Are meromorphic functions open? 
Is a non-constant meromorphic function on a region $G\subseteq\mathbb{C}$ an open mapping of $G$ into $\mathbb{C}?$ Is it an open mapping of $G$ into $\mathbb{C}_{\infty}$?

I believe the answer to the first question is no and the second is yes.
a) For the first part, I could only think in the easiest example of a meromorphic function: $1/z$. We may choose $G=\mathbb{C}\setminus\{0\}$, though I don't know how to prove it's not open from $G$ to $\mathbb{C}_{\infty}$.
b) I know we only have to see at the $z's$ in $G$ such that $f(z)=\infty$. Suppose $f$ has only one pole in $G$. The only hint I could find is if this happens, then we need to consider $1/f(z)$. So as $1/f:G\to\mathbb{C}$ is analytic and non-constant then it is open. But how can I conclude?
Any hint would be appreciated.

$1/f$ is not analytic. If $f$ has a zero?
To prove that is open to $\mathbb{C}_\infty$ consider a function $g(z)$, the extension of $f$ (ie, define g in poles as $\infty$)
