Complex Analysis - Laurent Series

I need to find the Laurent expansion of

\frac{1}{(z+1)(z+3)}

valid for |z|>3

I would have no problem if this was 1/(z+3) as then I would say

\frac{1}{z+3} = \frac{1}{z(1 + 3/z)} = \frac{1}{z}(1 - 3/z + 9/z^2 + ...) = 1/z - 3 + 9/z + ...

for |z| > 3

What do I do with the z+1 in the denominator though?

(edited 11 years ago)

I need to find the Laurent expansion of

\frac{1}{(z+1)(z+3)}

valid for |z|>3

I would have no problem if this was 1/(z+3) as then I would say

\frac{1}{z+3} = \frac{1}{z(1 + 3/z)} = \frac{1}{z}(1 - 3/z + 9/z^2 + ...) = 1/z - 3 + 9/z + ...

for |z| > 3

What do I do with the z+1 in the denominator though?

Split the expression by partial fractions

\frac{1}{(z+1)(z+3)} = \frac{A}{(z+1)} + \frac{B}{(z+3)}

and apply your method.

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