https://file.aiquickdraw.com/mathbot/file/1723083662749wv7qj71x.jpg

To determine the poles and their orders for the given function \( f(z) \), and to evaluate the residues at those points, we proceed as follows:

The function is given by:
\[ f(z) = \frac{z-1}{(z+1)^2 (z-2)} \]

### 1. Finding the Poles and Their Orders
Poles are the values of \( z \) for which the denominator of \( f(z) \) is zero, provided that the numerator does not also become zero at those points.

The denominator of \( f(z) \) is \( (z+1)^2 (z-2) \). Setting this equal to zero gives us:
\[ (z+1)^2 (z-2) = 0 \]

So, the poles are at:
\[ z = -1 \quad \text{and} \quad z = 2 \]

#### Order of the Poles:
- At \( z = -1 \): The factor \((z+1)^2\) indicates a pole of order 2.
- At \( z = 2 \): The factor \((z-2)\) indicates a pole of order 1.

### 2. Evaluating the Residues
The residue of a function \( f(z) \) at a pole \( z_0 \) is given by:
- For a simple pole (order 1): \(\text{Residue} = \lim_{z \to z_0} (z - z_0) f(z)\)
- For a pole of order \( m \): \(\text{Residue} = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right]\)

#### Residue at \( z = -1 \) (pole of order 2):
\[ \text{Residue} = \frac{1}{1!} \lim_{z \to -1} \frac{d}{dz} \left[ (z + 1)^2 f(z) \right] \]
\[ (z + 1)^2 f(z) = \frac{(z-1)}{z-2} \]
\[ \frac{d}{dz} \left[ \frac{(z-1)}{z-2} \right] = \frac{(z-2) \cdot 1 - (z-1) \cdot 1}{(z-2)^2} = \frac{(z-2) - (z-1)}{(z-2)^2} = \frac{-1}{(z-2)^2} \]
Evaluating this at \( z = -1 \):
\[ \text{Residue} = \frac{1}{1!} \left. \left( \frac{-1}{(z-2)^2} \right) \right|_{z = -1} = \frac{-1}{(-1-2)^2} = \frac{-1}{9} = -\frac{1}{9} \]

#### Residue at \( z = 2 \) (pole of order 1):
\[ \text{Residue} = \lim_{z \to 2} (z - 2) f(z) \]
\[ (z - 2) f(z) = \frac{(z-2)(z-1)}{(z+1)^2 (z-2)} = \frac{(z-1)}{(z+1)^2} \]
Evaluating this at \( z = 2 \):
\[ \text{Residue} = \left. \left( \frac{z-1}{(z+1)^2} \right) \right|_{z = 2} = \frac{2-1}{(2+1)^2} = \frac{1}{9} \]

### Summary
- The function \( f(z) \) has poles at \( z = -1 \) and \( z = 2 \).
  - The pole at \( z = -1 \) is of order 2.
  - The pole at \( z = 2 \) is of order 1.
- The residues are:
  - Residue at \( z = -1 \) is \( -\frac{1}{9} \).
  - Residue at \( z = 2 \) is \( \frac{1}{9} \).

Would you like any further details or have any other questions?

### Related Questions
1. How do you find the poles of a rational function?
2. What is the definition of a residue in complex analysis?
3. How do you determine the order of a pole?
4. What is the significance of a residue at a pole?
5. Can a function have an essential singularity? How is it different from a pole?
6. How do you compute residues for higher-order poles?
7. What are Laurent series and how do they relate to residues?
8. How can residues be used to evaluate complex integrals?

### Tip
Understanding how to compute residues and the behavior of poles is essential for evaluating complex integrals using the residue theorem.

Explore how to determine poles, evaluate residues, and understand their significance in complex analysis with detailed examples. Learn about the function f(z) and its poles at z = -1 and z = 2, including the order of each pole and their corresponding residues. Essential for students and practitioners of complex analysis.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation