Compute the residues at the singularities of $$f(z) = \frac{\cos z}{z^2(z- \pi )^3}$$
There are two poles: $z=0$ is of order 2 and $z= \pi$ is of order 3. We note that for poles $a$ of order $n$, we have
$$\mathrm{Res}_{z=a} f(z) = \frac{1}{(n-1)!} \lim_{z \to a} \frac{d^{n-1}}{dz^{n-1}} \left[(z-a)^n f(z)\right]$$
Using this tool, we have
\begin{align}
\begin{split}
\mathrm{Res}_{z=0} f(z) &= \frac{1}{(2-1)!} \lim_{z \to 0} \frac{d}{dz} \left[ \frac{\cos z}{z^2(z- \pi)^3} \right] \\
&= \lim_{z \to 0} – \frac{(z^2- \pi z)\sin z + (5z – 2\pi) \cos z}{z^3(z-\pi)^4}
\end{split}
\end{align}
I run into troubles here as I still cannot evaluate the above expression at $z=0$. We could use the Residue Theorem, i.e.
$$\int_C f(z) \hspace{0.2cm} dz = 2 \pi i \sum_{a_k} \mathrm{Res}_{z=a_k} f(z)$$
Maybe we could set $g(z) = \cos z / z^2$ and let $C$ be the circle $|z- \pi| = 1$. Note that $g(z)$ is analytic everywhere except $z=0$ and hence $g(z)$ is analytic on and interior to $C$. Now, by Cauchy's Integral Formula, we have
\begin{equation}
\int_C \frac{g(z)}{(z- \pi)^3} \hspace{0.2cm} dz = \frac{2\pi i }{2!} g''(\pi) = \pi i \cdot g''(\pi) \hspace{1cm} (\star)
\end{equation}
Now, we note that
\begin{align}
\begin{split}
& g''(z) = \frac{4z \sin z + (6-z^2) \cos z}{z^4} \\
\implies & g''(\pi) = \frac{4\pi \sin \pi + (6-\pi^2) \cos \pi}{\pi^4} \\
\end{split}
\end{align}
Thus, by $(\star)$, we have
\begin{equation}
\int_C \frac{g(z)}{(z- \pi)^3} \hspace{0.2cm} dz = i \frac{4\pi \sin \pi + (6-\pi^2) \cos \pi}{\pi^3} \hspace{1cm} (\star \star)
\end{equation}
A manipulation of the Residue Theorem reveals that
\begin{equation}
\sum_{a_k} \mathrm{Res}_{z= a_k} f(z) = \frac{1}{2\pi i} \int_C \frac{g(z)}{(z-\pi)^3} \hspace{0.2cm} dz
\end{equation}
And hence, by $(\star \star)$, we have
\begin{equation}
\mathrm{Res}_{z= \pi} f(z) = i \frac{4\pi \sin \pi + (6-\pi^2) \cos \pi}{2\pi^4}
\end{equation}
Is this all looking good? If not, how do I proceed? Also, what would be the best way to find $\mathrm{Res}_{z=0} f(z)$?
EDIT: There is an error in the line
Using this tool, we have \begin{align}
\begin{split}
\mathrm{Res}_{z=0} f(z) &= \frac{1}{(2-1)!} \lim_{z \to 0} \frac{d}{dz} \left[ \frac{\cos z}{z^2(z- \pi)^3} \right] \\
\end{split}
\end{align}
It should have been corrected to
\begin{align}
\begin{split}
\mathrm{Res}_{z=0} f(z) &= \frac{1}{(2-1)!} \lim_{z \to 0} \frac{d}{dz} \left[ \color{red}{z^2} \cdot \frac{\cos z}{z^2(z- \pi)^3} \right] \\
&= \lim_{z \to 0} \frac{d}{dz} \left[ \frac{\cos z}{(z- \pi)^3} \right]
\end{split}
\end{align}
Best Answer
I always find it strange to use integration to find residues. It normally goes the other way, that residues help us perform integrals. In general, whenever we are given well known functions with Taylor series and Lauren series, residues are typically easy to read from the function directly. A residue is the coefficient of $1/z$. So the residue is $a_{-1}$ whenever
$$ f(z) = \cdots + \frac{a_{-2}}{z^2} + \frac{a_{-1}}{z} + a_{0} + a_{1}z + a_{2}z^2 + \cdots $$
This is because the integration of each term is zero. So, when we are dealing with something like $$\frac{\cos(z)}{z^2(z-\pi)^3}$$ I typically just expand this at the pole of interest, for example, expand around $z=0$:
$$\frac{1}{z^2}\left(\frac{-1}{\pi}\left(1+\frac{z}{\pi}+\frac{z^2}{\pi^2}+\cdots\right)\right)^3 \left(1-\frac{1}{2}z^2+\frac{1}{24}z^4-\cdots\right)$$
the only terms relevant to $1/z$ reduce this to
$$\frac{1}{z^2}\left(\frac{-1}{\pi}\left(1+\frac{z}{\pi}\right)\right)^3 \left(1-\frac{1}{2}z^2\right)$$
from here we can easily see the coefficient of $1/z$ is
$$ \frac{-1}{\pi^3}\left(\frac{3}{\pi}\right)$$
In a similar way we can expand around $z=\pi$:
$$\frac{1}{(z-\pi)^3}\left(\frac{1}{\pi}\left(1+\frac{\pi-z}{\pi} + \left(\frac{\pi-z}{\pi}\right)^2+\cdots\right)\right)^2\left(-1+\frac{1}{2}(z-\pi)^2-\frac{1}{24}(z-\pi)^4+\cdots\right)$$
and again we can reason the same way, by only look at terms relevant to $1/(z-\pi)$.