To determine the poles of the given function $f(z)$ and evaluate the residues, let's first clearly write down the function:

$f(z) = \frac{2z}{(z-1)(z^2 - 4)}$

The poles of a function are the values of $z$ that make the denominator zero. Let's identify these values.

The denominator of $f(z)$ is:

$(z-1)(z^2 - 4)$

Set the denominator equal to zero and solve for $z$:

$(z-1)(z^2 - 4) = 0$

This equation can be factored further:

$(z-1)(z-2)(z+2) = 0$

So, the poles are at $z = 1$, $z = 2$, and $z = -2$.

Next, we need to evaluate the residues at each of these poles.

Residue at $z = 1$

For a simple pole at $z = a$, the residue is given by:

$\text{Res}(f, a) = \lim_{z \to a} (z - a) f(z)$

For $z = 1$:

$\text{Res}(f, 1) = \lim_{z \to 1} (z - 1) \frac{2z}{(z-1)(z-2)(z+2)} = \lim_{z \to 1} \frac{2z}{(z-2)(z+2)}$

Substitute $z = 1$:

$\text{Res}(f, 1) = \frac{2 \cdot 1}{(1-2)(1+2)} = \frac{2}{-1 \cdot 3} = -\frac{2}{3}$

Residue at $z = 2$

For $z = 2$:

$\text{Res}(f, 2) = \lim_{z \to 2} (z - 2) \frac{2z}{(z-1)(z-2)(z+2)} = \lim_{z \to 2} \frac{2z}{(z-1)(z+2)}$

Substitute $z = 2$:

$\text{Res}(f, 2) = \frac{2 \cdot 2}{(2-1)(2+2)} = \frac{4}{1 \cdot 4} = 1$

Residue at $z = -2$

For $z = -2$:

$\text{Res}(f, -2) = \lim_{z \to -2} (z + 2) \frac{2z}{(z-1)(z-2)(z+2)} = \lim_{z \to -2} \frac{2z}{(z-1)(z-2)}$

Substitute $z = -2$:

$\text{Res}(f, -2) = \frac{2 \cdot (-2)}{(-2-1)(-2-2)} = \frac{-4}{-3 \cdot -4} = \frac{-4}{12} = -\frac{1}{3}$

Summary

The poles of $f(z)$ are at $z = 1$, $z = 2$, and $z = -2$. The residues at these poles are:

• Residue at $z = 1$: $-\frac{2}{3}$
• Residue at $z = 2$: $1$
• Residue at $z = -2$: $-\frac{1}{3}$

Would you like more details or have any questions?

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Here are some additional related questions to consider:

1. How do you determine the order of a pole for a given function?
2. What is the significance of residues in complex analysis?
3. Can you explain the Residue Theorem and its applications?
4. How would you find the residues for higher-order poles?
5. How can the concept of residues be applied in evaluating real integrals?
6. What are the properties of meromorphic functions?
7. How do residues relate to the Laurent series expansion of a function?
8. Can you find the residues for a function with essential singularities?

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Tip: When finding residues for functions with more complicated denominators, partial fraction decomposition can often simplify the process.