- Prove that all entire functions that are also injective take the form \(f ( z ) = a z + b\) with \(a , b \in \mathbb { C }\) , and \(a \neq 0\)
[Hint: Apply the Casorati-Weierstrass theorem to \(f ( 1 / z ) . ]\)
Solution
Let \(f\) be entire and injective, and define \(g(z)=f(1/z)\) on a punctured neighborhood of \(0\).
The singularity of \(g\) at \(0\) cannot be removable: otherwise \(f(z)\) would have a finite limit as \(z\to\infty\), hence would be bounded outside a large disc and therefore bounded on all of \(\mathbb C\); Liouville would make \(f\) constant, contradicting injectivity.
It also cannot be essential. By Casorati-Weierstrass, if \(0\) were an essential singularity of \(g\), then for any \(w\in\mathbb C\) and every punctured neighborhood of \(0\), the values of \(g\) would come arbitrarily close to \(w\). Choose \(z_0\ne0\) and set \(w=f(z_0)\). Then there exists a sequence \(\zeta_n\to0\), with \(\zeta_n\ne 1/z_0\), such that \[ f(1/\zeta_n)=g(\zeta_n)\to f(z_0). \] Because \(f\) is injective and entire, the inverse function theorem applies at \(z_0\) once we note \(f'(z_0)\ne0\) (otherwise the local power-series expansion would have multiplicity at least \(2\), contradicting local injectivity). Hence \(f\) is locally one-to-one with a continuous local inverse near \(f(z_0)\), forcing \(1/\zeta_n\to z_0\), impossible since \(|1/\zeta_n|\to\infty\).
Therefore \(g\) has a pole at \(0\). Equivalently, \(f\) has a pole at infinity, so \(f\) is a polynomial. An injective polynomial must have degree \(1\): if \(\deg f=d\ge2\), then \(f'\) has degree \(d-1\) and hence has a zero, contradicting injectivity as above. Thus \[ f(z)=az+b, \qquad a\ne0. \]