Question

$$ \text { Berechnen Sie die Residuen der Funktion } f(z)=\frac{\sin z}{z\left(z^{2}-1\right)} \text {. } $$

Short answer

The residues are 0 at \( z=0 \), \( \frac{\sin 1}{2} \) at \( z=1 \), and \( -\frac{\sin 1}{2} \) at \( z=-1 \).

Step-by-step solution

Identify Singularities

The function is given by \( f(z) = \frac{\sin z}{z(z^2 - 1)} \). To find the singularities, we need to determine where the function is undefined. The singularities occur at the zeros of the denominator, which are \( z = 0 \), \( z = 1 \), and \( z = -1 \). These are the complex points where the denominator becomes zero.

Determine the Order of Singularities

The singularity at \( z = 0 \) occurs because of the term \( z \) in the denominator, making it a simple pole. At \( z = 1 \) and \( z = -1 \), the singularity is due to the terms \( z^2 - 1 = (z-1)(z+1) \), each contributing a simple pole. Therefore, \( z = 0, 1, -1 \) are all simple poles.

Calculate the Residue at \( z = 0 \)

For a simple pole at \( z = 0 \), the residue is found by taking the limit: \[ \text{Res}(f,z=0) = \lim_{z \to 0} z \cdot f(z) = \lim_{z \to 0} \frac{\sin z}{z^2 - 1}. \] Simplifying using \( \sin z \approx z \) near zero, we have: \( \lim_{z \to 0} \frac{z}{z^2-1} = \frac{0}{-1} = 0 \), so the residue at \( z = 0 \) is 0.

Calculate the Residue at \( z = 1 \)

For the simple pole at \( z = 1 \), the residue is: \[ \text{Res}(f,z=1) = \lim_{z \to 1} (z-1) \cdot \frac{\sin z}{z(z+1)}. \] Evaluating this limit: \( \lim_{z \to 1} \frac{\sin z}{z(z+1)} = \frac{\sin 1}{1 \cdot 2} = \frac{\sin 1}{2}. \) The residue at \( z = 1 \) is \( \frac{\sin 1}{2} \).

Calculate the Residue at \( z = -1 \)

For the simple pole at \( z = -1 \), the residue is: \[ \text{Res}(f,z=-1) = \lim_{z \to -1} (z+1) \cdot \frac{\sin z}{z(z-1)}. \] This limit evaluates to: \( \lim_{z \to -1} \frac{\sin z}{z(z-1)} = \frac{\sin(-1)}{-1 \cdot (-2)} = \frac{-\sin 1}{2}. \) Thus, the residue at \( z = -1 \) is \( -\frac{\sin 1}{2} \).

Key concepts

Complex Analysis

Complex Analysis is a branch of mathematics that deals with functions of a complex variable. These functions are differentiable and integrate over the complex plane, which gives rise to intricate and beautiful theorems. One of the main tools used in complex analysis is contour integration, which involves integrating a function along a specified path in the complex plane.

The function in our exercise, \( f(z) = \frac{\sin z}{z(z^2 - 1)} \), is defined over the complex plane except at points where the denominator is zero. Understanding how to handle these undefined points is key in complex analysis, as they significantly influence the behavior of the function.

• Complex functions can be continuous, differentiable, and integrable just like real functions, but they reside in a two-dimensional space when considering both real and imaginary components.
• Residue Theorem is a powerful tool within complex analysis, allowing us to evaluate integrals over complex paths by considering the residues of singularities inside the path.

Mastering the principles of complex analysis provides a greater appreciation for the dynamics of functions in different planes, and is foundational for deeper mathematical studies.

Singularities

Singularities are points at which a complex function is not defined or does not have a well-behaved limit. These points are crucial because they dictate many properties of the function, such as where it might blow up or become infinite.

In the function \( f(z) = \frac{\sin z}{z(z^2 - 1)} \), singularities occur where the denominator equals zero, namely at \( z = 0 \), \( z = 1 \), and \( z = -1 \), which are termed as points of failure in this context. Recognizing these singularities is often one of the first steps in performing complex analysis.

• Singularities can be categorized into removable, poles, or essential singularities, and they all require different treatment.
• Poles, especially, are a common type of singularity where the function goes to infinity.

Identifying and classifying these singularities is essential for executing residue calculations, as it impacts how we approach the overall solution.

Simple Poles

Simple poles are a specific type of singularity where the function behaves like \( \frac{1}{z-a} \) near the point \( z = a \). They are characterized by the function having a first-order zero in the denominator and no zero in the numerator at that point.

For our function, \( z = 0 \), \( z = 1 \), and \( z = -1 \) are all simple poles. That means, at these points, the function becomes undefined because it has a zero of degree one in the denominator and none in the numerator. Addressing these requires special residue calculations.

• Simple poles imply that the residue can be found using the limit formula: \( \text{Res}(f,z=a) = \lim_{z \to a} (z-a)f(z) \).
• This property simplifies calculations as compared to higher-order poles, where more elaborate derivations might be required.

Identification and accurate handling of simple poles are vital for correctly applying the Residue Theorem.

Limit Calculations

Limit calculations often come into play when dealing with poles and residues in complex analysis. They allow us to evaluate the behavior of a complex function as it approaches singularities.

In our exercise, calculating the residues at simple poles involves using limits to determine the finite value that the function approaches at points like \( z = 0 \), \( z = 1 \), and \( z = -1 \). For instance, to find the residue at \( z = 0 \), the limit \( \lim_{z \to 0} z \cdot f(z) \) simplifies to \( \frac{0}{-1} = 0 \), indicating a zero residue.

• The calculation of residues at simple poles often leverages straightforward limits, as these poles contribute a linear factor to the form of the complex function.
• Understanding the underlying behavior of \( \sin z \) and other complex expressions close to singular points is crucial for simplifying limit operations.

Performing accurate limit calculations ensures that you identify correct numerical residues, which are vital for applications of the Residue Theorem in evaluating contour integrals.