[proofplan]
For the forward direction, we show that a meromorphic function on $\widehat{\mathbb{C}}$ has finitely many poles (by compactness and discreteness), then subtract the principal parts at each pole to reduce to an entire bounded function, which must be a polynomial by Liouville's theorem. For the reverse direction, we verify directly that any rational function defines a meromorphic function on $\widehat{\mathbb{C}}$ by checking holomorphicity in the standard chart and the chart at infinity.
[/proofplan]
[step:Show that a meromorphic $f$ on $\widehat{\mathbb{C}}$ has finitely many poles]
Suppose $f: \widehat{\mathbb{C}} \to \widehat{\mathbb{C}}$ is meromorphic on $\widehat{\mathbb{C}}$. The set of poles of $f$ is a discrete subset of $\widehat{\mathbb{C}}$. Since $\widehat{\mathbb{C}}$ is compact (it is homeomorphic to the 2-sphere $S^2$), every discrete subset of a compact space is finite. Therefore $f$ has finitely many poles in $\widehat{\mathbb{C}}$.
[guided]
A meromorphic function on a domain has poles forming a discrete set — no accumulation point of poles can exist within the domain. On $\widehat{\mathbb{C}}$, the domain is the entire Riemann sphere, so the poles form a discrete subset of $\widehat{\mathbb{C}}$.
Now we use a topological fact: a discrete subset of a compact space must be finite. Why? If it were infinite, it would have an accumulation point (by the Bolzano-Weierstrass property of compact spaces), contradicting discreteness.
So $f$ has finitely many poles. Let us enumerate them as $z_1, \dots, z_N \in \widehat{\mathbb{C}}$ (some of which may be the point $\infty$).
[/guided]
[/step]
[step:Reduce to the case where $\infty$ is not a pole]
Let $z_1, \dots, z_N \in \widehat{\mathbb{C}}$ be the poles of $f$. If $\infty$ is a pole of $f$ of order $d$, then in the chart $w = 1/z$ near $\infty$, the function $g(w) := f(1/w)$ has a pole of order $d$ at $w = 0$. Define
\begin{align*}
\tilde{f}: \widehat{\mathbb{C}} &\to \widehat{\mathbb{C}} \\
z &\mapsto f(z) - \sum_{k=1}^{d} a_{-k} z^k
\end{align*}
where $a_{-1}, \dots, a_{-d}$ are the coefficients of the principal part of $g$ at $w = 0$, i.e., the Laurent expansion of $f$ near $\infty$ takes the form $f(z) = a_{-d} z^d + \dots + a_{-1} z + (\text{holomorphic at } \infty)$. Subtracting the polynomial $\sum_{k=1}^d a_{-k} z^k$ removes the pole at $\infty$ without introducing new poles. So we may assume WLOG that $f$ has poles only at finite points $z_1, \dots, z_N \in \mathbb{C}$.
[/step]
[step:Subtract the principal parts at each finite pole to obtain an entire function]
For each pole $z_j$ (with $j = 1, \dots, N$), let $m_j \geq 1$ denote the order of the pole and let the principal part of the Laurent expansion of $f$ at $z_j$ be
\begin{align*}
P_j(z) := \sum_{k=1}^{m_j} \frac{c_{j,-k}}{(z - z_j)^k}
\end{align*}
where $c_{j,-k} \in \mathbb{C}$ are the Laurent coefficients with $c_{j,-m_j} \neq 0$. Define
\begin{align*}
g: \mathbb{C} &\to \mathbb{C} \\
z &\mapsto f(z) - \sum_{j=1}^{N} P_j(z).
\end{align*}
At each $z_j$, the function $f - P_j$ has a removable singularity (since we subtracted the entire principal part), and the remaining terms $P_i$ for $i \neq j$ are holomorphic at $z_j$ (since $z_i \neq z_j$). Therefore $g$ extends to an entire function $g: \mathbb{C} \to \mathbb{C}$.
[guided]
The idea is classical: if we know the principal parts at all poles, subtracting them leaves a function with no poles — i.e., an entire function. This is the same principle behind the [Mittag-Leffler Theorem](/theorems/???), but here we have finitely many poles so no convergence issues arise.
For each pole $z_j$ of order $m_j$, the Laurent expansion of $f$ near $z_j$ takes the form
\begin{align*}
f(z) = \frac{c_{j,-m_j}}{(z - z_j)^{m_j}} + \dots + \frac{c_{j,-1}}{z - z_j} + (\text{holomorphic at } z_j).
\end{align*}
We define the principal part $P_j(z) = \sum_{k=1}^{m_j} c_{j,-k} (z - z_j)^{-k}$. Then $f(z) - P_j(z)$ is holomorphic at $z_j$ by construction.
Now set $g = f - \sum_{j=1}^N P_j$. At any pole $z_j$, we have:
- The term $f - P_j$ is holomorphic at $z_j$ (principal part cancelled).
- Each $P_i$ for $i \neq j$ is a rational function with pole only at $z_i \neq z_j$, hence holomorphic at $z_j$.
Therefore $g$ is holomorphic at every point of $\mathbb{C}$, i.e., $g$ is entire.
[/guided]
[/step]
[step:Apply Liouville's theorem to show $g$ is a polynomial]
We now show that $g$ is bounded near $\infty$ and hence extends to a holomorphic function on $\widehat{\mathbb{C}}$. Since $f$ is meromorphic on $\widehat{\mathbb{C}}$ and (after the reduction in the second step) has no pole at $\infty$, the function $f$ is holomorphic at $\infty$. Each principal part $P_j(z) = \sum_{k=1}^{m_j} c_{j,-k}(z - z_j)^{-k}$ satisfies $P_j(z) \to 0$ as $|z| \to \infty$. Therefore $g(z) = f(z) - \sum_j P_j(z)$ is holomorphic at $\infty$ with $g(\infty) = f(\infty)$.
An entire function that is also holomorphic at $\infty$ (i.e., meromorphic on $\widehat{\mathbb{C}}$ with no poles anywhere) must be a polynomial. More precisely, in the chart $w = 1/z$, the function $g(1/w)$ is holomorphic at $w = 0$, meaning $g$ has at most polynomial growth as $|z| \to \infty$. By [Liouville's Theorem](/theorems/???), an entire function of polynomial growth of degree at most $d$ is a polynomial of degree at most $d$. Therefore $g \in \mathbb{C}[z]$.
[guided]
Why must $g$ be a polynomial? We have established that $g$ is entire (holomorphic on all of $\mathbb{C}$) and holomorphic at $\infty$ on the Riemann sphere. In the coordinate $w = 1/z$ near $\infty$, the function $h(w) := g(1/w)$ is holomorphic at $w = 0$. This means $h$ has a Taylor expansion $h(w) = a_0 + a_1 w + a_2 w^2 + \dots$ converging near $w = 0$.
Translating back: $g(z) = h(1/z) = a_0 + a_1/z + a_2/z^2 + \dots$ for large $|z|$. But $g$ is entire, so its Taylor series at the origin $g(z) = \sum_{n=0}^\infty b_n z^n$ converges everywhere. For large $|z|$, we need $g(z) = a_0 + a_1/z + \dots$, which means $g(z) \to a_0$ as $|z| \to \infty$. An entire function that is bounded must be constant by Liouville's theorem.
Note that this argument gives $g$ constant only when $f$ has no pole at $\infty$. After the reduction step, $f$ is holomorphic at $\infty$, so $g(z) \to f(\infty)$ as $|z| \to \infty$, i.e., $g$ is bounded. By [Liouville's Theorem](/theorems/???), $g$ is constant.
If we did not perform the reduction (i.e., if we allow $f$ to have a pole of order $d$ at $\infty$), then $g(z) = f(z) - \sum_j P_j(z)$ grows like a polynomial of degree $\leq d$ as $|z| \to \infty$. By the generalised Liouville theorem (an entire function bounded by $C|z|^d$ for large $|z|$ is a polynomial of degree $\leq d$), we conclude $g$ is a polynomial.
Either way, $g \in \mathbb{C}[z]$.
[/guided]
[/step]
[step:Conclude that $f$ is a rational function]
From the previous step, $g \in \mathbb{C}[z]$ is a polynomial. Since $f = g + \sum_{j=1}^N P_j$, we have
\begin{align*}
f(z) = g(z) + \sum_{j=1}^{N} \sum_{k=1}^{m_j} \frac{c_{j,-k}}{(z - z_j)^k}.
\end{align*}
Bringing this to a common denominator $Q(z) := \prod_{j=1}^N (z - z_j)^{m_j} \in \mathbb{C}[z]$:
\begin{align*}
f(z) = \frac{g(z) \cdot Q(z) + \sum_{j=1}^N \sum_{k=1}^{m_j} c_{j,-k} \cdot \frac{Q(z)}{(z-z_j)^k}}{Q(z)} = \frac{P(z)}{Q(z)}
\end{align*}
where $P(z) := g(z) Q(z) + \sum_{j=1}^N \sum_{k=1}^{m_j} c_{j,-k} \cdot Q(z)/(z - z_j)^k \in \mathbb{C}[z]$ is a polynomial (each term $Q(z)/(z - z_j)^k$ is a polynomial since $(z - z_j)^k$ divides $Q(z)$ for $k \leq m_j$). Therefore $f = P/Q$ is a rational function.
[/step]
[step:Verify the converse: every rational function is meromorphic on $\widehat{\mathbb{C}}$]
Conversely, let $f = P/Q$ where $P, Q \in \mathbb{C}[z]$ with $Q \not\equiv 0$. We verify that $f$ defines a meromorphic function on $\widehat{\mathbb{C}}$.
**In the standard chart** ($z \in \mathbb{C}$): The function $f(z) = P(z)/Q(z)$ is a ratio of holomorphic functions. It is holomorphic at every $z_0 \in \mathbb{C}$ where $Q(z_0) \neq 0$, and has a pole of finite order at each root of $Q$ (since $P$ and $Q$ are polynomials, one can factor out the common zeros and the remaining denominator vanishes to finite order). Thus $f$ is meromorphic on $\mathbb{C}$.
**In the chart at $\infty$** ($w = 1/z$): Let $\deg P = p$ and $\deg Q = q$. In the coordinate $w = 1/z$:
\begin{align*}
f(1/w) = \frac{P(1/w)}{Q(1/w)} = \frac{a_p w^{-p} + \dots + a_0}{b_q w^{-q} + \dots + b_0} = \frac{w^q}{w^p} \cdot \frac{a_p + a_{p-1}w + \dots + a_0 w^p}{b_q + b_{q-1}w + \dots + b_0 w^q} = w^{q-p} \cdot \frac{\tilde{P}(w)}{\tilde{Q}(w)}
\end{align*}
where $\tilde{P}(w) = a_p + a_{p-1}w + \dots + a_0 w^p$ and $\tilde{Q}(w) = b_q + b_{q-1}w + \dots + b_0 w^q$ are polynomials with $\tilde{P}(0) = a_p \neq 0$ and $\tilde{Q}(0) = b_q \neq 0$. Therefore $f(1/w) = w^{q-p} \cdot (\text{holomorphic and non-vanishing at } w = 0)$, which shows that $f$ has a zero of order $q - p$ at $\infty$ if $q > p$, is holomorphic and non-zero at $\infty$ if $q = p$, and has a pole of order $p - q$ at $\infty$ if $p > q$. In all cases, $f$ is meromorphic at $\infty$.
Therefore $f$ is meromorphic on $\widehat{\mathbb{C}}$.
[guided]
We must check meromorphicity in both charts of the Riemann sphere. Recall that $\widehat{\mathbb{C}}$ has an atlas with two charts: the standard chart $z$ covering $\mathbb{C}$, and the chart $w = 1/z$ covering $\widehat{\mathbb{C}} \setminus \{0\}$ (in particular, a neighbourhood of $\infty$ corresponds to a neighbourhood of $w = 0$).
**In the standard chart:** $f(z) = P(z)/Q(z)$ is a ratio of entire functions. The zeros of $Q$ are isolated (finitely many, since $Q$ is a polynomial), and at each zero $z_0$ of $Q$, we can write $Q(z) = (z - z_0)^k Q_1(z)$ with $Q_1(z_0) \neq 0$. If $z_0$ is also a zero of $P$ of order $l$, then $f$ has a pole of order $k - l$ at $z_0$ if $k > l$, or is holomorphic there if $l \geq k$. In either case, the singularity is a pole or removable — not essential — so $f$ is meromorphic on $\mathbb{C}$.
**At $\infty$:** We substitute $z = 1/w$ and examine the behaviour as $w \to 0$. Write $P(z) = a_p z^p + \dots + a_0$ and $Q(z) = b_q z^q + \dots + b_0$. Then:
\begin{align*}
f(1/w) = \frac{a_p w^{-p} + \dots + a_0}{b_q w^{-q} + \dots + b_0}.
\end{align*}
Multiplying numerator and denominator by $w^p$ and $w^q$ respectively (i.e., multiply top and bottom by $w^{\max(p,q)}$), we can rewrite as $f(1/w) = w^{q-p} \cdot \tilde{P}(w)/\tilde{Q}(w)$ where $\tilde{P}(0) = a_p \neq 0$ and $\tilde{Q}(0) = b_q \neq 0$. The ratio $\tilde{P}/\tilde{Q}$ is holomorphic and non-vanishing at $w = 0$, so the factor $w^{q-p}$ determines the behaviour:
- If $p > q$: pole of order $p - q$ at $w = 0$ (i.e., at $\infty$).
- If $p = q$: holomorphic with value $a_p/b_q$ at $w = 0$.
- If $p < q$: zero of order $q - p$ at $w = 0$.
All three cases give meromorphic behaviour at $\infty$, completing the proof.
[/guided]
[/step]
