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QUESTION


Show that when $0&lt;|z-1|&lt;2$, then we have the Laurent expansion
$${z\over (z-1)(z-3)}=-3\sum_{n=0}^{\infty}{(z-1)^n\over
2^{n+2}}-{1\over 2(z-1)}.$$



ANSWER


Put $w=z-1$ and expand, as usual, about $w=0$. We get ${z\over
(z-1)(z-3)}={w+1\over w(w-2)}$ and now expand about $w=2$.




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