The problem is asking to find the Laurent series expansion of the function

$\frac{1}{1 - z^2}$

about $z = 0$.

Step 1: Factorize the Denominator

We can rewrite the denominator $1 - z^2$ as $(1 - z)(1 + z)$. Thus, the given function becomes: $\frac{1}{1 - z^2} = \frac{1}{(1 - z)(1 + z)}.$

Step 2: Expand Using Partial Fractions

The partial fraction decomposition of $\frac{1}{(1 - z)(1 + z)}$ is: $\frac{1}{1 - z^2} = \frac{1}{2(1 - z)} + \frac{1}{2(1 + z)}.$

Step 3: Series Expansion

Each term in the decomposition can now be expanded as a geometric series about $z = 0$, assuming $|z| < 1$.

Expansion of $\frac{1}{1 - z}$:

$\frac{1}{1 - z} = 1 + z + z^2 + z^3 + \cdots = \sum_{n=0}^\infty z^n.$

Expansion of $\frac{1}{1 + z}$:

$\frac{1}{1 + z} = 1 - z + z^2 - z^3 + \cdots = \sum_{n=0}^\infty (-1)^n z^n.$

Step 4: Combine the Expansions

Now, combine the expansions of the two terms: $\frac{1}{2(1 - z)} = \frac{1}{2} \sum_{n=0}^\infty z^n,$ $\frac{1}{2(1 + z)} = \frac{1}{2} \sum_{n=0}^\infty (-1)^n z^n.$

Adding these together: $\frac{1}{1 - z^2} = \frac{1}{2} \sum_{n=0}^\infty z^n + \frac{1}{2} \sum_{n=0}^\infty (-1)^n z^n.$

Step 5: Simplify the Series

The coefficients for even powers of $z$ will add up, and the coefficients for odd powers of $z$ will cancel. This results in: $\frac{1}{1 - z^2} = \sum_{n=0}^\infty z^{2n}.$

Final Answer:

The Laurent series expansion of $\frac{1}{1 - z^2}$ about $z = 0$ is: $\sum_{n=0}^\infty z^{2n}.$

Let me know if you want further clarification or have questions! Here are some related queries:

1. How is the convergence of Laurent series determined?
2. Can Laurent series include terms with negative powers?
3. How does the Laurent series differ from the Taylor series?
4. What are the applications of Laurent series in complex analysis?
5. Can we compute the Laurent series if $|z| > 1$?

Tip: Always check the radius of convergence for the series, as this determines where the expansion is valid!