Proving that all entire and injective functions take the form $f = ax + b$?

Question

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 $a \ne 0$.

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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:

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

2. 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?

Answer 1

The first part of what you say is right; that's 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 takes 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.

Answer 2

1. If $0$ is a removable singularity of $g$, Riemann'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).

2. 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.

Answer 3

Suppose that $f$ is non constant entire.

Case $1): f$ is a polynomial.

If $m=$deg $f>1,$ then $f$ has $m$ roots. If there are two distinct roots, then $f$ is not injective so suppose that all these $m$ roots are the same, say $u$, then $f(z)= (z-u)^m$ (wlog assume $f$ to be monic); and clearly $f(1+u)= f(-1+u)$ if $m$ is even. So suppose that $m$ is odd and note that $f(1+u)= f(\omega_m+u),$ where $\omega_m$ is the principal $m-$ th root of $1$. Hence, $f$ can't be $1-1$ in this case. So $m\le 1$.

Case $2): f$ is not a polynomial.

Lemma: $f$ does not have a pole at $\infty$.

Pf: If not, then $f(1/z)$ has a pole at $0$, say of order $k$. It follows that $\lim_{z\to \infty}z^k f(1/z)$ exists finitely and is non zero, whence it follows that $|f(z)|\le A|z|^k$ for some $r_0>0$ and for all $|z|\ge r_0$. Writing power series of $f$ around $0$ to get: $f(z)= \sum_{n=0}^\infty a_n z^n,$ it follows by Cauchy's inequality that $a_n=0$ for all $n>k$. Hence $f$ is a polynomial hence a contradiction. $\large\square$.

So $f$ must have either an isolated essential singularity at $\infty$ or a removable singularity. But in case of removable singularity, it follows that $f$ is bounded in a neighbourhood of $\infty$, hence bounded on the whole plane whence it follows by Liouville's that $f$ is a constant, which contradicts the fact that $f$ is injective. So $f$ must have an isolated essential singularity at $\infty$. Consider nbd. $1<|z|$ of $\infty$, in the nbd. By great Picard's theorem,$f$ attains every complex value with one possible exception , say $w$ in this case. Choose $u: |u|\le 1$ and $f(u)\ne w$. (Such $u$ can be chosen else $f$ is not injective). So by great Picard's theorem, there exists some $v: |v|>1$ such that $f(v)= f(u)$ hence $f$ can't be $1-1$.

It follows that $f$ is a polynomial and by case $1)$, it follows that deg $f\le 1.$