= Solution
{c}
The possible singular points are $i,-i,1$. Near $i$, put $h(z)=1/((z+i)(z-1)^2)$; then
$$
f(z)=\frac{h(z)-1/4}{z-i},\qquad h(i)=\frac14.
$$
The numerator vanishes at $i$, so the apparent pole cancels and \b[$i$ is removable]. Its extension value is $h'(i)=1/4+3i/8$, obtained from $h'/h=-1/(z+i)-2/(z-1)$.
At $-i$, the first term has a simple <pole> with nonzero <residue>
$$
\lim_{z\to-i}(z+i)f(z)=\frac1{(-2i)(-i-1)^2}=\frac14.
$$
The subtracted term is holomorphic there, so \b[$-i$ is a simple pole]. At $1$,
$$
\lim_{z\to1}(z-1)^2f(z)=\frac1{1^2+1}=\frac12\ne0,
$$
so \b[$1$ is a double pole]. There are no other finite singularities.
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