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Prove that all entire functions that are also injective take the form $f(z)=az+b$ with $a,b\in\Bbb C$.

Solution:

We take $g : \Bbb C^* \to \Bbb C$, $g( z) = f(1/z)$, 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 containing the origin. But $f$ is bounded on this closed circle, because $f$ is continuous, therefore, $f$ is bounded. Since $f$ is entire and bounded, by Liouville'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 Casorati-Weierstrass Theorem, if we chose a punctured disk centred at the origin $D^*$, then $g ( D^*)$ is dense in $\Bbb C$. This implies $f (\{ \lvert z\lvert > r\})$ is dense in $\Bbb C$. But $f (\{ \lvert z\lvert < r\})$ is open because any holomorphic mapping is an open mapping. Then $f (\{ \lvert z\lvert > r\})\cap f (\{ \lvert z\lvert < r\})\ne \emptyset$, 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. Since $f$ 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 $a, b \in \Bbb C$ and $a \ne 0$.

I have a couple questions:

  1. Doesn't it need to be the other way around: $f (\{ \lvert z\lvert > r\})$ is open and $f(\{ \lvert z\lvert < r\})-\{0\}$ is dense? Because Casorati Weierstraß states that the image of a surrounding of the singularity is dense in $\mathbb C$.
  2. "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" Why is this?
  3. "Since f is injective, the polynomial can have at most one root." Why is this? The polynomial $z^3$ is injective too.
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  1. Let $D^*$ a punctured disk centered at the origin. Then $g(D^*)$ is dense and, if $D^*=\left\{z\in\mathbb{C}\,:\,0<|z|<\frac1r\right\}$, then $$ g(D^*)=f\left(\left\{z\in\mathbb{C}\,:\,|z|>r\right\}\right). $$ And, since $g(D^*)$ is dense, $f\left(\left\{z\in\mathbb{C}\,:\,|z|>r\right\}\right)$ is dense.
  2. If $f(z)=a_0+a_1z+a_2^2+a_3z^3+\cdots$, then $g(z)=a_0+\frac z{a_1}+\frac{z^2}{a_2}+\frac{z^3}{a_3}+\cdots$. But this sum is a finite sum, which means that $a_n=0$ if $n\gg0$.
  3. Actually, $z\mapsto z^3$ is not injective since, for instance $$ 1^3=\left(-\frac12+\frac{\sqrt3}2i\right)^3=1. $$
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