We can do this by factoring out the term $(z-1)$ in the denominator and writing the function as a series in powers of $(z-1)$:
\begin{align*}
f(z) &= \frac{2}{(z-1)(z+4)} \\
&= \frac{2}{5} \frac{1}{(z-1)} \frac{1}{1+\frac{z-1}{5}} \\
&= \frac{2}{5} \frac{1}{(z-1)}
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