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I am looking at the proof of the following statement: Let $f$ be an entire function, that is also injective. Then $f$ take the form $f(z) = az+b$ with $a, b \in \Bbb C$ and $a \neq 0$.

The proof is given as follow: Assume $f$ is an entire injective function. Then $f$ is non-constant, so $g(z) :=f(\frac{1}{z})$ has either a pole or an essential singularity at $z = 0$.

1) Why we have that I can't be a removable zero too?

We will show first that the singularity at $0$ cannot be an essential singularity. If it were an essential singularity, then the Cazorati-Weierstrass theorem would imply that the set $g(B(0, 1) \setminus {0})$ is dense in $\Bbb C$. However, $g(B(2,\frac{1}{2}))$ is an open set by the open mapping theorem. Therefore these two sets intersect, which shows that $g(z)$ and hence $f(z)$ is not injective.

2) I don't see why we need to dense set to conclude that $g$ and hence $f$ is not injective.

Therefore, the singularity at $z = 0$ must be a pole, implying that $f(z)$ is a polynomial. Suppose $f(z)$ is a polynomial of degree $m$. Then $f$ has $m$ roots, counting multiplicity. Evidently, if $f$ has two distinct roots, then $f$ is not injective. Thus $f(z) = c(z − z_0)^m$ for some complex numbers $c$ and $z_0$. However, for $m \geq 2$ such functions are also non-injective: $f(z_0 + 1) = c = f(z_0 + e^{ 2πi/m})$. Thus $m = 1$ and $f(z)$ is a linear polynomial (evidently $c \neq 0$ since $f$ is non-constant).

I am pointed out my 2 questions. Thanks for some help!

Frederick Manfred
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2 Answers2

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  1. If it were removable $g$ would be bounded in a neighborhood of $0$, hence $f$ would be bounded and therefore constant.
  2. A dense set intersects any non-empty open set non-trivially, hence there are $x\in B(0,1)\setminus \{0\}, y\in B(2,1/2)$ with $g(x)=g(y)$, i.e. $f(1/x)=f(1/y)$. As these two sets are disjoint we have $x\ne y$. If $g(B(0,1)\setminus 0)$ were not dense there would be no reason at all why it should intersect with $g(B(2,1/2))$
leoli1
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First question: If it is removable singularity then $f$ is bounded and hence constant by Liouville's Theorem. That contradicts injectivity.

Second question: $B(0,1)\setminus \{0\}$ and $B(2,\frac 1 2 )$ are disjoint. Since $g$ is injective the images of these two sets under $g$ have no point in common. But the dense set $g(B(0,1)\setminus \{0\})$ intersects every non-empty open set, in particular $g(B(2,\frac 1 2))$. So we have a contradiction.

Kavi Rama Murthy
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