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To solve the problem of determining the poles of the function \( f(z) = \frac{z^2 + 1}{(z+1)^2 (z-2)} \) and evaluating the residues at these poles, we need to follow these steps:

### Step 1: Identify the Poles
Poles of a function are the values of \( z \) where the function becomes unbounded (i.e., where the denominator of the fraction is zero, provided the numerator is not zero at those points).

The denominator of the given function is \((z+1)^2 (z-2)\).

Set the denominator equal to zero to find the poles:
\[ (z+1)^2 (z-2) = 0 \]

This equation gives us the poles:
1. \( (z+1)^2 = 0 \) ⟹ \( z = -1 \) (double pole)
2. \( z-2 = 0 \) ⟹ \( z = 2 \) (simple pole)

### Step 2: Evaluate the Residues
Residues at poles can be evaluated using different methods depending on whether the pole is simple or of higher order.

#### Residue at \( z = -1 \) (Double Pole)
For a pole of order \( m \), the residue can be computed using the formula:
\[ \text{Res}(f, z_0) = \frac{1}{(m-1)!} \lim_{z \to z_0} \frac{d^{m-1}}{dz^{m-1}} \left[ (z - z_0)^m f(z) \right] \]

Here, \( z = -1 \) is a pole of order 2 (\( m = 2 \)):

\[ \text{Res}(f, -1) = \lim_{z \to -1} \frac{d}{dz} \left[ (z + 1)^2 \frac{z^2 + 1}{(z+1)^2 (z-2)} \right] \]
\[ = \lim_{z \to -1} \frac{d}{dz} \left[ \frac{z^2 + 1}{z-2} \right] \]
\[ = \lim_{z \to -1} \frac{(2z)(z-2) - (z^2 + 1)}{(z-2)^2} \]
\[ = \lim_{z \to -1} \frac{2z(z-2) - (z^2 + 1)}{(z-2)^2} \]
\[ = \lim_{z \to -1} \frac{2z^2 - 4z - z^2 - 1}{(z-2)^2} \]
\[ = \lim_{z \to -1} \frac{z^2 - 4z - 1}{(z-2)^2} \]
\[ = \frac{(-1)^2 - 4(-1) - 1}{(-3)^2} = \frac{1 + 4 - 1}{9} = \frac{4}{9} \]

Therefore,
\[ \text{Res}(f, -1) = \frac{5}{9} \]

#### Residue at \( z = 2 \) (Simple Pole)
For a simple pole, the residue can be computed directly using:
\[ \text{Res}(f, z_0) = \lim_{z \to z_0} (z - z_0) f(z) \]

Here, \( z = 2 \) is a simple pole:

\[ \text{Res}(f, 2) = \lim_{z \to 2} (z-2) \frac{z^2 + 1}{(z+1)^2 (z-2)} \]
\[ = \lim_{z \to 2} \frac{z^2 + 1}{(z+1)^2} \]
\[ = \frac{2^2 + 1}{(2+1)^2} \]
\[ = \frac{4 + 1}{9} \]
\[ = \frac{5}{9} \]

### Conclusion
The poles of the function \( f(z) = \frac{z^2 + 1}{(z+1)^2 (z-2)} \) are at \( z = -1 \) (double pole) and \( z = 2 \) (simple pole). The residues at these poles are:
- Residue at \( z = -1 \): \( \frac{5}{9} \)
- Residue at \( z = 2 \): \( \frac{5}{9} \)

Would you like more details or have any questions?

### Further Related Questions
1. What is the significance of residues in complex analysis?
2. How do you identify higher-order poles in a given function?
3. Can you provide an example of a function with a pole of order 3 and find its residue?
4. What is the difference between simple and higher-order poles?
5. How do Laurent series help in understanding residues?
6. Can residues be negative? Provide an example.
7. What are the applications of residue theory in physics and engineering?
8. How does the residue theorem simplify the evaluation of certain integrals?

### Tip
When dealing with higher-order poles, always remember to carefully apply the differentiation formula for residues to ensure accuracy.

Learn how to determine the poles of the function f(z) = (z^2 + 1) / ((z+1)^2 (z-2)) and evaluate the residues at these poles with detailed step-by-step explanations. This problem covers complex analysis concepts, including residue theory and pole identification. Suitable for advanced undergraduate students.

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