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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| is open because any holomorphic mapping is an open mapping. Then $f(\{ |z|>r \}) \cap f(\{|z|, 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 $a \neq 0$.