How to find Laurent Series for $z/(z-1)(z+4)$ How do I find the Laurent series for $$\frac{z}{(z-1)(z-4)}?$$ on:
i) $0<|z-1|<5$;
ii) $5<|z-1|$.
I broke it up into $$\frac{1}{5}\left(\frac{4}{z+4}+\frac{1}{z-1}\right)$$ but now I am stuck.
 A: The quickest way is to find binomial expansions of each fraction, which converge in the required regions. Let's first of all put $w=z-1$, so that the thing we're expanding in is only one symbol. Then the function is
$$ \frac{1}{5} \left( \frac{4}{w+5} + \frac{1}{w} \right). $$
Ah, so now we see $1/w$, which means we have nothing to do to the second term: it's already just a power of $w$. Now, in the first region, $0<\lvert w \rvert < 5$, so if we rearrange the first term to
$$ \frac{1}{5}\frac{4}{w+5} = \frac{4}{25} \left( 1+\frac{w}{5} \right)^{-1}, $$
we know that $\lvert w/5 \rvert < 1$ so the binomial series with $w/5$ as the variable will converge:
$$ \frac{4}{25} \left( 1+\frac{w}{5} \right)^{-1} = \frac{4}{25} \sum_{n=0}^{\infty} (-1)^n\frac{w^n}{5^n}. $$
Hence the entire expansion is
$$ \frac{1}{5w} + \sum_{n=0}^{\infty} 4(-1)^n\frac{w^n}{5^{n+2}}. $$
For the second region, the $1/(5w)$ works in the same way, but you have to rearrange the first term differently: now $\lvert 5/w \rvert <1$, so you can do a binomial expansion in $5/w$. A little thought will show you that the correct rearrangement is
$$ \frac{1}{5}\frac{4}{w+5} = \frac{4}{5w} \left( 1+\frac{5}{w} \right)^{-1}, $$
and I'm sure you can do the rest. (Note that the first term is the same order as the $1/(5w)$, so you need to add those separately.)
