Problem 13 - Advanced Engineering Mathematics

Question

$$\begin{array}{l} \operatorname{Res}(f(z), 0)=\frac{1}{1 !} \lim _{z \rightarrow 0} \frac{d}{d z}\left[z^{2} \cdot \frac{\cos z}{z^{2}(z-\pi)^{3}}\right]=\lim _{z \rightarrow 0} \frac{-(z-\pi) \sin z-3 \cos z}{(z-\pi)^{4}}=-\frac{3}{\pi^{4}} \\\ \operatorname{Res}(f(z), \pi)=\frac{1}{2 !} \lim _{z \rightarrow \pi} \frac{d^{2}}{d z^{2}}\left[(z-\pi)^{3} \cdot \frac{\cos z}{z^{2}(z-\pi)^{3}}\right]=\frac{1}{2} \lim _{z \rightarrow \pi} \frac{-z^{2} \cos z+4 z \sin z+6 \cos z}{z^{4}}=\frac{\pi^{2}-6}{2 \pi^{4}} \end{array}$$

Step-by-Step Solution

Answer

Residue at \( z=0 \) is \( -\frac{3}{\pi^4} \), and at \( z=\pi \) it is \( \frac{\pi^2-6}{2\pi^4} \).

Step 1: Identify function form

The function given is a composition involving factors like \( z^2 \), \( \cos z \), and denominators \( z^2(z-\pi)^3 \). We are asked to find residues at points 0 and \( \pi \), indicating poles at these points.

Step 2: Finding Residue at \( z=0 \)

Residue at \( z=0 \) for function \( f(z) = \frac{\cos z}{z^2 (z-\pi)^3} \) involves first calculating the limit \( \lim_{z \to 0} \frac{-(z-\pi)\sin z - 3\cos z}{(z-\pi)^4} \). This is derived using the first derivative after extracting \( z^2 \) from numerator and denominator canceling terms.

Step 3: Calculate limit at \( z=0 \)

After simplifications, we recognize that as \( z \rightarrow 0 \), \( \cos z = 1 \) and \( \sin z \rightarrow z \). Performing the limit operation, we substitute these approximations to evaluate the limit expression leading to \( -\frac{3}{\pi^4} \).

Step 4: Finding Residue at \( z=\pi \)

For \( z=\pi \), the residue is calculated using a second derivative since it's a pole of order 3. This involves the limit \( \frac{1}{2} \lim_{z \rightarrow \pi} \frac{-z^2 \cos z + 4z \sin z + 6\cos z}{z^4} \).

Step 5: Calculate limit at \( z=\pi \)

As \( z \rightarrow \pi \), \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \), simplifying the expression. Evaluating this limit gives \( \frac{\pi^2 - 6}{2\pi^4} \).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis, particularly useful for evaluating complex integrals. It connects the residues of a function's poles to the contour integral around these poles. The essence of the theorem is that if you have a function with isolated singularities, the integral around a closed curve is determined by the sum of the residues within that curve.

• Residue: It's the coefficient of \( \frac{1}{z-a} \) in the function's Laurent series expansion around a singularity \( a \).
• Importance: This simplifies complex integral calculations because instead of computing integrals directly, observing the residues gives you the answer.
• Application: The theorem is often applied in calculating line integrals in the presence of singularities.

Understanding residues and discerning their values at given poles contributes significantly to simplifying complex integration tasks.

Poles of a Function

Poles are particular types of singularities where a function becomes unbounded. In the context of our example, the function \( f(z) = \frac{\cos z}{z^2 (z-\pi)^3} \) has poles at \( z = 0 \) and \( z = \pi \).

• Order of Poles: The order of a pole is key to determining how to find a residue. A pole of order \( n \) means \( (z-a)^n f(z) \) is analytic at \( a \) but not as \( n-1 \).
• Simple vs Higher Order: Poles can be simple (order 1) or higher. Simple poles are easier to handle as residues are directly related to limits. Higher order poles need derivatives.
• Finding Residues: For first-order poles, you get the residue by \( \lim_{z \to a} (z-a) f(z) \), while for higher-order, derivatives and limit calculations are involved.

Being able to identify and categorize poles is fundamental for calculating residues correctly.

Limit Calculations

Limit calculations play a critical role when determining residues for poles of higher order. In the given problem, this method helps in evaluating residues at the poles of higher order or when the analytic manipulation is not straightforward.

• Simplification: As with the residue at \( z = 0 \) and \( z = \pi \), simplifications often include factoring out terms and using known limits, like \( \lim_{z \to 0} \frac{\sin z}{z} = 1 \).
• Using Derivatives: When dealing with higher order poles, derivatives are essential. The limit of derivatives, as the poles tend towards the singularity, give the residue.
• Practical Significance: Limits are not only a mathematical curiosity; they ensure that calculations are handled rigorously, leading to accurate residue determination.

Familiarizing oneself with common limit forms and their calculations streamlines the process significantly.

Trigonometric Functions

Understanding how trigonometric functions behave in analytic contexts is important when working with residues and poles. In the given problem, functions like \( \cos z \) and \( \sin z \) are part of the analysis.

• Key Properties: \( \cos z \) and \( \sin z \) exhibit periodic behavior, which means they repeat values over intervals, playing a decisive role in convergence and divergence near singularities.
• Approximations: For small values of \( z \), \( \cos z \approx 1 \) and \( \sin z \approx z \). These simplifications are crucial when evaluating limits.
• Derivatives: The derivatives of trigonometric functions often simplify or alter the expressions considerably when computing limits at specific points, essential when dealing with higher order poles.

Comprehending the subtleties of trigonometric functions is necessary because their unique characteristics can heavily influence the analytic solution to functions in complex analysis.