I.* Determine all the isolated singularities of each of the following functions and compute the residue at each singularity. * z-1 sin z
I.* Determine all the isolated singularities of each of the following functions and compute the residue at each singularity. * z-1 sin z
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Isolated singularities and residue
![### Complex Analysis: Singularity and Residue Calculations
#### Task 1:
**Objective:** Determine all the isolated singularities of each of the following functions and compute the residue at each singularity.
Consider the function:
\[ \frac{z - 1}{\sin z} \]
**Steps to Solve:**
1. **Identify Isolated Singularities:**
- An isolated singularity is a point where a function is not analytic, but analytic at some deleted neighborhood of the point.
- For the function \(\frac{z - 1}{\sin z}\), singularities occur when the denominator \(\sin z\) is zero because division by zero is not defined.
2. **Zeros of \(\sin z\):**
- The sine function \(\sin z\) is zero at \(z = n\pi\) for \(n \in \mathbb{Z}\) (the set of all integers). These points are \(0, \pm\pi, \pm2\pi, \ldots\).
3. **Compute Residues:**
- The residue at a singularity \(z=a\) is a coefficient \(A_{-1}\) in the Laurent series expansion of \(f(z)\) around \(a\), where \(f(z)\) can be expressed as:
\[ f(z) = \frac{g(z)}{h(z)} \]
for \(g(z) = z - 1\) and \(h(z) = \sin z\).
- **Use the formula:**
\[ \text{Res}(f, a) = \frac{g(a)}{h'(a)} \]
4. **Specific Residue Calculation:**
- For \( f(z) = \frac{z - 1}{\sin z}\):
- At \(z = n\pi, n \neq 0\):
\[ \text{Res}\left(\frac{z - 1}{\sin z}, n\pi\right) = \frac{n\pi - 1}{(-1)^{n} \cdot \cos(n\pi)} \]
Since \(\cos(n\pi) = (-1)^n\), this simplifies to:
\[ \text{Res}\left(\frac{z - 1}{\sin z}, n\pi\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7fd47556-f3ce-4f39-818d-be563d9523c8%2F3e8c9db2-98e7-43d8-94a9-66e5bd174dee%2Fl1d0z3e_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Complex Analysis: Singularity and Residue Calculations
#### Task 1:
**Objective:** Determine all the isolated singularities of each of the following functions and compute the residue at each singularity.
Consider the function:
\[ \frac{z - 1}{\sin z} \]
**Steps to Solve:**
1. **Identify Isolated Singularities:**
- An isolated singularity is a point where a function is not analytic, but analytic at some deleted neighborhood of the point.
- For the function \(\frac{z - 1}{\sin z}\), singularities occur when the denominator \(\sin z\) is zero because division by zero is not defined.
2. **Zeros of \(\sin z\):**
- The sine function \(\sin z\) is zero at \(z = n\pi\) for \(n \in \mathbb{Z}\) (the set of all integers). These points are \(0, \pm\pi, \pm2\pi, \ldots\).
3. **Compute Residues:**
- The residue at a singularity \(z=a\) is a coefficient \(A_{-1}\) in the Laurent series expansion of \(f(z)\) around \(a\), where \(f(z)\) can be expressed as:
\[ f(z) = \frac{g(z)}{h(z)} \]
for \(g(z) = z - 1\) and \(h(z) = \sin z\).
- **Use the formula:**
\[ \text{Res}(f, a) = \frac{g(a)}{h'(a)} \]
4. **Specific Residue Calculation:**
- For \( f(z) = \frac{z - 1}{\sin z}\):
- At \(z = n\pi, n \neq 0\):
\[ \text{Res}\left(\frac{z - 1}{\sin z}, n\pi\right) = \frac{n\pi - 1}{(-1)^{n} \cdot \cos(n\pi)} \]
Since \(\cos(n\pi) = (-1)^n\), this simplifies to:
\[ \text{Res}\left(\frac{z - 1}{\sin z}, n\pi\
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