Using the Cauchy Integration Formula to Evaluate Complex Integrals
This guide outlines the process of using the Cauchy Integration Formula to evaluate complex integrals. We will walk through a specific example and explain the steps involved in applying this powerful tool in complex analysis.
Introduction to the Cauchy Integration Formula
The Cauchy Integration Formula is a fundamental theorem in complex analysis that allows us to evaluate integrals of holomorphic (analytic) functions over simple closed contours. The formula states that for a function (f(z)) that is holomorphic inside and on a simple closed contour (gamma), and (a) being a point inside (gamma), the following holds:
[int_{gamma} frac{f(z)}{z-a} , dz 2pi i f(a)]Problem and Setup
Let's consider the following integral:
[ int_{B_{20}} frac{3z}{z^3-1} , dz ]where ( B_{20} {z in mathbb{C} : |z|
Step 1: Identify the Function and the Contour
The function we are integrating is given by:
[ f(z) frac{3z}{z^3-1} ]The contour ( B_{20} ) is the open disk of radius 2 centered at the origin. We need to determine if there are any singularities inside this contour.
Step 2: Find the Singularities
The singularities of ( f(z) ) occur where the denominator is zero:
[ z^3 - 1 0 implies z^3 1 implies z 1, e^{2ipi/3}, e^{-2ipi/3} ]The roots of the equation ( z^3 1 ) are the three cube roots of unity. Among these, only ( z 1 ) lies inside the contour ( B_{20} ). The other two roots ( e^{2ipi/3} ) and ( e^{-2ipi/3} ) lie outside the contour.
Step 3: Applying the Cauchy Integration Formula
Since ( f(z) ) is analytic inside and on the contour ( B_{20} ) except for the singularity at ( z 1 ), we can apply the Cauchy Integration Formula:
[int_{B_{20}} frac{f(z)}{z-1} , dz 2pi i f(1)]Now, we need to evaluate ( f(1) ):
[ f(1) frac{3 cdot 1}{1^3 - 1} frac{3}{0} ]Note that the term ( 1^3 - 1 ) simplifies to 0, which makes ( f(1) ) undefined. However, we can rewrite the function to isolate the term containing the singularity:
[ f(z) frac{3z}{(z-1)(z^2 z 1)} ]Therefore, the integral becomes:
[int_{B_{20}} frac{3z}{z^3-1} , dz int_{B_{20}} frac{3z}{(z-1)(z^2 z 1)} , dz]Thus, applying the Cauchy Integration Formula:
[int_{B_{20}} frac{3z}{(z-1)(z^2 z 1)} , dz 2pi i cdot frac{3z}{z^2 z 1} bigg|_{z1} ]Step 4: Evaluating the Function at the Singularity
Substitute ( z 1 ) into the simplified function:
[ frac{3 cdot 1}{1^2 1 1} frac{3}{3} 1 ]Thus, the integral evaluates to:
[int_{B_{20}} frac{3z}{z^3-1} , dz 2pi i cdot 1 2pi i ]Step 5: Correcting the Sign and Final Result
Note that we have a negative sign in the denominator of the original function, so the final result should be:
[int_{B_{20}} frac{3z}{z^3-1} , dz 2pi i cdot (-3) -6pi i]Therefore, the evaluated integral is:
[boxed{-6pi i}]Conclusion
This example demonstrates how to use the Cauchy Integration Formula to evaluate complex integrals over bounded regions. By recognizing and isolating the singularities inside the contour, we can apply the formula effectively, leading to accurate evaluations of the integrals.