Injective holomorphic functions on $\mathbb{C} \setminus \{0\}$ anyone can give me a hint, please.
Show that if $f \   \colon \mathbb{C} \setminus \{0\} \to \mathbb{C}$ is holomorphic and injective, then for some $ c \in \mathbb{C}$  $f(\mathbb{C} \setminus \{0\})= \mathbb{C} \setminus \{c\}$
 A: Well-know fact: If an analytic function $f$ has a zero of order $m$ at $z_0,$ then there is a deleted neighborhood of $z_0$ in which $f$ is $m$-to-$1.$ Corollary: If $g$ has a pole of order $m$ at $z_0,$ then there is a deleted neighborhood of $z_0$ in which $g$ is $m$-to-$1.$ 
To our problem: The injectivity of $f$ implies $f$ has a removable singularity or a pole at $0$ (Casorati-Weierstrass + open mapping theorem). I'll leave the rem0vable singularity case to you. Suppose, then, $f$ has a pole of order $m$ at $0.$  Then by the exercise, $m=1.$ Thus there is a constant $a\ne 0$ such that
$$f(z) = \frac{a}{z} + g(z),\,\, z\in \mathbb C\setminus \{0\},$$
where $g$ is entire. It follows that $f(1/z) = az +g(1/z).$ But note $f(1/z)$ is also injective on the same domain. Using the same reasoning as above, we see $g$ is a polynomial of degree $\le 1.$ It follows that there are constants $a,b,c$ such that
$$f(z) = \frac{a}{z} + b + cz.$$
Suppose both $a,c\ne 0.$ Then $f'(z) = -a/z^2 + c.$ Thus $f'$ has zeros in $\mathbb C\setminus \{0\},$ which implies $f$ is not injective there, contradiction. Therefore either $a=0$ or $c=0$ (but not both). From this the conclusion follows.
