Suppose that $f:U\subset\mathbb{C}\to\mathbb{C}$, where $U$ is a region in the complex plane, is a holomorphic function. 


Of course, if $c\in\mathbb{R}$ is a regular value for $\text{Re}(f(z))$ then $\text{Re}(f(z))^{-1}(c)$ is an union of differentiable curves in the plane. 


 **Question:** 


If $c$ is not a regular value and $\text{Re}(f(z))^{-1}(c)$ have at least one cluster point is this set also a piece-wise differentiable curve ?



What $Re(f(z))=c$ can be if $f$ is a holomorphic function ?
Yes. The set of points in $\mathbb C$ having real part equal to $c$ form a line, i.e. a smooth simple curve $l$. The counterimage $f^{-1}(l)$ of any smooth simple curve $l$ via a holomorphic function $f$ is always piecewise smooth. 


To prove the last sentence, take any point $z\_0\in U$ with $f(z\_0) \in l$. There are two local diffeomorphisms at $z\_0$ and $f(z\_0)$ that move $z\_0$ to $0$ and transform $f$ locally into $g(z) = z^n$. Since $f$ is not constant, we have $n>0$. Local diffeomorphisms send smooth curves to smooth curves. The counterimage along $g$ of a smooth curve passing through $0$ is the union of $n$ smooth curves exiting from 0. Therefore $f^{-1}(l)$ is a piecewise smooth curve.


