How can we compute the Laurent expansion of
$f(z)=\frac{1}{z(z-i)^2}$
about $0$ in $\Omega=B_1(0)\setminus\{0 \}$?
Thoughts:
By partial fractions we have that
$f(z)=\frac{1}{z-i}-\frac{i}{(z-i)^2}-\frac{1}{z}$
but I'm unsure is to which expansions will be valid in $\Omega$. Any help would be very appreciated.
Best, MM.