If $f$ is analytic, never $0$, and $g^2 = f$ with $g$ continuous, then $g$ is analytic Let $f: \Omega \to \mathbb{C} \setminus \{0\}$ be analytic.
Let $g: \Omega \to \mathbb{C}$ be continuous with $g^2 = f$ on $\Omega$.
I'd like to show that $g$ must be analytic.
My attempt was centered on these ideas:
If I take a point $z_0 \in \Omega$, I can find neighborhoods of it where the functions $f$, $g$ do not intersect, respectfully $\{e^{i\alpha r}| r\geq 0\}, \{e^{i\beta r}| r\geq 0\}$, since $f$, $g$ are never zero.
Taking the intersection of these neighborhoods, and on it getting continuous logarithms for $f$ and $g$. Since the function $ln$ restricted to the image of $f$, $g$ in the above intersection is well defined and analytic, we get that on the intersection the continuous logarithm of $f$ is analytic.
$g^2 = f$ ties the the logarithm of $g$ to that of $f$ (on the intersection) and so $g$ is analytic on the intersection.
Are these ideas correct? Is there a simpler way?
 A: I can not follow your prof ! What do you mean by $\{e^{i\alpha r}| r\geq 0\}, \{e^{i\beta r}| r\geq 0\}$ ?
I suppose that $\Omega$ is a region.
My proof: let $z_0 \in \Omega$. Then there is an open dic $K$ with center $z_0$, which is contained in $ \Omega$. Since $K$ is simply connected, there is an analytic function $h:K \to \mathbb C$ such that
$h^2=f$ on $K$. Hence
$h^2=g^2$ on $K$.
If $z \in K$, then $h(z)=g(z)$ or $h(z)=-g(z)$.
Now define $A=\{z \in K:h(z)=g(z)\}$ and $B=\{z \in K:h(z)=-g(z)\}$.
Since $h$ and $g$ have no zeroes, we have $A \cap B= \emptyset$. Furthermore we have that $A \cup B=K$.
Your task: prove that $A$ and $B$ are open. Since $K$ is connected, we derive:
$A=K$ or $B=K$.
Hence: $g=h$ on $K$ or $g=-h$ on K. In both cases we find that $g$ is analytic on $K$.
Since $z_0$ was arbitrary, $g$ is analytic on $ \Omega$.
A: Let $z_0\in\Omega$ and let $r>0$ be such that $D(a,r)\subset\Omega$ and that $f\bigl(D(a,r)\bigr)\subset D\bigl(f(a),\bigl|f(a)\bigr|\bigr)$. Since $D\bigl(f(a),\bigl|f(a)\bigr|\bigr)$ is a simply connected subset of $\mathbb{C}\setminus\{0\}$, there is an analytic square root function $s$ defined there (or you can just consider the analytic branch of the square root defined on $\mathbb{C}\setminus\{\lambda f(a)\,|\,\lambda\in(-\infty,0]\}$). Therefore, $g|_{D(a,r)}=s\circ f|_{D(a,r)}$ or $g|_{D(a,r)}=-s\circ f|_{D(a,r)}$ (this is where the continuity of $g$ is used). In each case, $g$ is analytic on $D(a,r)$ and, since being analytic is a local property, $g$ is analytic.
