Suppose $D$ is a domain in $\Bbb{C}$ such that $f$ is holomorphic on $D$. Let $U \subset D$ be an open set with $f$ identically zero on $U$.
Suppose that $U$ is not the whole of $D$. Show that $\exists w \in D$ such that $f(w)=0$, and $\forall \epsilon \gt0$ there are points $z_1, z_2 \in B_{\epsilon}(w)$ such that $f(z_1)=0$ and $f(z_2) \neq 0$
Hint: $D - \bar{U}$ is open, non empty and that $\forall z \in D - \bar{U}$, every neighborhood of $z$ contains a point $\alpha$ such that $f(\alpha) \neq 0$.
My attempt:
Let $z$ be a point on the closure of $U$. Thus, every neighborhood of $z$ contains a point $z_1$ in $U$ and a point $z_2$ in $D-\bar{U}$. Since $f$ is identically zero in $U$, $f(z_1)=0$. For $z_2$, by the hint, there is a ball around $z_2$ contained within the neighborhood of $z$. Within the ball around $z_2$, there exists a point $\alpha$ such that $f(\alpha) \neq 0$. This $\alpha$ is in the neighborhood of $z$, as desired.
Of course the question requires a point $w$ s.t. $f(w) = 0$. What I have shown is for a point on the closure of $U$. I'm not quite sure how to show this for a $w$ s.t. $f(w) = 0$.