Evaluating Contour Integrals of Complex Functions

Contour integrals are a fundamental concept in complex analysis, which deals with complex-valued functions of a complex variable. In this article, we will explore the process of evaluating a specific contour integral involving a complex function with singularities. This example will illustrate the application of the Residue Theorem, a powerful tool in complex analysis for evaluating contour integrals with complex residues.

The Contour Integral We Aim to Solve

Let us consider the contour integral:

[int_{|z-i|sqrt{2}}frac{z^2}{cos z^2} dz]

The integrand has singularities when the denominator, (cos z^2), is equal to zero. Solving the equation (cos z^2 0) gives us the singularities:

[z pmsqrt{frac{2kpi}{2}} quad text{for any } k in mathbb{Z}]

We now need to determine which of these singularities lie inside the contour (|z-i|sqrt{2}). By checking, we find that the only singularities inside the contour are:

[z isqrt{frac{pi}{2}}quadtext{and} quad z -isqrt{frac{pi}{2}}]

Both of these points are simple poles.

Computing the Residues

To evaluate the contour integral, we need to compute the residues of the function at these singularities. The residue of a simple pole at (z z_0) is given by:

[text{Res}(f, z_0) lim_{z to z_0} (z - z_0) f(z)]

For the residue at the pole (z isqrt{frac{pi}{2}}), we have:

[text{Res}left(frac{z^2}{cos z^2}, isqrt{frac{pi}{2}}right) lim_{z to isqrt{frac{pi}{2}}} left(z - isqrt{frac{pi}{2}}right) frac{z^2}{cos z^2}]

This can be simplified as:

[ lim_{z to isqrt{frac{pi}{2}}} z^2 cdot lim_{z to isqrt{frac{pi}{2}}} frac{z - isqrt{frac{pi}{2}}}{cos z^2} -frac{pi}{2} cdot lim_{z to isqrt{frac{pi}{2}}} frac{1}{-2zsin z^2} frac{i}{2} sqrt{frac{pi}{2}}]

Similarly, for the residue at the pole (z -isqrt{frac{pi}{2}}), we have:

[text{Res}left(frac{z^2}{cos z^2}, -isqrt{frac{pi}{2}}right) lim_{z to -isqrt{frac{pi}{2}}} left(z isqrt{frac{pi}{2}}right) frac{z^2}{cos z^2}]

This can be simplified as:

[ lim_{z to -isqrt{frac{pi}{2}}} z^2 cdot lim_{z to -isqrt{frac{pi}{2}}} frac{z isqrt{frac{pi}{2}}}{cos z^2} frac{pi}{2} cdot lim_{z to -isqrt{frac{pi}{2}}} frac{1}{2zsin z^2} frac{1}{2} sqrt{frac{pi}{2}}]

Applying the Residue Theorem

The Residue Theorem states that the value of a contour integral over a closed contour is equal to (2pi i) times the sum of the residues of the function inside the contour. Therefore, we have:

[int_{|z-i|sqrt{2}}frac{z^2}{cos z^2} dz 2pi i cdot left(frac{i}{2} sqrt{frac{pi}{2}} frac{1}{2} sqrt{frac{pi}{2}}right)]

Let's simplify this expression:

[ 2pi i cdot left(frac{i}{2} sqrt{frac{pi}{2}} frac{1}{2} sqrt{frac{pi}{2}}right) pi cdot left(-1 i sqrt{frac{pi}{2}}right)]

Hence, the value of the contour integral is:

[pi cdot left(-1 i sqrt{frac{pi}{2}}right)]

Conclusion

The Residue Theorem provides a powerful method for evaluating contour integrals in complex analysis. By finding the residues at the singularities within the contour and applying the Residue Theorem, we can efficiently compute the integral.

There is no "equality" symbol in the modulus, as the integral is over the contour (|z-i|sqrt{2}). This contour represents a circle with radius (sqrt{2}) centered at the point (i) in the complex plane.