$f\left( z\right) =\dfrac {e^{-z}}{\left( z-1\right) \left( z+2\right) ^{2}}$
$\dfrac {1}{z-1}=\dfrac {1}{z+2-3}=\dfrac {1}{3\left( \dfrac {z+2}{3}-1\right) }$
$=-\dfrac {1}{3}\dfrac {1}{1-\dfrac {z+2}{3}}$
$=-\displaystyle\sum ^{\infty }_{n=0}\dfrac {\left( z+2\right) ^{n}}{3^{n+1}}$
$f(z)=-e^{-z}\displaystyle\sum ^{\infty }_{n=0}\dfrac {(z+2)^{n-2}}{3^{n+1}}=-e^{-z}\sum ^{\infty }_{n=2}\dfrac {(z+2)^{n}}{3^{n-1}} $
$0<|z+2|<3$