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Expand the function $$f(z) = \frac{1}{(z + 1)(z + 3)}$$ in a Laurent series valid for $1 < |z| < 3$


My attempt:

$$\frac{1}{(z + 1)(z + 3)}=\frac{1}{4}.\frac{1}{1+z}-\frac{1}{4}.\frac{1}{3+z}$$

$$=\frac{1}{4}.\frac{1}{1-(-z)}-\frac{1}{4}.\frac{1}{3-(-z)}$$

$$=\frac{1}{4}.\frac{1}{z}\frac{1}{\frac{1}{z}-(-1)}-\frac{1}{4}.\frac{1}{3-(-z)}$$

$$=\frac{1}{4}.\frac{1}{z}\frac{1}{1-(-\frac{1}{z})}-\frac{1}{4}.\frac{1}{3}.\frac{1}{1-(-\frac{z}{3})}$$

and now both fraction can be expanded using the geometric series

$$=\frac{1}{4}.\frac{1}{z}(1-\frac{1}{z}+\frac{1}{z^2}...)-\frac{1}{12}.(1-\frac{z}{3}+\frac{z^2}{3^2}...)$$


Is the expansion correct?

NOTE: What is written above is the entire question.

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    $\begingroup$ It is correct except that $\frac 1 4$ should be |frac 1 2$ in the first step. $\endgroup$ – Kavi Rama Murthy Apr 23 at 23:31
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    $\begingroup$ $$\frac{1}{(z + 1)(z + 3)}=\frac{1}{2 (1 + z)} - \frac{1}{2 (3 + z)}$$ $\endgroup$ – E.H.E Apr 23 at 23:32
  • $\begingroup$ Thank you very much $\endgroup$ – Mohamad Moustafa Apr 23 at 23:53

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