How can $f(\frac{1}{z})$ not have a singularity at 0? Suppose we have $f(z)$ with a Laurent Expansion with an infinite analytic part. Then $f(1/z)$ should have an infinite principal part of the form $\sum a_k (\frac{1}{z})^k$ and thus what appears to be an essential singularity at 0? 
This seems too broad to be true, is there something wrong with it?
 A: You have to be careful about where the Laurent expansion is valid, but there is some truth to what you're saying.
If $f(z)$ is a meromorphic function on the Riemann sphere, then it is a rational function. So this tells us that "most" meromorphic functions on $\mathbb{C}$ do in fact have an essential singularity at $\infty$.
A: The problem with this argument is that your Laurent series for $f(1/z)$ may not converge in any deleted neighborhood of $0$, so that you don't get $0$ as an isolated singularity.  For instance, if you take $f(z)=1/(1-z)$ with its expansion $f(z)=\sum_{n=0}^\infty z^n$, the expansion for $f(1/z)$ you get only converges for $|z|>1$.  But the function $f(1/z)$ is actually analytic at $0$.
In general, if your Laurent series for $f(z)$ converges on the annulus $r<|z|<R$, then the corresponding Laurent series for $f(1/z)$ will converge for $1/R<|z|<1/r$.  So if $R<\infty$ (i.e., the radius of convergence of the analytic part of the Laurent expansion is finite), then you cannot conclude that $f(1/z)$ has a singularity at $0$ (it might have a singularity, or it might be analytic, or it might not be defined in any deleted neighborhood of $0$ at all).  But if $R=\infty$ then your argument is correct, and you can conclude that $f(1/z)$ has an essential singularity at $0$.
