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What will be the Laurent series for above function?

Bernard
  • 175,478

1 Answers1

1

Hint:

Set $z=1+u$. This function can be rewritten as

$$\frac{1}{1-z^2}=\frac 1{(1-z)(1+z)}=-\frac1{2u}\,\frac1{1+\cfrac u2}$$ and you can expand the second fraction with the geometric series.

Bernard
  • 175,478