Complex analysis: Prove a meromorphic function to be rational. I come across a problem about complex analysis:
Show that a meromorphic function on the complex plane, which achieves any complex number no more than fixed given times, must be rational.
The only way I know to prove a meromorphic function being rational is to show the infinity is not an essential singularity of the function (thus it can be controlled by polynomial). Following this, we can use Picard's Great Theorem to solve the problem.
I wonder if anyone can help me think of another method. (Without using Picard's Great Theorem.)
 A: 
Show that a meromorphic function on the complex plane, which achieves any complex number no more than fixed given times, must be rational.

If available, Picard's theorem is probably the quickest way to swat this fly [and it yields the result without assuming a fixed bound on the number of times a value is attained]. If Picard's theorem is not available or shall be avoided:
Let's call the meromorphic function $f$, and let's call the fixed bound on the number of times any complex number is attained $n$.
If $f$ has at least $k$ poles, then every complex number of sufficiently large modulus is attained at least $k$ times, so $k \leqslant n$. Let $\{\pi_1,\dotsc, \pi_p\}$ be the set of poles of $f$.
Let $m$ be the maximum number any complex value is attained at all (counting multiplicities). Then $m \leqslant n$. Pick $w_0 \in \mathbb{C}$ such that $w_0$ is attained $m$ times (counting multiplicities), and choose $R$ so large that $f(z) \neq w_0$ for $\lvert z\rvert \geqslant R$, and that $\lvert \pi_k\rvert < R$ for all poles of $f$. Choose $\varepsilon > 0$ small enough that $0 < \lvert z - \pi_k\rvert \leqslant \varepsilon \implies \lvert f(z)\rvert \geqslant 1 + \lvert w_0\rvert$ and $\lvert \pi_k\rvert + \varepsilon < R$. Let $G = \{ z\in \mathbb{C} : \lvert z\rvert < R, \lvert z-\pi_k\rvert > \varepsilon\}$. Then look at
$$N(w) := \frac{1}{2\pi i} \int_{\partial G} \frac{f'(z)}{f(z) - w_0}\,dz.$$
By construction, $N(w_0) = m$, and $N$ is constant on each component of $\mathbb{C}\setminus f(\partial G)$, so we have $N(w) \equiv m$ on some neighbourhood $U$ of $w_0$, which means that every $w\in U$ is attained $m$ times (counting multiplicities) in $G$. By the maximality of $m$, that means no $w\in U$ is attained outside $G$, in particular
$$U \cap f\bigl(\mathbb{C}\setminus \overline{D_R(0)}\bigr) = \varnothing.$$
By the Casorati-Weierstraß theorem, it follows that $\infty$ is not an essential singularity of $f$. Then the rationality of $f$ follows by subtracting the principal parts of $f$ at the $\pi_k$ and (possibly) at $\infty$.
A: Take the extended complex plane $\mathbb{C}\cup\{\infty \}$. Introduce a function $f$ that has a punctured neighbourhood at the origin (hence is analytic in a punctured neighbourhood at infinity). The remaining region has finitely many poles hence the collection of poles of $f$ (denoted $a_n$) is finite. Introduce Laurent expansions at one of the poles $a_m$ (for fixed $m$) and at infinity. Look at the nagtive power sections of Laurent expansions at the pole and infinity (call these $l_k$ and $l_{\infty}$ respectively) to derive polynomials of finite length at each pole, and subtract them from $f$. This will give something like
\begin{equation*}
f-\sum l_k-l_{\infty}.
\end{equation*}
Conclude that this result is entire and bounded to get that this is rational.
