Suppose \(f:\mathbb C\to\mathbb C\) is injective and entire. Prove that there are \(a,b\in\mathbb C\) with \(a\ne0\) such that \[ f(z)=az+b \] for every \(z\in\mathbb C\).
Solution
1. The derivative \(f'(z)\) is nonzero for every \(z\in\CC\).
Suppose \(f'(z_0)=0\). Since \(f\) is not constant, its Taylor expansion gives an integer \(m\geq2\) and a holomorphic function \(u\) with \(u(z_0)\neq0\) such that \[ f(z)-f(z_0)=(z-z_0)^m u(z). \] On a sufficiently small disk about \(z_0\), the nonvanishing function \(u\) has a holomorphic \(m\)-th root \(v\). Put \[ H(z)\coloneqq(z-z_0)v(z). \] Then \(H'(z_0)=v(z_0)\neq0\), so after shrinking the disk, \(H\) is a biholomorphism onto a neighborhood of \(0\), while \[ f(z)-f(z_0)=H(z)^m. \] Two sufficiently small numbers differing by a nontrivial \(m\)-th root of unity have the same \(m\)-th power, and their distinct preimages under \(H\) therefore have the same image under \(f\). This contradicts injectivity. Thus \(f'(z_0)\neq0\).
2. For\[ g(\zeta)\coloneqq f(1/\zeta), \qquad \zeta\neq0, \]the isolated singularity at \(0\) is a pole.
The map \(g\) is injective on \(\CC\sm\{0\}\).
Suppose first that \(0\) were an essential singularity. Choose \(\zeta_0\neq0\). By step 1 and the chain rule, \(g'(\zeta_0)\neq0\), so the holomorphic inverse function theorem gives a neighborhood \(U\) of \(\zeta_0\), with closure disjoint from \(0\), such that \(g(U)\) contains an open neighborhood \(V\) of \(g(\zeta_0)\). By the Casorati–Weierstrass theorem, the image under \(g\) of every punctured neighborhood of \(0\) is dense in \(\CC\). Taking such a punctured neighborhood disjoint from \(U\), there is a point \(\zeta\) in it with \(g(\zeta)\in V\). Since \(V\subseteq g(U)\), some \(u\in U\) has \(g(u)=g(\zeta)\), contradicting injectivity of \(g\).
Suppose instead that the singularity were removable. Then \(g\) would be bounded near \(0\), so \(f\) would be bounded outside a sufficiently large disk. It is also bounded on that disk by continuity. Liouville’s theorem would make \(f\) constant, contradicting injectivity. Hence the singularity is neither essential nor removable, so it is a pole.
3. The entire function \(f\) is a polynomial.
If the pole of \(g\) at \(0\) has order \(m\), then for some constants \(C,R>0\), \[ \abs{f(z)}\leq C\abs{z}^m \] whenever \(\abs{z}\geq R\). Enlarging the constant gives \[ \abs{f(z)}\leq C(1+\abs{z}^m) \] for all \(z\in\CC\). Let \[ f(z)=\sum_{k=0}^\infty a_kz^k. \] For \(k>m\), Cauchy’s estimate on the circle \(\abs{z}=r\) gives \[ \abs{a_k} \leq \frac{C(1+r^m)}{r^k}. \] Letting \(r\to\infty\) yields \(a_k=0\). Thus only finitely many Taylor coefficients are nonzero.
4. The polynomial \(f\) has degree \(1\).
1. Hence \(\deg f=1\).
5. Therefore there are \(a,b\in\CC\) with \(a\neq0\) such that\[ \boxed{f(z)=az+b} \]for every \(z\in\CC\).
By step 4, \(f\) is a degree-one polynomial, so it has the displayed form with nonzero leading coefficient.
6. Q.E.D.
Step 5 is the required conclusion.