evaluate 1/(z-1)(z-2) is valid for the region mod(z-1)<1 using laurents series 

To evaluate the function \( \frac{1}{(z-1)(z-2)} \) using a Laurent series that is valid in the region \( |z-1| < 1 \), follow these steps:

### Step 1: Express the Function as a Sum of Partial Fractions
The given function can be decomposed into partial fractions:
\[
\frac{1}{(z-1)(z-2)} = \frac{A}{z-1} + \frac{B}{z-2}
\]
To find the constants \( A \) and \( B \), multiply both sides by \( (z-1)(z-2) \) and set the equation equal to the numerator:
\[
1 = A(z-2) + B(z-1)
\]
Expanding and equating coefficients:
\[
1 = A(z) - 2A + B(z) - B
\]
\[
1 = (A + B)z - (2A + B)
\]
From this, we equate the coefficients of \( z \) and the constant term:
\[
A + B = 0
\]
\[
-2A - B = 1
\]

### Step 2: Solve for \( A \) and \( B \)
From \( A + B = 0 \), we have \( B = -A \).

Substitute \( B = -A \) into the second equation:
\[
-2A - (-A) = 1 \quad \Rightarrow \quad -2A + A = 1 \quad \Rightarrow \quad -A = 1 \quad \Rightarrow \quad A = -1
\]
Then, \( B = 1 \).

So, the partial fraction decomposition is:
\[
\frac{1}{(z-1)(z-2)} = \frac{-1}{z-1} + \frac{1}{z-2}
\]

### Step 3: Expand Each Term in the Region \( |z-1| < 1 \)
We can now expand each term in a Laurent series. Since \( |z-1| < 1 \), we consider the series expansion around \( z = 1 \):

For the first term:
\[
\frac{-1}{z-1}
\]
This is already in a simple form and is the first term of the Laurent series.

For the second term \( \frac{1}{z-2} \), rewrite it by factoring out \( (z-1) \):
\[
\frac{1}{z-2} = \frac{1}{(z-1) - 1}
\]
\[
= \frac{1}{-(1 - (z-1))} = -\frac{1}{1-(z-1)}
\]
Expand this using the geometric series for \( |z-1| < 1 \):
\[
-\frac{1}{1-(z-1)} = -\sum_{n=0}^{\infty} (z-1)^n
\]
This is valid for \( |z-1| < 1 \).

### Step 4: Combine the Series
Now combine the two series:
\[
\frac{1}{(z-1)(z-2)} = -\frac{1}{z-1} - \sum_{n=0}^{\infty} (z-1)^n
\]
This is the Laurent series expansion of the function \( \frac{1}{(z-1)(z-2)} \) valid in the region \( |z-1| < 1 \).

### Final Answer
\[
\frac{1}{(z-1)(z-2)} = -\frac{1}{z-1} - 1 - (z-1) - (z-1)^2 - (z-1)^3 - \cdots
\]

Would you like more details or have any questions?

---

1. How would the Laurent series change for a different region, such as \( |z-2| < 1 \)?
2. What are the general steps to perform partial fraction decomposition?
3. How can the geometric series expansion be generalized for other functions?
4. What are the conditions for the convergence of the Laurent series in this region?
5. How would the series expansion look if the region were \( 1 < |z-1| < |z-2| \)?

**Tip:** When working with Laurent series, always carefully consider the region of convergence to determine the correct expansion.

Learn how to evaluate the function 1/(z-1)(z-2) using a Laurent series valid in the region |z-1| < 1. This problem involves complex analysis techniques, including partial fraction decomposition and geometric series expansions. Suitable for advanced undergraduate students.

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