meromorphic functions on the extended complex plane

Question

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?

Answer 1

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.

Answer 2

$e^z$ has an essential singularity at $\infty$, since $$ e^{1/z} = 1 + \frac1z +\frac1{2!z^2} + \cdots$$ has infinitely many terms with negative powers in its Laurent series.

Alternatively, the singularity is essential since $\lim_{z\to\infty} e^z$ doesn't exist (even as $\infty$)

Answer 3

The function $f(z)=e^{z}$ does not have a removable singularity at $\infty$.

Assuming otherwise, it would be bounded in a neighbourhood of infinity, meaning that $|f(z)|$ is bounded on $\{z\in\mathbb{C}: |z|>R\}$ for some $R>0$. But $|f(z)|$ is also bounded on $\{z\in\mathbb{C}: |z|\leqslant R\}$ as continuous function is bounded on a compact set. We conclude that $e^{z}$ is bounded on $\mathbb{C}$, so it is constant by Liouville's theorem, contradiction.