# help with laurent series

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

• At which points is $X$ singular? Commented Sep 16, 2014 at 21:48
• z = 1 and z = -1 Commented Sep 16, 2014 at 21:51
• What is your Laurent series? Commented Sep 16, 2014 at 22:16
• The region of convergence depends on the point of expansion. Is it in powers of $(z-2)$ or in powers of $z$ -- the answers are different.
– user147263
Commented Sep 16, 2014 at 22:20
• Yes what @Thursday said is right. It depends upon the center of your annulus. Commented Sep 16, 2014 at 22:21

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.