22 -3z + 2 5. f(z) = |3D (z 1)2(z - 3)3
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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![### Complex Analysis: Identifying Poles and Residues
#### Problem Set Instructions:
For Problems 4-6, find the poles for the indicated function. Identify the order of each pole and the residue of the function at the pole.
#### Problem 4:
\[ f(z) = \frac{\sin z}{z(2z - \pi)} \]
#### Problem 5:
\[ f(z) = \frac{z^2 - 3z + 2}{(z - 1)^2 (z - 3)^3} \]
#### Problem 6:
\[ f(z) = \frac{z + 1}{z^2 + 9} \]
### Solution Guidelines:
1. **Identify the poles** by determining the values of \( z \) that make the denominator of the function zero.
2. **Determine the order** of each pole by the power to which the term corresponding to the pole is raised in the denominator.
3. **Find the residue** at each pole using appropriate residue formulas for simple and higher-order poles.
#### Example Steps for Problem 4:
1. **Identify Poles:**
- Denominator: \( z(2z - \pi) = 0 \)
- Poles: \( z = 0 \) and \( z = \frac{\pi}{2} \)
2. **Determine Order:**
- Both poles \( z = 0 \) and \( z = \frac{\pi}{2} \) are simple poles (order 1).
3. **Find Residue:**
- Use the formula for residue at simple pole \( z = z_0 \):
\[
\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)
\]
- For \( z = 0 \):
\[
\text{Res}(f, 0) = \lim_{z \to 0} \frac{\sin z}{2z - \pi} = \frac{\sin 0}{2 \cdot 0 - \pi} = 0
\]
- For \( z = \frac{\pi}{2} \):
\[
\text{Res}(f, \frac{\pi}{2}) = \lim_{z \to \frac{\](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F4f812bf4-4e37-4075-b786-04672d25a0a0%2F9bfef485-ce88-4f08-bfa4-f57dee66e606%2Fwaioe0d.jpeg&w=3840&q=75)
Transcribed Image Text:### Complex Analysis: Identifying Poles and Residues
#### Problem Set Instructions:
For Problems 4-6, find the poles for the indicated function. Identify the order of each pole and the residue of the function at the pole.
#### Problem 4:
\[ f(z) = \frac{\sin z}{z(2z - \pi)} \]
#### Problem 5:
\[ f(z) = \frac{z^2 - 3z + 2}{(z - 1)^2 (z - 3)^3} \]
#### Problem 6:
\[ f(z) = \frac{z + 1}{z^2 + 9} \]
### Solution Guidelines:
1. **Identify the poles** by determining the values of \( z \) that make the denominator of the function zero.
2. **Determine the order** of each pole by the power to which the term corresponding to the pole is raised in the denominator.
3. **Find the residue** at each pole using appropriate residue formulas for simple and higher-order poles.
#### Example Steps for Problem 4:
1. **Identify Poles:**
- Denominator: \( z(2z - \pi) = 0 \)
- Poles: \( z = 0 \) and \( z = \frac{\pi}{2} \)
2. **Determine Order:**
- Both poles \( z = 0 \) and \( z = \frac{\pi}{2} \) are simple poles (order 1).
3. **Find Residue:**
- Use the formula for residue at simple pole \( z = z_0 \):
\[
\text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0)f(z)
\]
- For \( z = 0 \):
\[
\text{Res}(f, 0) = \lim_{z \to 0} \frac{\sin z}{2z - \pi} = \frac{\sin 0}{2 \cdot 0 - \pi} = 0
\]
- For \( z = \frac{\pi}{2} \):
\[
\text{Res}(f, \frac{\pi}{2}) = \lim_{z \to \frac{\
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