Compute the Integral via Residue Theorem and Laurent Expansion Let $f(z)=\dfrac{z^{15}}{(z^2+1)^2(z^4+2)^3}$ . My goal is to compute $\displaystyle I=\int_{|z|=4}f(z) \,dz$.
Since all the singular points (except $∞$) lie in the circle $|z|<4$, we obtain $I=-2\pi i\operatorname{Res}(f,∞)$. Thus, we just need to find the Laurent coefficient $a_{-1}$ at $∞$.
$f\left(\dfrac 1 w\right) = \dfrac w {(1+w^2)^2(1+2w^4)^3}$. Any convenient ways to find $a_{-1}$?

You want to find $Res_{w = 0} \left[\frac 1{w^2}f\left(\frac 1w \right)\right] = Res_{w = 0} \left[\dfrac 1{w(1+w^2)^2(1+2w^4)^3}\right]$ . Now we have:
$$a_{-1} = \frac{1}{2\pi i} \int_{C_0} \frac {dw}{w(1+w^2)^2(1+2w^4)^3} = \frac{1}{(1+0^2)^2(1+2\cdot 0^4)^4} = 1$$
by Cauchy Integral Formula, as $\frac{1}{(1+w^2)^2(1+2w^4)^3}$ is analytic at $C_0$, which can be taken as any curve inside a disk $|z| \le \frac 14$, for example.

A direct way of finding the residue is to use partial fraction decomposition. We have:
$$\dfrac 1{w(1+w^2)^2(1+2w^4)^3} = \frac Aw + \frac{F(w)}{(1+w^2)^2(1+2w^4)^3}$$ Now it's not har to see that $A = a_{-1}$, so multiply both sides by $w$ and evaluate it at $w=0$
