The book says:
The Laurent series converges and represents $F(z)$ in the open annulus obtained by continuously increasing the radius of $C_2$ and decreasing the radius of $C_1$ until each of $C_1$ and $C_2$ reaches a point where $F(z)$ is singular.
How can I find the region of convergence when the function is:
$$X(z) = \frac{1}{1-z^2}$$
edit: there are 2 regions: $|z| < 1$ and $|z| > 1$ ?
Thanks
There are two Laurent series for $f(z)=\frac1{1-z^2}$ centered at $z=0$. There is the normal Taylor series: $$ \frac1{1-z^2}=1+z^2+z^4+z^6+\dots $$ which converges for $|z|\lt1$. Then there is the series $$ \begin{align} \frac1{1-z^2} &=\frac1{z^2}\frac1{\frac1{z^2}-1}\\ &=-\frac1{z^2}-\frac1{z^4}-\frac1{z^6}-\frac1{z^8}-\dots \end{align} $$ which converges for $|z|\gt1$. The second series can be simply derived from the first.
