help with laurent series

Question

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

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. — , Sep 16 '14 at 22:20
Yes what @Thursday said is right. It depends upon the center of your annulus. — amcalde, Sep 16 '14 at 22:21
Sorry, forgot to put that on the question. I have to find all laurent series for this function with center in z=0. — bruno, Sep 16 '14 at 22:27

score 3 · Accepted Answer · answered Sep 16 '14 at 22:55

3

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.

answered Sep 16 '14 at 22:55

robjohn

345,667

I'm working on a great solution of it. – user 1591719 Sep 19 '14 at 20:17

help with laurent series

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