If we are given a function such as $e^{z+\frac{1}{z}}=e^ze^{\frac{1}{z}},$ and we are trying to classify it's singularities. Is it correct to say, that because $e^{\frac{1}{z}}$ has a singularity at $z_0=0$, and $e^z$ is entire, that f(z) has singularities only at $z_0=0$ ?
Then are we allowed to only consider $e^{\frac{1}{z}}$ when classifying the singularity ?
I.e. during the next step, where we state that because $f(z)=e^{\frac{1}{z}}$ is known to be analytic on all of $\Bbb C /\{0\}$, then by Laurent's theorem we know that because f is differentiable on this set that it can be expressed as
$$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$$
where
$$a_n=\frac{1}{2\pi i}\int_{C(z_0,r)}\frac{f(z)}{(z-z_0)^{n+1}}dz$$
which in the particular case I was asking about would become
$$a_n=\frac{1}{2\pi i } \int_{C(z_0,r)} \frac{e^{\frac{1}{z}}}{z^{n}}dz$$
which we would then parametrise. So is this okay, or should I be considering the entire original function $f(z)=e^{z+\frac{1}{z}}$?