19.8k views
4 votes
[ Complex Analysis ]

Find the laurent series for 1/(1-Z2) about Z0=1.

1 Answer

1 vote

Recall that for
|z|<1, we have


\frac1{1-z}=\displaystyle\sum_(n\ge0)z^n

Then


\frac1{1-z}=-\frac1z\frac1{1-\frac1z}=-\frac1z\displaystyle\sum_(n\ge0)z^(-n)=-\sum_(n\ge0)z^(-n-1)

valid for
|z|>1, so that


\frac1{1-z^2}=-\frac1{z^2}\frac1{1-\frac1{z^2}}=-\frac1{z^2}\displaystyle\sum_(n\ge0)z^(-2n)=-\sum_(n\ge0)z^(-2n-2)

also valid for
|z|>1.

User Jimmy Zhang
by
5.1k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.