0=r e^{i\theta}$ (which gives $d\zeta=ire^{i\theta}d\theta$ ) on $\partial\mathcal B_r(z_0)$ and the bound on $f$ to have $|f^{(k)}(z_0)| \le \dfrac{k!}{2\pi} \int_0^{2\pi} \dfrac{M}{r^{k+1}} d\theta= \dfrac{k!}{2\pi} \dfrac{M}{r^{k+1}} 2\pi=\dfrac{k! M}{r^{k}}\,.$ This completes the proof. ◻

It is easy to see that the Cauchy estimates in the above theorem can be derived for every point inside the disk $\mathcal B_r(z_0)$ , not just at $z_0$ . Here is a modified version of the theory.

Theorem 10 (The Modified Cauchy Estimates). Let $f(z)$ be analytic in an open set $\Omega$ that contains the disk $\mathcal B_r(z_0)$ ( $r>0$ ). Assume that $|f(z)|\le M$ $\forall z\in\overline{\mathcal B_r(z_0)}$ . Then for any $z\in\mathcal B_r(z_0)$ , we have $|f^{(k)}(z)| \le \dfrac{k!r\,M}{(1-|z-z_0|)^{k+1}},\ \ k=1, 2, \cdots\,.$

Proof. By Cauchy’s Deriative Formula, we have $|f^{(k)}(z)| =|\dfrac{k!}{2\pi i}\int_{\partial \mathcal B_r(z_0)} \dfrac{f(\zeta)}{(\zeta-z)^{k+1}} d\zeta| \le \dfrac{k!}{2\pi} \int_{\partial \mathcal B_r(z_0)} |\dfrac{f(\zeta)}{(\zeta-z)^{k+1}} ||d\zeta|.$ Using the parameterization $\zeta-z_0=r e^{i\theta}$ , the bound on $f$ , and the fact that $|\zeta-z|=|\zeta-z_0+z_0-z| \ge |\zeta-z_0|-|z-z_0|=1-|z-z_0|$ (since $|z-z_0|<1$ ), we have $|f^{(k)}(z)| \le \dfrac{k!}{2\pi} \int_0^{2\pi} \dfrac{r\, M}{(1-|z-z_0|)^{k+1}} d\theta= \dfrac{k!r\,M}{(1-|z-z_0|)^{k+1}}\,.$ This completes the proof. ◻

Guided review and additional examples

Learning goals

Identify the analytic numerator and the point enclosed by a contour.
Apply Cauchy’s integral formula without unnecessary parameterization.
Use the higher-derivative form of the formula.
Connect contour data with values and derivatives inside the contour.

Concept connection: boundary values determine interior values

Cauchy’s integral formula says that an analytic function’s value at an interior point is a weighted average of its boundary values. Its derivative formulas go further: the same boundary data determine every derivative. This rigidity is one of the central differences between analytic and merely differentiable real functions.

Worked example 1: direct use of the formula

For the positively oriented circle $|z|=2$ , the point $1$ lies inside and $e^z$ is entire. Therefore

$\int_{|z|=2}\frac{e^z}{z-1}\,dz=2\pi i\,e.$

If the contour did not enclose $1$ , the integrand would be analytic throughout its interior and the integral would be zero.

Worked example 2: a higher-order denominator

The derivative formula is

$\int_C\frac{f(z)}{(z-a)^{n+1}}\,dz =\frac{2\pi i}{n!}f^{(n)}(a).$

Taking $f(z)=e^z$ , $a=1$ , and $n=2$ gives

$\int_{|z|=2}\frac{e^z}{(z-1)^3}\,dz =\frac{2\pi i}{2!}e=\pi i e.$

Check your understanding

What three checks should you make before applying Cauchy’s integral formula?

Show the answer

Check that the contour is closed with known orientation, that the evaluation point lies inside it, and that the numerator is analytic on and throughout the contour. Also verify that no other denominator singularities invalidate the setup.