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Magazine Chapter 18: Problem 12
\(\oint_{C}\left(z+\frac{1}{z^{2}}\right) d z ;|z|=2\)
Short Answer
Expert verified
The integral evaluates to \( -2 \pi i \).
Step by step solution
01
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.
02
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.
03
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.
04
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 \( 2 \pi i (-1) \).
05
Compute the Integral
Finally, we calculate the integral using the residue found: \( 2 \pi i (-1) \), which results in \( -2 \pi i \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
Contour Integration
Contour integration is a method of evaluating certain types of integrals along a path or contour in the complex plane.
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.
- 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.
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.
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.
