Cauchy’s Integral Theorem and Formula 

Cauchy’s Integral Theorem and Cauchy’s Integral Formula are fundamental results in complex analysis that link the values of analytic functions to contour integrals. The Integral Theorem states that if a function is analytic over a simply connected region, its integral over any closed contour is zero. The Integral Formula goes further, providing the exact value of a function at a point inside a contour using an integral. These tools are powerful for evaluating complex integrals and studying function behavior.

1.0What is Cauchy’s Integral Theorem?

Cauchy’s Integral Theorem states:

If a complex-valued function f(z) is analytic (holomorphic) on and inside a simple closed curve C, then the line integral of f(z) over C is zero.

Mathematically: ${\int }_{c}f(z)dz=0$

Conditions:

• f(z) must be analytic on and inside C
• C must be a simple (non-intersecting), closed curve traversed in the positive (counter-clockwise) direction

2.0Cauchy Integral Formula

Cauchy's Integral Formula gives the value of an analytic function inside a closed curve in terms of an integral over that curve.

Statement:

Let f(z) be analytic inside and on a simple closed curve C, and let a be a point inside C. Then,

$f(a)=\frac{1}{2\pi i}{\int }_c\frac{f(z)}{z-a}dz$

This formula not only evaluates the function at a point but also is the foundation for deriving higher-order derivatives and solving complex integrals.

3.0What is Cauchy's Integral Formula for Derivatives?

Cauchy's formula can be extended to find derivatives of an analytic function:

General Form: $f^n(a)=\frac{n!}{2\pi i}{\int }_c\frac{f(z)}{(z-a{)}^{n-1}}dz$

This is known as the Cauchy's Integral Formula for Derivatives. It helps evaluate any order derivative of a function f(z) at a point a, using only values of f(z) around a contour.

4.0Application of Cauchy's Integral Theorem and Formula

• Solving complex integrals quickly
• Verifying analyticity of a function
• Evaluating real definite integrals using contour integration
• Deriving Taylor and Laurent series
• Computing residues in advanced complex analysis

5.0What is Cauchy’s Integral Problem?

Cauchy's integral problem typically refers to problems involving evaluating contour integrals using Cauchy’s Theorem or Formula. These are common in exams like GATE, JEE Advanced, and undergraduate analysis.

Example: Evaluate ${\int }_c\frac{e^x}{z-2}dz$, where C is a circle centered at the origin with radius 3.

Solution:

Since $f(z)=e^x$ is analytic everywhere, and z = 2 lies inside C, by Cauchy’s Integral Formula:

6.0Solved Examples on Cauchy’s Integral Theorem and Formula

Example 1: Evaluate ${\int }_c\frac{z^2+3z+2}{z-1}dz$, where C is a circle of radius 2 centered at origin.

Solution:

Since z = 1 is inside C, and $f(z)=z^2+3z+2$ is analytic,

$\begin{array}{rl}& {\int }_c\frac{f(z)}{z-1}dz=2\pi i\cdot f(1)\\ & =2\pi i\cdot (1+3+2)\\ & =2\pi i\cdot 6=12\pi i\end{array}$

Example 2: Find f''(0) where $f(z)=\frac{1}{z^2+1}$, and C is the circle |z| = 2.

Solution:
Use the derivative form of Cauchy’s Integral Formula:

$f"(0)=\frac{2!}{2\pi i}{\int }_c\frac{f(z)}{z^3}dz$

$f"(0)=\frac{2}{2\pi i}{\int }_c\frac{1}{(z^2+1)z^3}dz$

Now use residue theorem or partial fractions to evaluate.

Example 3: Evaluate the integral ${\int }_c\frac{e^z}{z-1}dz$ where C is the circle |z| = 2 traversed counterclockwise.

Solution:
This is directly an application of Cauchy’s Integral Formula. The integrand is of the form $\frac{f(z)}{z-a}$, where:

• $f(z)=e^x$, analytic in and on C
• a = 1, and $1ϵInt(C)$

By Cauchy’s Integral Formula:

${\int }_c\frac{f(z)}{z-a}dz=2\pi if(a)$

$\Rightarrow {\int }_c\frac{e^z}{z-a}dz=2\pi i{e}^1=2\pi ie$

Answer: $2\pi ie$

Example 4 : Let $f(z)=\frac{1}{z^2+1}$. Evaluate ${\int }_{c}f(z)dz$ where C is the circle |z| = 2.

Solution:

Factor the denominator:$f(z)=\frac{1}{(z-i)(z+i)}$

Poles at z = i and z = -i, both inside the circle |z| = 2. Since f(z) is not analytic inside C, Cauchy's theorem does not apply directly. Instead, compute via Residue Theorem or decompose:

Let’s use Partial Fraction Decomposition:

$f(z)=\frac{1}{2i}(\frac{1}{z-i}-\frac{1}{z+i})$

Apply Cauchy’s Integral Formula for each:

${\int }_c\frac{1}{z-i}dz=2\pi i,{\int }_c\frac{1}{z+i}dz=2\pi i$

So, ${\int }_{c}f(z)dz=\frac{1}{2i}(2\pi i-2\pi i)=0$

Answer: 0

Example 5 : Evaluate the integral ${\int }_c\frac{sinz}{z}dz$ where C is any simple closed contour enclosing the origin.

Solution:

$f(z)=\frac{sinz}{z}$ is analytic at all points including at z = 0, since:

$\underset{x\to 0}{\lim }\frac{sinz}{z}=1$

Thus, f(z) is entire.

By Cauchy’s Theorem, the integral over a closed contour is:

Answer: 0

Example 6 : Let $f(z)=z^3$, and let C be the circle |z| = 2. Evaluate  $f"(0)=\frac{2!}{2\pi i}{\int }_c\frac{f(z)}{z^3}dz$

Solution:

Cauchy’s Integral Formula for derivatives states:

$f"(a)=\frac{n!}{2\pi i}{\int }_c\frac{f(z)}{(z-a{)}^{n+1}}dz$

Given:

• $f(z)=z^3$
• a = 0
• n = 2

Apply:

$f"(0)=\frac{2!}{2\pi i}{\int }_c\frac{z^3}{z^3}dz$

$\Rightarrow f"(0)=\frac{2!}{2\pi i}{\int }_{c}1dz$

$\Rightarrow f"(0)=\frac{2}{2\pi i}.0=0$

(Since the integral of a constant over a closed contour is 0.)

Answer: 0

Example 7: Evaluate: ${\int }_c\frac{1}{z}dz$where C is the unit circle |z| = 1 oriented counterclockwise.

Solution:

This is a standard result:${\int }_c\frac{1}{z}dz=2\pi i$

Answer: 2πi 

7.0Practice Questions on Cauchy’s Integral Theorem and Formula

1. Evaluate ${\int }_c\frac{z^3+2z}{z-i}dz$, where C is a circle of radius 2 centered at origin.
2. Use Cauchy’s Integral Formula to evaluate ${\int }_c\frac{e^z}{(z-a{)}^2}dz$.
3. Prove that the integral of any analytic function around a closed curve is zero using Cauchy's Integral Theorem.
4. State and prove the Cauchy's Integral Formula for Derivatives.
5. Find $f^{(3)}(0)$ for $f(z)=\frac{sinz}{z^2}$

8.0Sample Questions on Cauchy’s Integral Theorem and Formula

Q1: State Cauchy’s Integral Formula.

Ans:  $f(a)=\frac{1}{2\pi i}{\int }_c\frac{f(z)}{z-a}dz$. It evaluates the function f(a) using an integral over the curve C.

Q2: What is Cauchy’s Integral for Derivatives?

Ans:  $f^{(n)}(a)=\frac{n!}{2\pi i}{\int }_c\frac{f(z)}{(z-a{)}^{n+1}}dz$