Suppose $f^2$ is analytic in a domain $\Omega$ and $f$ is continuous in $\Omega$, show that $f$ is also analytic in $\Omega$.
I know several solutions already exist here but there is one part of a proof that confuses me so I will sketch out the proof and highlight where I am confused.
Proof: Let $g = f^2$. If $g$ is identically zero then the result is clear. So assume $g$ is not identically zero. Then its zeros must be isolated and thus the zeros of $f = {g}^{1/2}$ are isolated. Let $z_0 \in \Omega$ such that $g(z_0) \neq 0$. Then there exists a neighborhood $U$ of $z_0$ such that $g$ is non-zero in $U$ and hence we can define its analytic square root in $U$ which shows that $f = g^{1/2}$ is analytic at every $z_0$ such that $g(z_0) \neq 0$.
Now here is the part that confuses me: If $z_0$ is a zero of $g$ then we can define the square root in a continuous way that winds around the zero so that after a full circle, we get back to the same branch we started with. I think the reasoning behind this is because if we let $\gamma: [0,1] \rightarrow \Omega$ be a path around $z_0$ then from what was said above, if $z_1 \in \gamma$ we can define $f_1 = g^{1/2}$ in a neighborhood $U_1$ of $z_1$. Then for $z_2 \in \gamma$ sufficiently close to $z_1$ we can define $f_2 = g^{1/2}$ in a neighborhood $U_2$ of $z_2$ such that $U_1 \cap U_2 \neq \emptyset$. Since $f_1$ and $f_2$ agree on the intersection of $U_1$ and $U_2$, by the identity theorem we see that $f_2$ is just an analytic extension of $f_1$. Since $\gamma$ is compact, we can cover it with finitely many $U_k$. Then by continuity, we must have that $f_1(\gamma(0)) = f_k(\gamma(1))$. Ok, so if that is all correct, why is this enough to justify that $f = g^{1/2}$ is analytic at the zeros of $g$?
Some of my details may not be correct so please let me know if I made an error.
No, that is not enough, otherwise just let $g(z)=z$ and you have constructed a holomorphic square-root of $z$ around $0$!
You need to use the fact that $g$ was defined to be $f^2$ so $g\circ\gamma$ has even winding number about $0$ as $\gamma$ circles around $z_0$ once.