Let $D$ be a region and $f_1(z),f_2(z)$ be two different analytic branches of $\sqrt{z-z_0}$. Show that $f_1(D)\cap f_2(D)=\varnothing$.
My Attempts. Prove by contradiction. Say $f_1(z_1)=f_2(z_2)$, then easy to see $z_1=z_2:=w$. Note that $(f_1(z)-f_2(z))(f_1(z)+f_2(z))\equiv 0$. Let $h(z)=f_1(z)-f_2(z)$. If $h(z)$ is constant then we are done. If not, $w$ must be isolated as a zero point of $h(w)$, i.e. there exists $D_0(w,\delta)$ such that $\forall z\in D_0,f_1(z)+f_2(z)=0$. Using isolation of zero points again, $f_1(z)+f_2(z)\equiv 0$. But I don't know how to proceed (or maybe there'll be a better solution).
I would be extremely appreciative of any assistance!
You are on the right track. First note that $w \ne 0$ because there is no analytic branch of $\sqrt{z-z_0}$ in a neighborhood of $z=z_0$.
From $$ (f_1(z)-f_2(z))(f_1(z)+f_2(z))\equiv 0 $$ it follows that one of the factors must have an accumulation point of zeros in $D$, and therefore is identically zero. (Here we use that $D$ is a region, i.e. a connected open set.)
So we have $f_1 \equiv f_2$ or $f_1 +f_2 \equiv 0$. The latter is not possible because $f_1(z_1) + f_2(z_1) = 2w \ne 0$.
In the same manner one can show that two holomorphic branches of $(z-z_0)^{1/n}$ ($n\ge 2$ an integer) on a (connected) domain are either identical, or have disjoint images.