If $f(z)g(z)=0$ then $f(z)=0$ or $g(z)=0$ Let $D$ be a domain and let $f,g$ be analytic in $D$. I need to prove that if $f(z)g(z)=0$ for all $z\in D$, then $f(z)=0$ for all $z\in D$ or $g(z)=0$ for all $z\in D$.This is my answer. 
Assume there exists $z_0 \in D$ such that $f(z_0)\neq 0$ and $g(z_0)\neq 0$. Since $f,g$ are continuous in $D$ there exists $r_1,r_2>0$ such that $f(z)\neq 0$ for all $z\in D_{r_1}(z_0)$ and $g(z)\neq 0$ for all $z\in D_{r_2}(z_0)$. Ler $r=min${$r_1,r_2$}. Then, $$f(z)g(z)\neq0,z\in D_r(z_0)$$
This is a contradiction. Thus $f(z)=0$ for all $z\in D$ or $g(z)=0$ for all $z\in D$. Is this correct? Any improvements?Better answers? Thanks
 A: The idea is to note the discreetness of the zero set of a nonzero analytic function. Let $f \not\equiv 0$.If $z_0 \in D$ is such that $f(z_0) = 0$, then the zero at $z_0$ of $f$ must have a finite order $m$. ( Otherwise, $f$ is zero on an open set, implying $f \equiv 0$). Hence, $f(z) = (z-z_0)^m h(z)$ where $h(z_0) \ne 0$. So $f$ is nonzero on a punctured disk around $z_0$. Remember that $fg \equiv 0$. So, $g \equiv 0$ on the punctured disk, which is an open set, implying that $g \equiv 0$. 
So the important thing to note is that an analytic function on a domain is never zero on an open set. To prove it, note that if $z_0$ is a zero of an entire function, then either the function is zero on an open set containing $z_0$, or it is nonzero in a punctured disk around $z_0$.Then use connectedness of a domain. 
In fact, consider the set $U \subset D$ consisting of all  points $p$ such that $f(p) = 0$ and $f$ is zero on an open disk containing $p$. Then $U$ is clearly open. By the observation made above, $U$ is also closed. QED
A: When you negate "$f(z) = 0$ for all $z$ or $g(z) = 0$ for all $z$", you don't get to assume that there is a single point $z_0$ with $f(z_0) \neq 0$ and also $g(z_0) \neq 0$. Instead, you only get points $z_1$ and $z_2$ with $f(z_1) \neq 0$ and $g(z_2) \neq 0$. A priori, it's possible that $f(z) = 0$ whenever $g$ is nonzero and vice versa, which would still give $f(z)g(z) = 0$ for all $z$.
