Proving that all entire and injective functions take the form $f = ax + b$? 
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 $a \ne 0$.

Original text image
I'm a little confused at both the overall logic in this proof. Are we simply using $g(z)$ to make conclusions about $f(z)$, because $g(z)$ is the reciprocal of $f$? Is the proof assuming that $f$ is injective and entire (all the while knowing that it has some sort of singularity at $z = 0$), and then trying to reach contradictions in the essential singularity and removable singularity cases? Then, once it concludes that $z = 0$ is a pole singularity, it reaches the conclusion that $f$ must be of the form $f(z) = az + b$?
Also, more specific questions about the different cases:

*

*Removable singularity case: Why is $f$ bounded on the closed circle if $f$ is continuous? Am I missing something simple?


*Essential singularity case: Why exactly is $f(\{|z| > r \} \cap f(\{|z|<r\}) \neq \emptyset$? $f(\{|z| > r \})$ is dense, but how does $f(\{|z|<r\}$ being open guarantee that their union is non-empty?
 A: The first part of what you say its right, thats what the author is trying to do.
For the other 2 specific questions.
1.$f$ Is bounded because the closed circle is a compact set and $f$ is continuous.
2.Because the set $f(\{|z|<r\})$ is open it means that there is some ball inside $f(\{|z|<r\})$ that only take points from $f(\{|z|<r\})$, now because Cassorati-Weierstrass assures you that $f(\{|z| > r \}$ is dense in all $\mathbb{C}$, it must have some point inside that ball, thus the intersection, $f(\{|z| > r \} \cap f(\{|z|<r\})$ is non-empty.
A: *

*If $0$ is a removable singularity of $g$, Riemman's theorem shows that there is an open disk $D:=D(0,\frac{1}{r})$ in which $g$ is bounded, for some $r>0$. Note that
$$
\textstyle \{|z| < \frac{1}{r}\} = \{\frac{1}{|z|} > r\},
$$
and so $g(D)$ is bounded if and only if $f(\mathbb{C} - D)$ is bounded. But $f$ is also bounded on $D[0,r]$, because this is a compact set and we may use the extreme value theorem (any continuous function $K\to\mathbb R$ from a compact set attains maxima and minima).


*Denseness is equivalent to intersecting any non-empty set. The open mapping theorem assures that $f(\{
|z| < r)$ is open, and therefore intersects any dense set.
