If $f:\Omega \to C$ is analytic on $\Omega$ and we know that $f$ is not constant, how can I show that if a point $z_o$ is in $\Omega$ and $f(z_0) = w$ then there is some disc around $z_0$ where $f(z) \ne w$ for any $z$ in the disc.
I know that this statement is true if $w=0$, however I'm trying to prove it for any $w$. I also know that if $f,g:\Omega \to C$ are analytic on $\Omega$ and the set {$z : f(z)=g(z)$} has an accumulation point in $\Omega$ then $f(z) \equiv g(z)$ in $\Omega$.
My attempt: I'm thinking that I can try to prove this by contradiction. So if I assume that there is no disc around $z_0$ where $f(z) \ne w$ for all $z$ in the disc, that means that I can continuously shrink the radius of my disc, and then the sequence of points $z$, such that $f(z) = w$ converges to $z_o$, making $z_0$ a cluster point. Would this then imply that $f(z) = w$ for all $z$ in $\Omega$?