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To find the Laurent series expansion of the given function \( f(z) \) in the specified annulus \( 1 < |z| < 2 \), we will proceed step by step.

### Problem Statement
The function is given by:
\[ f(z) = \frac{1}{(z - 2)(z + 4)} \]
and we need to find its Laurent series expansion in the region \( 1 < |z| < 2 \).

### Steps to Solve:

1. **Identify Singularities**: Determine the points where the function \( f(z) \) is not analytic.
2. **Decompose the Function**: Express the function in a form suitable for series expansion.
3. **Expand Each Term**: Find the series expansion for each term in the region \( 1 < |z| < 2 \).
4. **Combine Series**: Sum the series to obtain the Laurent series for \( f(z) \).

#### 1. Identify Singularities

The function \( f(z) \) has singularities where the denominator is zero. Thus, we solve:
\[
(z - 2)(z + 4) = 0
\]
This gives the singularities at \( z = 2 \) and \( z = -4 \).

#### 2. Decompose the Function

We use partial fraction decomposition to express \( f(z) \) as:
\[
f(z) = \frac{1}{(z - 2)(z + 4)} = \frac{A}{z - 2} + \frac{B}{z + 4}
\]

To find \( A \) and \( B \), we multiply both sides by \( (z - 2)(z + 4) \) and equate coefficients:
\[
1 = A(z + 4) + B(z - 2)
\]

Plugging in suitable values of \( z \), we get:
- For \( z = 2 \):
  \[ 1 = A(2 + 4) + B(0) \]
  \[ A = \frac{1}{6} \]

- For \( z = -4 \):
  \[ 1 = A(0) + B(-4 - 2) \]
  \[ B = -\frac{1}{6} \]

Thus:
\[
f(z) = \frac{1/6}{z - 2} - \frac{1/6}{z + 4}
\]

#### 3. Expand Each Term

Now, we expand each term as a Laurent series in the region \( 1 < |z| < 2 \):

- For \( \frac{1}{z - 2} \), since \( |z| < 2 \), we can write:
  \[ \frac{1}{z - 2} = -\frac{1}{2}\left(\frac{1}{1 - \frac{z}{2}}\right) = -\frac{1}{2} \sum_{n=0}^\infty \left(\frac{z}{2}\right)^n = -\sum_{n=0}^\infty \left(\frac{z}{2}\right)^n \]

- For \( \frac{1}{z + 4} \), since \( |z| > 1 \), we can write:
  \[ \frac{1}{z + 4} = \frac{1}{4}\left(\frac{1}{1 + \frac{z}{4}}\right) = \frac{1}{4} \sum_{n=0}^\infty \left(-\frac{z}{4}\right)^n = \sum_{n=0}^\infty (-1)^n \left(\frac{z}{4}\right)^n \]

Thus:
\[
f(z) = \frac{1}{6}\left[-\sum_{n=0}^\infty \left(\frac{z}{2}\right)^n\right] - \frac{1}{6}\left[\sum_{n=0}^\infty (-1)^n \left(\frac{z}{4}\right)^n\right]
\]

#### 4. Combine Series

Combining the series, we get:
\[
f(z) = -\frac{1}{6}\sum_{n=0}^\infty \left(\frac{z}{2}\right)^n - \frac{1}{6}\sum_{n=0}^\infty (-1)^n \left(\frac{z}{4}\right)^n
\]

This series is the Laurent series expansion of \( f(z) \) in the region \( 1 < |z| < 2 \).

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### Relative Questions:
1. What is a Laurent series and how does it differ from a Taylor series?
2. What are the conditions for the existence of a Laurent series expansion?
3. Why do we consider the annulus \( 1 < |z| < 2 \) for the series expansion?
4. How can singularities affect the convergence of a Laurent series?
5. What is the role of partial fraction decomposition in finding Laurent series?
6. How can one determine the radius of convergence for each series in the Laurent expansion?
7. Can the Laurent series have both positive and negative powers of \( z \)?
8. How is the Laurent series useful in complex analysis, especially in residue theory?

**Tip:** When dealing with series expansions in complex analysis, always check the region of convergence for the validity of the series.

Learn how to find the Laurent series expansion of f(z) = 1 / ((z - 2)(z + 4)) in the annulus 1 < |z| < 2. Detailed step-by-step solution covering complex analysis concepts, including Laurent series, partial fraction decomposition, and series expansion. Suitable for advanced undergraduate students.

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