Question

Let \(C\) be the circle \(|z-i|=3\) integrated in the positive sense. Find the value of each of the following integrals. (a) \(\oint_{C} \frac{e^{z}}{z^{2}+\pi^{2}} d z\), (b) \(\oint_{C} \frac{\sinh z}{\left(z^{2}+\pi^{2}\right)^{2}} d z\), (c) \(\oint_{C} \frac{d z}{z^{2}+9}\), (d) \(\oint_{C} \frac{d z}{\left(z^{2}+9\right)^{2}}\), (e) \(\oint_{C} \frac{\cosh z}{\left(z^{2}+\pi^{2}\right)^{3}} d z\) (f) \(\oint_{C} \frac{z^{2}-3 z+4}{z^{2}-4 z+3} d z\).

Short answer

(a) 0, (b) 0, (c) \(\frac{2\pi i}{3}\), (d) \(-\frac{2\pi i}{27}\), (e) 0, (f) 0

Step-by-step solution

Analyzing the functions

We can start by examining each function to determine its analyticity on C and its antiderivative, if it exists.

(a) Integrating \(\oint_{C} \frac{e^{z}}{z^{2}+\pi^{2}} d z\)

This function is perfectly analytic on the circle C. We can find the antiderivative by recognizing that this is a standard integral form, and its antiderivative is \( F(z) = \frac{e^z}{z+i\pi} - \frac{e^z}{z-i\pi}\). Hence, the integral is 0 by Cauchy's Integral Formula.

(b) Integrating \( \oint_{C} \frac{\sinh z}{\left(z^{2}+\pi^{2}\right)^{2}} d z\)

This function is also perfectly analytic on the circle C. We can again find the antiderivative by recognizing that this is another standard integral form, and its antiderivative is \( F(z) = \frac{\cosh z}{z+i\pi}-\frac{\cosh z}{z-i\pi}\). Hence, the integral is 0 by Cauchy's Integral Formula.

(c) Integrating \(\oint_{C} \frac{d z}{z^{2}+9}\)

This function introduces a pole at z=3 and z=-3. However, given the circle C is |z-i|=3 on the positive sense, it contains the pole z=3. Thus, we need to use Residue Theorem. After calculating, we will find the integral value is \(\frac{2\pi i}{3}\).

(d) Integrating \(\oint_{C} \frac{d z}{\left(z^{2}+9\right)^{2}}\)

Just like the previous function, this one introduces poles at z=3 and z=-3. Using the same logic, we find that the integral value is \(-\frac{2\pi i}{27}\).

(e) Integrating \(\oint_{C} \frac{\cosh z}{\left(z^{2}+\pi^{2}\right)^{3}} d z\)

This function is perfectly analytic on the circle C, but calculating its antiderivative is not straightforward. However, we can still assert that, by Cauchy's Integral Formula, the integral is 0.

(f) Integrating \(\oint_{C} \frac{z^{2}-3 z+4}{z^{2}-4 z+3} d z\)

This function has a pole at z=2+\sqrt{2} which is not included in the circle C which means we don't need to use Residue Theorem. It's a simple rational function and because it's analytic on C its integral is 0.

Key concepts

Cauchy's Integral Formula

Cauchy's Integral Formula plays a crucial role in complex integration, especially when dealing with contours encompassing poles of a function. It essentially states that if a function is analytic (has a derivative at every point) within and on a simple closed contour, then the integral of that function around the contour is determined solely by the value of the function at the points within the contour.

Imagine you are drawing a loop around a point on a plane; the value of the integral of an analytic function over this loop depends on the function's behavior at the point encircled by the loop. For instance, in our exercise part (a), the function \( e^z / (z^2 + \pi^2) \) is analytic on and inside the given contour C. Because of this, Cauchy's Integral Formula allows us to swiftly conclude that the integral is zero without calculating the antiderivative explicitly.

In more complicated scenarios, like (b), where we have \( \sinh z / (z^2 + \pi^2)^2 \) and (e) with \( \cosh z / (z^2 + \pi^2)^3 \), we also lean on the formula's assurance that if the function is analytic inside the contour, these higher order integrals will also equal zero.

Residue Theorem

When encountering singularities within an integration contour, the Residue Theorem comes to the rescue. A central concept in the study of complex integration, the theorem applies to functions with isolated singularities, providing a method to evaluate integrals more efficiently by using residues at those singular points.

Residues are, simply put, coefficients of the \(1/(z-a)\) term in a function's Laurent series expansion around a point \(a\). In exercises (c) and (d), where we have integrals of functions with poles inside the contour C, the Residue Theorem tells us how to calculate the integral by identifying and utilizing these residues. For example, in part (c), we have a pole at \(z=3\), and the theorem helps us extract the residue at this point, yielding \(2\pi i/3\) as the integral's result because this pole lies within our contour.

Understanding and applying the Residue Theorem can greatly simplify the evaluation of complex integrals around curves that encircle one or more singularities, as we can focus on the singular points themselves rather than laboriously computing the integral around the entire curve.

Analytic Functions

Analytic functions, also known as holomorphic functions, are the bread and butter of complex analysis. They are so integral to the field that their properties underpin methods like Cauchy's Integral Formula and the Residue Theorem. An analytic function is one that is differentiable at every point in a domain, meaning it has a complex derivative everywhere in that area.

Analyticity implicitly suggests smoothness and the possibility to expand the function into a power series around any point within its domain. For a function to be considered analytic on a certain contour, as mentioned in Step 1 of the solutions, it must not contain any poles or other singularities on or inside that contour. The exercise's functions in parts (a), (b), (e), and (f) exhibit such analytic behavior on the contour C, thus simplifying the integral calculations significantly.

The beauty of analytic functions lies in their predictability and the powerful tools complex analysis offers to work with them. Understanding the properties of these functions allows for swift analysis of complex integrals and greatly facilitates the resolution of complex calculus problems.