Prove that all entire functions that are injective are of the form \(f(z) = az + b\) with \(a,b\in \CC\) and \(a\neq 0\).
Hint: Apply the Casorati-Weierstrass theorem to \(f(1/z)\).
Solution
Let \(f: \CC \to \CC\) be an injective entire function, with Taylor series \(f(w)=\sum_{n\ge0}b_nw^n\), and let \(g(z) = f(1/z)\) on the punctured disc \(\DD^* = \{z \in \CC : 0 < \abs{z} < 1\}\). Then \(g(z)=\sum_{n\ge0}b_nz^{-n}\) on \(\DD^*\), and \(z = 0\) is an isolated singularity of \(g\).
1. \(z=0\) is not an essential singularity of \(g\).
Suppose it were. By the Casorati–Weierstrass theorem, \(g(\DD^*) = f(\CC \setminus \overline{\DD})\) is dense in \(\CC\). Since \(f\) is injective, it is nonconstant, so by the open mapping theorem \(f(\DD)\) is a nonempty open set. Hence \(f(\DD)\) meets \(f(\CC \setminus \overline{\DD})\): there are \(z_1 \in \DD\) and \(z_2 \in \CC \setminus \overline{\DD}\) with \(f(z_1) = f(z_2)\). Since \(\abs{z_1} < 1 < \abs{z_2}\), \(z_1 \neq z_2\), contradicting injectivity.
2. \(z=0\) is not a removable singularity of \(g\).
If it were, \(g\) would be bounded near \(0\), so \(f\) would be bounded on \(\abs w>R\) for some \(R\), hence bounded on \(\CC\). By Liouville’s theorem \(f\) would be constant, contradicting injectivity.
3. \(f\) is a polynomial of degree \(m\ge1\).
By steps 1 and 2, \(z=0\) is a pole of \(g\), of some order \(m\ge1\). The Laurent expansion \(g(z)=\sum_{n\ge0}b_nz^{-n}\) then has \(b_n=0\) for \(n>m\) and \(b_m\neq0\), so \(f(w)=\sum_{n=0}^m b_nw^n\) has degree \(m\).
4. \(m = 1\).
Suppose \(m\ge2\). The critical values of \(f\) are the finitely many values \(f(c)\) with \(f'(c)=0\). Choose \(c_0\in\CC\) not among them. By the fundamental theorem of algebra, \(f(z) = c_0\) has \(m\) roots counted with multiplicity, and each is simple because \(f'\) does not vanish there. So \(f\) takes the value \(c_0\) at \(m\ge2\) distinct points, contradicting injectivity.
5. Q.E.D.
By steps 3 and 4, \(f(z) = \boxed{az + b}\) with \(a=b_1\neq0\) and \(b=b_0\).