Question

Let \(C\) be the boundary of a square whose sides lie along the lines \(x=\pm 3\) and \(y=\pm 3 .\) For the positive sense of integration, evaluate each of the following integrals. (a) \(\oint_{C} \frac{e^{-z} d z}{z-i \pi / 2}\), (b) \(\oint_{C} \frac{e^{z} d z}{z\left(z^{2}+10\right)}\), (c) \(\oint_{C} \frac{\cos z d z}{\left(z-\frac{\pi}{4}\right)\left(z^{2}-10\right)}\), (d) \(\oint_{C} \frac{\sinh z d z}{z^{4}}\), (e) \(\oint_{C} \frac{\cosh z d z}{z^{4}}\), (f) \(\oint_{C} \frac{\cos z d z}{z^{3}}\), (g) \(\oint_{C} \frac{\cos z d z}{(z-i \pi / 2)^{2}}\), (h) \(\oint_{C} \frac{e^{z} d z}{(z-i \pi)^{2}}\), (i) \(\oint_{C} \frac{\cos z d z}{z+i \pi}\), (j) \(\oint_{C} \frac{e^{z} d z}{z^{2}-5 z+4}\), (k) \(\oint_{C} \frac{\sinh z d z}{(z-i \pi / 2)^{2}}\), (l) \(\oint_{C} \frac{\cosh z d z}{(z-i \pi / 2)^{2}}\), (m) \(\oint_{C} \frac{\tan z d z}{(z-\alpha)^{2}}, \quad\) for \(-3<\alpha<3\), (n) \(\oint_{C} \frac{z^{2} d z}{(z-2)\left(z^{2}-10\right)}\).

Short answer

As there are no singularities inside the contour for integral (a), the result is zero. The procedure for the remaining integrals will follow similarly, by identifying singularities, computing residues at these points (if inside the contour), and summing them up

Step-by-step solution

Identify Singularities

Identify the points in the complex plane at which the integrand fails to be analytic, i.e., the singularities. Consider the function under the integral sign. In integral (a), the denominator is \(z - i\frac{\pi}{2}\), which is zero when \(z = i\frac{\pi}{2}\). So, the function has a single singularity at \(z = i\frac{\pi}{2}\). However, as this point lies outside of the square whose side lies along \(x= ±3\) and \(y= ±3\), we say that there are no singularities inside the contour.

Compute Residues

If there were singularities inside the contour, we would find the residues at these points. But since there are none, there are no residues to calculate.

Apply the Residue Theorem

Finally, we apply the residue theorem. Given that the sum of the residues is zero (since there are no residues inside the contour), the result of the integral (a) is zero.

Repeat for Other Integrals

Repeat Steps 1–3 for all other integrals. The key difference for each integral will be identifying the singularities which are the values of \(z\) for which the denominator is zero. If these singularities are within the boundaries of the square, their residues are calculated and totaled; otherwise, the result of the integral is zero as in integral (a).

Key concepts

Cauchy's Residue Theorem

Cauchy's Residue Theorem is a powerful tool in complex analysis. It's used to evaluate contour integrals, especially when direct integration is challenging. The theorem states that for a function that is analytic inside and on some closed contour, the contour integral of the function around this contour is given by \[2\pi i \times \text{(sum of residues inside the contour)}.\]
In simple terms, the key idea is that the value of the integral is determined by the "residues" of the function's singularities inside the contour. "Residues" are complex numbers that provide crucial information about the behavior of the function near its singular points.

• If a function has no singularities inside the contour, the integral evaluates to zero.
• This theorem simplifies complex contour integrations by converting them into a problem of finding residues.

Singularities in Complex Functions

Singularities are points where a complex function ceases to be analytic, which means they are where the function might "blow up" or fail to exist. Understanding singularities is crucial because their nature and position influence how integrals behave.

There are different types of singularities, but in the context of contour integration and Cauchy's Residue Theorem, we deal primarily with:

• Poles: These are like points of infinity for a function. At poles, a function behaves like some form of \(\frac{1}{(z-a)^n}\), where \(n\) is the order of the pole.
• Essential Singularities: More chaotic than poles, where the behavior is unpredictable as one approaches the singularity.

In contour integration exercises, the aim is to identify where singularities lie relative to the contour. If they lie inside, their residues impact the result; if not, they do not.

Contour Integration

Contour integration is a method of evaluating integrals of complex functions over certain "paths" in the complex plane. These paths, or contours, are usually closed loops.
The exercise you see here uses a square contour defined by boundaries along specific lines (i.e., \(x = \pm 3\) and \(y = \pm 3\)). This is a geometrical way of enclosing the complex plane to evaluate integrals.

• The choice of contour affects whether singularities are enclosed, which changes the integral's value.
• Understanding the shape and position of contours helps you determine what part of the complex plane you're integrating over.

Using contours allows for the application of the powerful theorems of complex analysis, like Cauchy's Integral Theorem and Residue Theorem, to solve integrals that would be difficult otherwise.

Analytic Functions

Analytic functions are the backbone of complex analysis. An analytic function is one that is complex differentiable everywhere in some region of the complex plane. This means it can also be expressed as a convergent power series in that area.

Key characteristics of these functions include:

• Being differentiable (not just in the calculus sense, but infinitely differentiable within its radius of convergence).
• Having no sudden jumps, breaks, or sharp turns within that region.
• Allowing for powerful results and theorems, such as Cauchy's Theorem, which establishes that if a function is analytic and closed within a contour, the integral around it is zero.

Analyticity and its loss at singularities are pivotal when applying concepts like the Residue Theorem to solve integrals. Identifying whether a function is analytic on and inside a contour determines the integration approach entirely.