Residue Complex Function I've applied the formula  $$\lim_{z \to i} \frac{d}{dz}{(z-z_0)^2.f(z)}$$ and I keep getting a value of 0. Could someone prove my sanity by starting this off?

Hint: Since $f(z)$ has a double pole at $z=i$, then we see that
\begin{align}
\operatorname{Res}_{z=i}f(z)=\lim_{z\rightarrow i}\frac{1}{2}\frac{d}{dz}\left((z-i)^2\frac{z^2-1}{(z^2+1)^2}\right).
\end{align}
Edit: To put you at ease, observe the power series expansion at $z=i$ of the function is given by
\begin{align}
\frac{z^2-1}{(z^2+1)^2} =&\ \frac{1}{(z-i)^2}\frac{z^2-1}{(z+i)^2} =\frac{-1}{4(z-i)^2}\frac{(z-i)^2+2i(z-i)-2}{\left(1+\frac{z-i}{2i}\right)^2}\\
=&\ \frac{-1}{4(z-i)^2}\left((z-i)^2+2i(z-i)-2\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\
=&\ \left(\frac{-1}{4} -\frac{i}{2(z-i)}+\frac{1}{2(z-i)^2}\right)\sum_{k=0}^\infty(-1)^k(k+1)\left(\frac{z-i}{2i} \right)^k\\
=&\ \frac{1}{2(z-i)^2}-\frac{1}{8}-\frac{1}{8}i(z-i)+\text{higher order terms}.
\end{align}
Hence it is clear that the residue is zero.
