Let $f$ be entire and injective. What can you say about the isolated singularity $0$ of $f(1/z)$? The theorem of Casorati-Weierstrass could help.
I would guess that $0$ is not an essential singularity. Is that right and if so, how can you prove it?
[Math] Type of the singularity in $0$ of $f(1/z)$
complex-analysissingularity
Best Answer
If $f$ is entire function, $f(z) = \sum\limits_{n=0}^{\infty} a_n z^n$ has infinity radius of convergence. Suppose that $f(\frac{1}{z})$ has a removable singularity at $z=0$.
Laurent expansion of $f(\frac{1}{z}) = \sum\limits_{n=-\infty}^{\infty} b_n z^n$ with $b_{-n} = a_n$ for $n \ge 0$ and $b_n = 0$ for $n > 0$. We know that a analytic function has a removable singularity at $z_0$ iff your laurent expansion has $b_n = 0$ $\forall n\le -1$. So $f(\frac{1}{z}) = \sum\limits_{n=-\infty}^{\infty} b_n z^n = b_0 = a_0$ is constant, i.e., is not injective.
In fact, we proved that if $f$ is entire function and $f(\frac{1}{z})$ has a removable singularity on $z=0$.
Analogously, if $f$ is entire function and $f(\frac{1}{z})$ has a pole of order $m$ then $f$ is polynomial of degree $m$, with $m$ zeros and therefore is no injective except by $m=1$. If $m=1$ then $f(z) = az + b$ for some $a,b \in \mathbb{C}$.
There remains only the case that $f(\frac{1}{z})$ has a essential singularity on $z=0$. Suppose $f$ is injective and $f(\frac{1}{z})$ has an essential singularity on $z=0$. By Casorati-Wierstrass, for every $\delta>0$, if $V_{\delta} = \{z \in \mathbb{C} : 0 < |z| < \delta\}$ and $g(z) = f(\frac{1}{z})$ then $ \overline{g(V_{\delta})} = \mathbb{C}$. Let $z\in\mathbb{C}$, $\exists (a_n)$ sequence on $V_{\delta}$ such that $z_n=g(a_n) \longrightarrow z$.
g continuous on $V_{\delta}$ $\implies$ if $a_n \longrightarrow a$ then $g(a) = z$. If $z \in \mathbb{C} - g(V_{\delta})$ then $a$ is limit point of $V_{\delta}$. Analogously for $V_{\frac{\delta}{2}}$, we get $b_n \longrightarrow b$ with $g(b) = z$. How $f$ is injective, we get $\frac{1}{a} = \frac{1}{b} \implies a = b$. So $a$ is a limit point of both, i.e., $a=0$. Absurd !