<s>[INST] Classifying singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$ 
Classify all the singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$

I know that the singularities of $f$ are exactly $z=2,z=3$. I just want to check that if my solution for the classification is correct:
$z=2$: since $1/(3-z)$ is analytic at $z=2$, then the composition $e^{\frac{1}{3-z}}$ is also analytic at this point. Analyitic functions admit Taylor expansion, hence the expansion of $e^{\frac{1}{3-z}}$ will only give "positive coeffcients" (coefficients with positive powers), so it suffices to only look for the Laurent expansion of $\frac{1}{(z-2)^2}$, which is already in this form. Thus, $z=2$ is a pole of order 2.
$z=3$: now $\frac{1}{(z-2)^2}$ is analytic at this point, so by the argument above we don't have to care about its series expansion. Now, since $e^z$ is entire, this series admits expansion in the region $3<|z|<\infty$, so we can write: 
$$e^{\frac{1}{3-z}} = \sum_{n=0}^{\infty}\frac{1}{n!(3-z)^n}.$$
The uniqueness of the laurent series guarantee us that this series above is the Laurent expansion of $e^{\frac{1}{3-z}}.$ Therefore, $z=3$ is an essential singularity.
Is it correct? Or I do have to find the series expansion in each case?
 [/INST] Your reasoning is correct.
$e^{-\frac{1}{z-3}}$ is analytic at  $z=2$, so it can be expanded in a Taylor series about that point in the region $0<|z-2|<1$:
\begin{align}e^{-\frac{1}{z-3}}=e+e(z-2)+\frac{3e}{2}(z-2)^2+\cdots\end{align}
Therefore, the Laurent series of $f(z)$ about $z=2$ in the region $0<|z-2|<1$ is given by:
\begin{align}f(z)=&\frac{1}{(z-2)^2}+e+e(z-2)+\frac{3e}{2}(z-2)^2\\&+\cdots\end{align}
Similarly, $\frac{1}{(z-2)^2}$ is analytic at $z=3$, so it can be expanded in a Taylor series about that point in the region $0<|z-3|<1$:
\begin{align}\frac{1}{(z-2)^2}=&1-2(z-3)+3(z-3)^2+\cdots\end{align}
Therefore, the Laurent series of $f(z)$ about $z=3$ in the region $0<|z-3|<1$ is given by:
\begin{align}f(z)=&\cdots+\frac{1}{2}\frac{1}{(z-3)^2}-\frac{1}{z-3}+2\\&-2(z-3)+3(z-3)^2+\cdots\,\end{align}
and we arrive at your conclusions.
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