Let $f$ and $g$ be two holomorphic functions on the disk $D$. If $f(z)g(z)$ $=$ $0$ for all z $\in$ $D$, show that either $f$ or $g$ in constantly zero in $D$.
So this is what I have gotten:
Suppose $f(z)g(z) = 0$ for all $z \in D$
Let $Z(fg) = \{z\in D: \, f(z)g(z)=0\}$ is the zero set of $(fg)$
Then $Z(fg)$ has a limit point in $D$ since $Z(fg) = D$, call the limit point as $\alpha$.
Let $$Z(f) = z\in D \mid f(z)=0$$ $$Z(g) = z\in D \mid g(z)=0$$ Choose $r>0$ such that $B_r$($\alpha$)$\subset$$D$
If $f(z) = 0$ for all $z$ in $B_r$($\alpha$), then $B_r(\alpha)\subset Z(f)$
and since $B_r$($\alpha$) has a limit point in $D$, so does $Z(f)$
hence $f = 0$
Now, if $f\not\equiv0$ in $B_r$($\alpha$), there is some $z_0\in B_r(\alpha)$
such that $f(z_0) \neq 0$
but then $g(z_0) = 0$
Now, let $0<r_1<r$ such that $z_0 \notin Br_1 (\alpha)$
If $f\equiv0$ on $Br_1 (\alpha)$ then, we are done.
If that is not the case, there is $z_1 \in Br_1 (\alpha)$ such that
$f(z_1)\neq 0$. Then, $g(z_1)=0$
Either this process will terminate ($f\equiv 0$ on $Br_c(\alpha)$ for some $i$)
(or) we will have a sequence $(z_c)_{c=0}^\infty$ converging to $\alpha$.
But this sequence is entirely in $Z(g)$
So $Z(g)$ has a limit point in $0$.
Hence by identity theorem, $g\equiv0$ in $D$.
It is correct. You did a writing mistake that I corrected: in fact, you wrote $g(z_1) \neq 0$, but it must be $g(z_1)=0$ in order to obtain the correct proof.