Classify the singularities and determine de radius of convergence of the function $f(z)=\frac{e^{-z^2}}{(z-2)}$ at $a=0$.
$f$ has a pole of order $m$ at $z_0$ if there exists a function $g$ that is holomorphic in a neighbourhood of $z_0$ with $g(z_0) \neq 0$ such that $ (z-z_0)^m f(z) = g(z) $
So $(z-2)f(z)=e^{-z^2}$ and $g(z)=e^{-z^2}$ is holomorphic in a neighbourhood of $z_0=2$, so $f$ has a pole of order $1$. I am not certain of this answer. Is anyone could comment a little bit my solution?
Yes, you're right. But a detail: you can't cancel $z-2$. Really, you must calculate $\lim_{z\to 2} (z-2)f(z)$ and, because is a nonzero real number, $2$ is a pole of degree 1