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

The denominator $d(z) = z^6+1$ has a simple zero at $z = i$, so the residue is simply
$$\frac{i}{d'(i)} = \frac{i}{6 i^5} = \frac{1}{6i^4} = \frac{1}{6}.$$
Whenever we have a quotient
$$\frac{f(z)}{g(z)}$$
where $g$ has a simple zero in $z_0$, the residue of $\frac{f}{g}$ in $z_0$ is
$$\frac{f(z_0)}{g'(z_0)},$$
as can be seen from writing $g(z) = (z-z_0)\cdot h(z)$ with $h(z_0) = g'(z_0) \neq 0$.
