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Injective Entire Functions


Prove that all entire functions that are also injective, take the form f(z)=az+b with a,b \in \mathbb{C} and a \neq 0.

Solution:
We take g: \mathbb{C^*} \to \mathbb{C}, g(z)=f(\frac{1}{z}), which is holomorphic everywhere except the origin. Now, we try to find out what type of singularity is the origin for g.
If the origin is a removable singularity for g, then g is bounded on a closed disk centred at the origin, which implies that f is bounded outside a closed circle containinf the origin. But f is bounded on this closed circle, because f is continuous, therefore, f is bounded. Because f is entire and bounded, by Liouiville’s Theorem, f is constant. This contradicts the injectivity of f. So the origin is not a removable singularity for g.
Suppose now that 0 is an essential singularity for g. Then, by Cassorati-Weierstrass Theorem, if we chose a punctured disk centred at the origin, D^* then g(D^*) is dense in \mathbb{C}. This implies f(\{ |z|>r \}) is dense in \mathbb{C}. But f(\{|z|<r \}) is open because any holomorphic mapping is an open mapping. Then f(\{ |z|>r \}) \cap f(\{|z|<r \})\neq \emptyset, which is again a contradiction with the injectivity of f.
Therefore 0 is a pole for g. Since the Laurent expansion is unique, and the principal part of g is the same as the analytic part of f, it follows that the analytic part of f has finitely many terms, which implies that f is a polynomial. Because f is injective, the polynomial can have at most one root. Because f is not constant, we conclude that the only expression of f can be of the form f(z)=az+b, where a,b \in \mathbb{C} and a \neq 0.

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