I'm highly confused by one theorem. Every meromorphic function on the extended complex plane is rational. But $e^z$ is analytic everywhere in the plane, and since it is analytic outside a bounded set, it has an isolated singularity at $\infty$. It seems to me that this singularity at $\infty$ is removable because the laurent expansion at $\infty$ has no positive terms. Therefore it is meromorphic on the extended complex plane. But $e^z$ cannot possibly be rational, it does not have any zero. What's wrong here?
[Math] meromorphic functions on the extended complex plane
complex-analysisrational-functions
Best Answer
To look at the behaviour of $e^z$ at infinity, you need to look at the behaviour of $e^{1/z}$ at $z=0$. If you look at the Laurent series for $e^{1/z}$ it has infinitely many terms of type $\displaystyle{1/z^n}$, so it cannot be a removable singularity or a pole. Thus we have an essential singularity at infinity, and your problem is resolved.