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    Surjective holomorphic maps between the right half-plane and the plane

    P-JHUFA05ANH

    Area
    Complex Analysis
    Topics
    Conformal Maps, Liouville's Theorem
    Status
    draft
    1. Let \(H=\{z\in\mathbb C:\operatorname{Re}z>0\}\) be the right half-plane.

    1. Does there exist a surjective holomorphic map \(f : H \to \mathbb { C } ?\) Either give an example or prove that one does not exist.

    2. Does there exist a surjective holomorphic map \(f : \mathbb { C } \to H ?\) Either give an example or prove that one does not exist.

    Solution

    The answer to (a) is yes; the answer to (b) is no.

    1. The polynomial \(p(z)=z^2-z\) maps \(H\) onto \(\mathbb C\).

    The restriction of this polynomial to \(H\) is holomorphic. For any \(w\in\mathbb C\), let \(z_1,z_2\) be the two roots, counted with multiplicity, of \(z^2-z-w=0\). The quadratic formula gives \(z_1+z_2=1\), hence \(\operatorname{Re}z_1+\operatorname{Re}z_2=1\). At least one root has real part at least \(1/2\), so lies in \(H\). That root maps to \(w\). This proves surjectivity onto \(\mathbb C\).

    2. Every holomorphic map from \(\mathbb C\) to \(H\) is constant.

    For such a map \(f\), put \[ g(z)=\frac{f(z)-1}{f(z)+1}. \] The denominator never vanishes because \(\operatorname{Re}f>0\). Moreover \(|f(z)+1|^2-|f(z)-1|^2=4\operatorname{Re}f(z)>0\), so \(g\) is entire and \(|g|<1\). Liouville’s theorem gives \(g=c\) for some \(|c|<1\) [@SS03]. Solving for \(f\) gives \(f=(1+c)/(1-c)\), a constant. It cannot be surjective onto \(H\), which contains more than one point. Thus no surjective map in direction (b) exists.

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