For $f(z)$ which is an analytic function of the complex variable $z$, ($z=x+iy$):
$$f(z(x,y)) = u(x,y)+iv(x,y)$$
that satisfies:
$$v(x,0)=0$$
for $x$ in some finite range:
$$a<x<b$$
Consider a point $x=c$ for $c<a$ where:
$$v(c,0)\ne 0$$
does this necessarily imply that there exists at least one point of non-analyticity of $f(z)$ on the real line in the interval $(c,a)$?
i.e. $f(z)=\ln(z)$ is purely real for $z=x$, $x>0$ and has a point of non-analyticity at $x=0$ and has non-zero imaginary part for $z=x$, $x<0$. (As does $f(z)=z^{1/2}$, I want to know if this will always be the case.)
In that situation - assuming the condition is not vacuous, i.e. assuming $a < b$ - there always must be a singular point of $f$ in the interval $(c,a]$. (We cannot assert a singular point in the open interval $(c,a)$ since $a$ may be the only singular point on the real axis.)
Say that $f$ is defined and holomorphic on the open set $U \subset \mathbb{C}$. Let $V = \{ \overline{z} : z \in U\}$ and define
$$g(z) = \overline{f(\overline{z})}$$
for $z \in V$. Then $g$ is holomorphic on $V$, and by assumption we have $g(z) = f(z)$ for all $z \in (a,b)$. Therefore $g(z) = f(z)$ for all $z$ belonging to the connected component $W$ of $U\cap V$ that contains $(a,b)$ by the identity theorem. But also by assumption we have $g(c) = \overline{f(c)} \neq f(c)$, hence $c \notin W$. Thus it is impossible that $[c,a] \subset U$; for if it were, then $[c,b) \subset U \cap V$ and hence $[c,b) \subset W$.