Classification of the singularity of the following function
$f(z)=e^{z+\frac{1}{z}}$
So $e^{z+\frac{1}{z}}=e^ze^{\frac{1}{z}}$
So we have a singularity at $z=0$, how do I classify this?
I think I can neglect $e^z$ as it is an entire function
Ok so I have $f(z)=e^{\frac{1}{z}}$ is known to be analytic on all of $\Bbb C /\{0\}$, then by Laurent's theorem I have
$$f(z)=\sum_{n=-\infty}^{\infty}a_n(z-z_0)^n$$ .
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
$$a_n=\frac{1}{2\pi i } \int_{C(z_0,r)} \frac{e^{\frac{1}{z}}}{z^{n}}dz$$
Can someone explain how I solve this to classify the singularity and does this tell me the order of the pole??
Along the sequence $z_n=n$ the function tends to $\infty$. Along the sequence $z_n=\frac i n$ we get $f(z_n)=e^{\frac i n+\frac n i}=e^{-i(n-\frac 1n)}$ which has modulus $1$. These facts imply that $0$ is neither a pole nor a removable singularity. Hence it is an essential singularity.