# 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:

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 $$f(\{|z| > r \} \cap f(\{|z|? $$f(\{|z| > r \})$$ is dense, but how does $$f(\{|z| being open guarantee that their union is non-empty?

• Can you give the reference to the picture you took? like the source of the problem? Commented Jul 12, 2018 at 11:39

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 $$f(\{|z| is open it means that there is some ball inside $$f(\{|z| that only takes points from $$f(\{|z|; now because Cassorati-Weierstrass assures you that $$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| is non-empty.

• Sorry, why is the closed circle a compact set? Commented Oct 9, 2014 at 0:52
• The Heine-Borel theorem tells us that in $\mathbb{R}^n$, and thus in $\mathbb{C}$, the compact sets are precisely those sets that are both closed and bounded.
– jxnh
Commented Oct 9, 2014 at 1:19
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.

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 $$f\le 1.$$