Residue of $\frac {z}{z^6+1}$ at $z=i$ Is there an easy way to find the residue of $$\frac {z}{z^6+1}$$
at $z=i$? the formula given by taylor theorem doesn't seem to help and evaluating $(z-i)\frac {z}{z^6+1}$ at $z=i$ seems to laborious. maybe I am wrong about this and it's not that hard to  to do this calculations, but if that's the case, I'm missing something.
The residue is $\frac{1}{6}$ by the way

Since the pole is first order we can multiply by $z-i$ and use L'Hospital to get the residue:
$$
\begin{align}
\lim_{z\to i}z\frac{z-i}{z^6+1}
&=\lim_{z\to i}z\lim_{z\to i}\frac1{6z^5}\\
&=i\frac1{6i}\\
&=\frac16
\end{align}
$$
