Introduction to Contour Integration

Contour integration is a powerful tool in complex analysis, allowing us to evaluate integrals by expressing them in terms of complex functions and contours. This article aims to guide you through the process of calculating the integral of a complex function using Cauchy’s Integral Theorem and the Residue Theorem. We will apply these theorems to find the integral of the function fz(z) 1/(z - 1) middot; (z - 3i) cos(2i / (z - 3i)) over the circle z 4.

Step 1: Identifying Singular Points

The first step involves identifying the singular points of the function. Our function contains two distinct parts: 1/(z - 1) and (z - 3i) cos(2i / (z - 3i)). Each of these parts has a singular point inside the contour z 4. Let’s find these points step by step:

Singular Point at z 1

For the first term, 1/(z - 1), the singular point is clearly at z 1. This function is analytic everywhere except at z 1. According to Cauchy’s Integral Theorem, for a function analytic in a simple closed contour, the integral of the function over that contour is zero. However, if there is a singularity inside the contour, the integral is given by the value of the function times 2πi at that point.

[int_{z4} frac{1}{z-1} dz 2pi i cdot 1 2pi i]

Singular Point at z 3i

For the second term, (z - 3i) cos(2i / (z - 3i)), there is a singularity at z 3i. We need to determine the nature of this singularity using the Residue Theorem. The function is not defined at z 3i but is analytic everywhere else. Therefore, we need to use the residue theorem to evaluate the integral around this singularity.

Using the Residue Theorem

The Residue Theorem states that the integral of a meromorphic function (a function that is analytic in a region except at a finite number of singular points) over a closed contour is 2πi times the sum of the residues of the function at the poles inside the contour. Here, we need to find the residue of the function at z 3i.

Finding the Residue

First, we need to express the integrand in a simpler form. Let's make the substitution u z - 3i. This shifts the contour from z 4 to u 1, and the integral becomes:

[int_{u1} (u cos(2i/u)) du]

As u approaches 3i, the function cos(2i/u) approaches a specific value, and we can analyze the behavior around this singularity. We need to find the coefficient of the 1/u term in the Laurent series of the function. To do this, let's rewrite the integrand using the power series expansion for cosine:

[cos(2i/u) 1 - frac{1}{2!} left(frac{2i}{u}right)^2 ldots]

Substituting this back into the integral, we get:

[int_{u1} u left(1 - frac{1}{2!} left(frac{2i}{u}right)^2 ldotsright) du int_{u1} u - frac{2}{u} ldots du]

The term of interest is the coefficient of 1/u, which is -2. This coefficient is the residue of the function at z 3i. Therefore, the integral is:

[int_{z4} (z - 3i) cos(2i / (z - 3i)) dz 2pi i cdot (-2) -4pi i]

Combining Results with Cauchy’s Integral Theorem

Now, we combine the results from both parts of the function using the integral property from Cauchy’s Integral Theorem. Adding the contributions from both singular points, we get:

[I int_{z4} left(frac{1}{z-1} (z - 3i) cos(2i / (z - 3i))right) dz 2pi i - 4pi i -2pi i]

Therefore, the final answer is:

I 6πi

Conclusion

Through this detailed process, we have successfully calculated the integral of the function 1/(z - 1) middot; (z - 3i) cos(2i / (z - 3i)) over the circle z 4 using Cauchy’s Integral Theorem and the Residue Theorem. This exercise demonstrates the power of complex analysis in evaluating contour integrals, providing a robust framework for solving problems in mathematical physics and engineering.