<s>[INST] Evaluate $\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}$ . Problem:
Evaluate
$$
\underset{z=0}{\text{Res}} \; \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)}.
$$
My question:
This is a question from a previous complex analysis qualifying exam that I am trying to work through. I know that the straightforward formula for calculating this residue would be
$$
\frac{1}{4!} \lim_{z\to 0} \frac{d^4}{dz^4} \left(z^5 \cdot \frac{(z^6-1)^2}{z^5(2z^4 -5z^2 + 2)} \right)
$$
However, I'm guessing there is a better method in this case than taking four derivatives or trying to turn this into a Laurent series. Can someone point me in the right direction here?
 [/INST] Let $$f(z) = \frac{(z^6-1)^2}{z^5(2z^4 - 5z^2 + 2)}, \quad g(z) = \frac{(z^3-1)^2}{2z^2 - 5z + 2},$$ so that we have $f(z) = g(z^2)/z^5$.  So by computing the series expansion of $g$ about $0$, we can obtain the Laruent expansion of $f$, since $g$ has no singularity at  $0$.
To this end, observe $$2z^2 - 5z + 2 = (2z-1)(z-2) = 2(1-2z)(1-z/2),$$ hence $$\begin{align}
g(z) &= \frac{1}{6}(z^3 - 1)^2 \left( \frac{4}{1-2z} - \frac{1}{1-z/2} \right) \\
&= \frac{1}{6} (z^3 - 1)^2 \sum_{k=0}^\infty 4(2z)^k - (z/2)^k \\
&= \frac{1}{6} (z^6 - 2z^3 + 1) \sum_{k=0}^\infty (2^{k+2} - 2^{-k}) z^k.
\end{align}$$
We specifically need the coefficient of $z^2$ in this expansion, since this yields the coefficient of $z^4$ in $g(z^2)$, which in turn is the coefficient of $1/z$ in the Laurent expansion of $f$.  This corresponds to the choice $k = 2$:  $$g(z) = \cdots + \frac{1}{6}(2^4 - 2^{-2}) z^2 + \cdots.$$  Thus the required coefficient is $21/8$, which is also the desired residue of $f$ at $0$.
</s>