Find all proper holomorphic maps \[ f:\mathbb C^*\to\mathbb C. \]
Find all proper holomorphic maps \[ g:\mathbb C\to\mathbb C^*. \]
Solution
For part (a), the proper maps are exactly the Laurent polynomials \[ \boxed{ f(z)=\sum_{k=-m}^{n}a_kz^k, \qquad m,n\ge1, \quad a_{-m}a_n\neq0.} \] Thus a proper map must have a genuine pole at both ends \(0\) and \(\infty\) of \(\mathbb C^*\).
For part (b), there are no proper holomorphic maps \(\mathbb C\to\mathbb C^*\).
1. If \(f:\mathbb C^*\to\mathbb C\) is proper, then\[ |f(z)|\longrightarrow\infty \quad\text{as }z\to0 \quad\text{and as }z\to\infty. \]
Fix \(R>0\). The closed disk \[ K_R=\{w\in\mathbb C:|w|\le R\} \] is compact. Properness implies \(f^{-1}(K_R)\) is compact in \(\mathbb C^*\). Hence it is contained in an annulus \[ \varepsilon_R\le |z|\le M_R \] for some \(0<\varepsilon_R<M_R<\infty\). Therefore \[ |z|<\varepsilon_R \quad\text{or}\quad |z|>M_R \quad\Longrightarrow\quad |f(z)|>R. \] Since \(R\) is arbitrary, \(|f(z)|\) tends to infinity at both ends.
2. Such an \(f\) has a pole at both \(0\) and \(\infty\).
By 1, \(f\) is nonzero near \(0\) and \[ \frac1{f(z)}\longrightarrow0 \qquad(z\to0). \] Thus \(1/f\) has a removable singularity at \(0\) with value \(0\), so \(f\) has a pole at \(0\).
For the other end, apply the same argument to \[ F(\zeta)=f(1/\zeta) \] near \(\zeta=0\). Since \(|F(\zeta)|\to\infty\), the point \(\infty\) is a pole of \(f\).
3. Therefore every proper \(f:\mathbb C^*\to\mathbb C\) is a Laurentpolynomial\[ f(z)=\sum_{k=-m}^{n}a_kz^k \]with \(m,n\ge1\) and \(a_{-m}a_n\neq0\).
A holomorphic function on \(\mathbb C^*\) has a Laurent expansion valid on the annulus \(0<|z|<\infty\). Because \(0\) is a pole, only finitely many negative powers occur; because \(\infty\) is a pole, only finitely many positive powers occur. Hence the Laurent series is finite.
The pole at \(0\) forces a nonzero negative extreme term \(a_{-m}z^{-m}\) with \(m\ge1\), and the pole at \(\infty\) forces a nonzero positive extreme term \(a_nz^n\) with \(n\ge1\).
4. Conversely, every Laurent polynomial of the form in 3 is properas a map \(\mathbb C^*\to\mathbb C\).
The extreme terms give \[ |f(z)|\longrightarrow\infty \qquad(z\to0) \] and \[ |f(z)|\longrightarrow\infty \qquad(z\to\infty). \] Let \(K\subset\mathbb C\) be compact and choose \(R\) with \(K\subset\{|w|\le R\}\). The two limits provide \(0<\varepsilon<M<\infty\) such that \[ f^{-1}(K)\subset\{\varepsilon\le|z|\le M\}. \] The set \(f^{-1}(K)\) is closed, and the displayed annulus is compact. Therefore \(f^{-1}(K)\) is compact, proving properness.
5. Now suppose \(g:\mathbb C\to\mathbb C^*\) is proper. For every\(R>1\), there is \(M_R>0\) such that for \(|z|>M_R\) one has either\[ |g(z)|<\frac1R \]for all such \(z\), or\[ |g(z)|>R \]for all such \(z\).
The annulus \[ A_R=\left\{w:\frac1R\le|w|\le R\right\} \] is compact in \(\mathbb C^*\). Hence \(g^{-1}(A_R)\) is compact in \(\mathbb C\), so it lies in some disk \(|z|\le M_R\).
Thus, on the connected exterior region \(|z|>M_R\), the continuous function \(|g|\) avoids the whole interval \([1/R,R]\). It cannot take values on both sides of that interval, by the intermediate value theorem on a path in the exterior region. Therefore exactly one of the two displayed alternatives holds throughout that region.
6. Consequently, as \(z\to\infty\), either\[ g(z)\longrightarrow0 \]or\[ |g(z)|\longrightarrow\infty. \]
Apply 5 first with \(R=2\). Outside a sufficiently large disk, either \(|g|<1/2\) everywhere or \(|g|>2\) everywhere.
In the first case, apply 5 with any \(R\ge2\). On a sufficiently far exterior region the alternative \(|g|>R\) is incompatible with the already valid inequality \(|g|<1/2\), so necessarily \(|g|<1/R\). Thus \(g(z)\to0\).
In the second case the alternative \(|g|<1/R\) is similarly impossible for every \(R\ge2\), so \(|g(z)|>R\) eventually. Thus \(|g(z)|\to\infty\).
7. Both alternatives in 6 are impossible for an entire nonvanishingfunction.
If \(g(z)\to0\) at infinity, then \(g\) is bounded outside a large disk and is bounded on the remaining compact disk. Hence \(g\) is a bounded entire function. Liouville’s theorem makes it constant, and its limit at infinity would force that constant to be \(0\), contradicting \(g(\mathbb C)\subset \mathbb C^*\).
If \(|g(z)|\to\infty\), then \(1/g\) is entire because \(g\) never vanishes, and \[ \frac1{g(z)}\longrightarrow0 \qquad(z\to\infty). \] The same Liouville argument forces \(1/g\equiv0\), impossible.
Therefore no proper holomorphic map \(\mathbb C\to\mathbb C^*\) exists.