Proof explanation: the codomain of an entire, non constant function is dense in $\mathbb{C}$

Question

By contraposition: suppose that $f(\mathbb{C})\subseteq \mathbb{C}\backslash B(z_0,r)$ for a certain $z_0\in\mathbb{C}$ and $r>0$. First, let's consider $z_0=0$, then $1/f:\mathbb{C}\to \bar{B}(0,1/r)$ is holomorphic on the whole set $\mathbb{C}$. ...

I don't see why we can conclude that $1/f$ is an entire function? What about possible zeroes of $f$? Isn't $1/f$ holomorphic on $\mathbb{C}\backslash\{\text{zeroes of }f\}$?

Thanks.

Answer 1

You assume that $f(\mathbb C) \subseteq \mathbb C \setminus B(0, r) \subseteq \mathbb C \setminus \{0\}$. So $f$ does not have any zeros, whence $1/f$ is an entire function.

Answer 2

First, note that a composition of holomorphic functions is holomorphic.
$f$ is holomorphic on $\mathbb{C}$ and $g(z)=\frac{1}{z}$ is holomorphic on $\mathbb{C} \setminus \{0\}$.
So if you know $f$ has no zeroes, because $f(\mathbb{C}) \subseteq \mathbb{C} \setminus \{0\}$, then the composition $g \circ f = \frac{1}{f}:\mathbb{C} \rightarrow \mathbb{C}$ is a composition of holomorphic functions, and as such holomorphic (which means it's entire beacause its domain is $\mathbb{C}$).