Let $M$ be a compact Riemann surface and $S \subset M$ is discrete. Suppose $f: M \setminus S \to \mathbb{C}$ is analytic and nonconstant. Show that the image of $M \setminus S$ under $f$ is dense in $\mathbb{C}$.
So if we try contradiction, we assume $f(M \setminus S)$ isn't dense in $\mathbb{C}$, then there exists an neighborhood $U\subset \mathbb{C}$ that does not intersect $f(M \setminus S)$. The problem hints that $S$ must then contain poles, but I don't see how this follows? Also I don't know how you could get a contradiction?
If it's not dense, e.g. if $a\in\mathbb C$ is not in the closure of the image, then $1/(f-a)$ is holomorphic on $M\backslash S$ and bounded, so by removable singularity thm we can extend $1/(f-a)$ to a holomorphic (and still bounded) function on $M$, but its image is both compact and open.