What will be the Laurent series for above function?
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Welcome to stackexchange. To get help here, please [edit] the question to show us what you tried and where you are stuck. – Ethan Bolker Dec 06 '19 at 11:23
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What have you tried? – MANI Dec 06 '19 at 11:35
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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
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