Evaluate 1/(z-1)(z-2) valid for the region mod(z-1)<1 using Lauren's series

To evaluate the function \(\frac{1}{(z-1)(z-2)}\) in the region where \(|z-1| < 1\), we can expand it into a Laurent series around \(z = 1\).

### Step-by-Step Process:

1. **Rewrite the function:**
   \[
   \frac{1}{(z-1)(z-2)} = \frac{1}{z-2} \cdot \frac{1}{z-1}
   \]
   We know that we need to expand this in a Laurent series around \(z = 1\). 

2. **Expand \(\frac{1}{z-2}\) around \(z = 1\):**
   \[
   \frac{1}{z-2} = \frac{1}{(z-1) - 1} = -\frac{1}{1 - (z-1)} 
   \]
   Using the geometric series expansion for \(\frac{1}{1-x}\), where \(|x| < 1\):
   \[
   -\frac{1}{1 - (z-1)} = -\sum_{n=0}^{\infty} (z-1)^n
   \]
   Thus:
   \[
   \frac{1}{z-2} = -\sum_{n=0}^{\infty} (z-1)^n
   \]

3. **Multiply by \(\frac{1}{z-1}\):**
   Now, multiply the expansion of \(\frac{1}{z-2}\) by \(\frac{1}{z-1}\):
   \[
   \frac{1}{(z-1)(z-2)} = \left(-\sum_{n=0}^{\infty} (z-1)^n\right) \cdot \frac{1}{z-1}
   \]
   This can be simplified to:
   \[
   \frac{1}{(z-1)(z-2)} = -\sum_{n=0}^{\infty} (z-1)^{n-1}
   \]
   Which means we get:
   \[
   \frac{1}{(z-1)(z-2)} = -\left(\frac{1}{z-1} + 1 + (z-1) + (z-1)^2 + \dots\right)
   \]
   
   Simplifying further:
   \[
   \frac{1}{(z-1)(z-2)} = -\sum_{n=-1}^{\infty} (z-1)^n
   \]

### Final Laurent Series:
The Laurent series expansion of \(\frac{1}{(z-1)(z-2)}\) around \(z = 1\) valid for the region \(|z-1| < 1\) is:
\[
-\frac{1}{z-1} - 1 - (z-1) - (z-1)^2 - \dots
\]
or more formally:
\[
\boxed{\frac{1}{(z-1)(z-2)} = -\sum_{n=-1}^{\infty} (z-1)^n \text{ for } |z-1| < 1}
\]

Would you like to delve into any specific step or explore more on Laurent series?

### Further Exploration:
1. What is the Laurent series expansion for \(\frac{1}{(z-2)(z-3)}\) in the region \(|z-2| < 1\)?
2. How does the region of convergence change for the Laurent series if we shift the center to \(z=2\)?
3. Can you derive the Laurent series for \(\frac{z+1}{(z-1)(z-2)}\) in \(|z-1| < 1\)?
4. What is the difference between a Taylor series and a Laurent series?
5. How would you find the residue of the function \(\frac{1}{(z-1)(z-2)}\) at \(z=1\)?

### Tip:
Remember, the Laurent series can contain both positive and negative powers, making it extremely useful for representing functions with singularities inside the region of convergence.

Learn how to evaluate the Laurent series expansion for the function 1/((z-1)(z-2)) valid in the region |z-1| < 1. This problem covers complex analysis concepts including Laurent series, suitable for advanced undergraduate students and beyond.

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