Problem 4
Question
Welche Werte kann das Kurvenintegral \(\oint_{K} \frac{1}{z\left(z^{2}-1\right)} d z\) haben, wenn \(K\) eine stückweise glatte, geschlossene und doppelpunktfreie Kurve ist, die nicht durch die Punkte 0,1 und \(-1\) läuft ?
Step-by-Step Solution
Verified Answer
Curve K contains0, 1, or -1; then it’s \ \ theoretical residue principal ofcalculated residues from originalresidues.
1Step 1: Identify singularities
Recognize that the function \(\frac{1}{z(z^{2} - 1)} \) has singularities at the points where the denominator is zero. This occurs at z=0, z=1, and z=-1.
2Step 2: Determine curve properties
Given that K is a piecewise smooth, closed, and simple curve that does not pass through the points 0, 1, and -1.
3Step 3: Apply Cauchy's Residue Theorem
Cauchy's Residue Theorem states that \(\frac{1}{2\pi \iota} \oint_{K} \frac{f(z)}{z-z_{0}}dz = n \) where n is the number of poles inside K. The function has poles of order 1 at z=0, z=1, and z=-1.
4Step 4: Calculate residues
The residues are computed as follows: At z=0, Res=f(z)=-1, At z=1, Res=f(z)=1/2, At z=-1, Res=f(z)=-1/2
5Step 5: Sum residues inside K
Add the residues of the poles that lie inside the curve K=
6Step 6: Combine results
Identify whether individual residues should be summed; if all are within z, then all residues should be summed.
Key Concepts
Cauchy's Residue TheoremSingularities in Complex FunctionsContour Integration
Cauchy's Residue Theorem
Cauchy's Residue Theorem is a powerful tool in complex analysis. It allows us to evaluate complex line integrals using residues. Residues are essentially the coefficients of the \(\frac{1}{z}\) term in a Laurent series expansion of a complex function at a singularity.
This theorem states that if we have a function \(f(z)\) which is analytic within and on a simple closed contour K, except at a finite number of singularities inside K, then:
\[\frac{1}{2ullptrpi \rightarrow} \rightöint_{K} f(z) dz = \text {Sum of residues of \(f\) at singularities}\right \] This simplifies the integration process significantly since evaluating residues is easier than directly computing the integrals.
To apply this theorem, we follow these steps:
This theorem states that if we have a function \(f(z)\) which is analytic within and on a simple closed contour K, except at a finite number of singularities inside K, then:
\[\frac{1}{2ullptrpi \rightarrow} \rightöint_{K} f(z) dz = \text {Sum of residues of \(f\) at singularities}\right \] This simplifies the integration process significantly since evaluating residues is easier than directly computing the integrals.
To apply this theorem, we follow these steps:
- Identify the singularities of the function.
- Determine if these singularities lie within the contour K.
- Calculate the residues at these singularities.
- Sum these residues to find the value of the integral.
Singularities in Complex Functions
Singularities are points where a complex function becomes unbounded or undefined. These points are crucial since they dictate the behavior of the function nearby.
There are different types of singularities:
There are different types of singularities:
- Poles: Points where the function goes to infinity. For instance, a pole of order n at z=\( z_0 \) means near \( z_0 \), f(z) behaves like \(\frac{1}{(z-z_0)^n}\).
- Essential Singularities: Points where the function exhibits chaotic behavior. Near these points, the function does not have a power series expansion.
- Removable Singularities: Points where the function is undefined but can be defined in a way that makes the function analytic.
Contour Integration
Contour integration is the process of integrating a complex function along a specified path or contour in the complex plane. Here’s a structured way to grasp it:
In the original exercise, given the curve K is piecewise smooth and closed, and does not pass through singularities (0, 1, -1), you can apply Cauchy's Residue Theorem to find the integral's value.
- Piecewise Smooth Contour: A contour that can be decomposed into a finite number of smooth segments.
- Closed Contour: A path that starts and ends at the same point.
In the original exercise, given the curve K is piecewise smooth and closed, and does not pass through singularities (0, 1, -1), you can apply Cauchy's Residue Theorem to find the integral's value.