# Understanding proof for holomorphic function

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.

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 $$1^3=\left(-\frac12+\frac{\sqrt3}2i\right)^3=1.$$