Problem 12 - Advanced Engineering Mathematics

Question

\(\oint_{C}\left(z+\frac{1}{z^{2}}\right) d z ;|z|=2\)

Step-by-Step Solution

Answer

The integral evaluates to \( -2 \pi i \).

Step 1: Understand the Problem

We are asked to evaluate the integral \( \oint_{C}\left(z+\frac{1}{z^{2}}\right) d z \) where \( C \) is the contour defined by \( |z|=2 \). This is a closed contour integral in the complex plane, and we will use the residue theorem to solve it.

Step 2: Identify Singular Points

Identify the singularities of the function \( f(z) = z + \frac{1}{z^2} \). The term \( \frac{1}{z^2} \) has a pole of order 2 at \( z = 0 \). Since \( \frac{1}{z^2} \) is the only term with singularities and they lie inside the contour \( |z| = 2 \), we only have to consider this pole.

Step 3: Compute the Residue at the Pole

Since \( \frac{1}{z^2} \) has a pole of order 2 at \( z = 0 \), we can find the residue by considering the limit \( \lim_{z \to 0} \frac{d}{dz}((z-0)^2 f(z)) \). Calculating this gives the residue as -1.

Step 4: Apply the Residue Theorem

According to the residue theorem, the contour integral \( \oint_{C} f(z) \, dz \) over a closed curve \( C \) is \( 2 \pi i \) times the sum of residues inside \( C \). Since the residue at \( z = 0 \) is -1, the integral \( \oint_{C} (z + \frac{1}{z^2}) \, dz \) is \( 2 \pi i (-1) \).

Step 5: Compute the Integral

Finally, we calculate the integral using the residue found: \( 2 \pi i (-1) \), which results in \( -2 \pi i \).

Key Concepts

Residue Theorem

The Residue Theorem is a powerful tool in complex analysis. It simplifies the evaluation of complex integrals over closed contours in the complex plane. The theorem states that:

• The integral of a function around a closed contour is directly related to the residues of the function inside that contour.
• The integral is equal to \(2 \pi i\) times the sum of all residues within the contour.

Residues are like the special fingerprints of functions at points called singularities. By knowing these residues, we can easily calculate complicated integrals without evaluating the entire contour directly. This process is especially useful in evaluating integrals in complex domains where direct computation is impractical.

Contour Integration

Contour integration is a method of evaluating certain types of integrals along a path or contour in the complex plane.

• Instead of a straight line, the path is often a loop, like a circle or rectangle, which closes back on itself.
• These integrals are expressed in the form \( \oint_{C} f(z) \, dz \), where \(C\) is the path.

One significant advantage of contour integration is the ability to reduce complex areas of integration to simpler paths. This allows solutions that leverage the properties of holomorphic functions, which are nicely behaved.
For example, if a function is analytic everywhere inside a contour except at isolated singularities, contour integration and the Residue Theorem can be particularly beneficial.

Singularities

In complex analysis, singularities are points where a function does not behave nicely, or is not well-defined.

• These are the spots where the function can become infinite or undetermined.
• Common types of singularities include poles and essential singularities.

Identifying singularities is crucial before applying the residue theorem, as we must locate these points to determine where residues occur. For instance, in the exercise, the function \(\frac{1}{z^2}\) has a singularity at \(z = 0\). Recognizing such points is the first step in effectively using advanced techniques like the residue theorem.

Poles

Poles are a specific kind of singularity where a function goes to infinity at a certain rate. They are commonly encountered in complex analysis and are defined as such:

• A pole is an isolated singularity where a function can be expressed in the form \( (z-a)^{-n} \) for \( n > 0 \).
• A simple pole is when \( n = 1 \); if \( n > 1 \), the pole is called higher order.

In the context of the given exercise, we deal with a pole of order 2 at \( z = 0 \). In such cases, evaluating the residue involves a derivative calculation, making the process slightly more involved than with simple poles. Understanding poles and their properties is key to finding residues, and thus solving complex integrals effectively.