3 added 81 characters in body

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.

Take a

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.

2 added 47 characters in body

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.

Take a 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.

1

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.

Take a 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 $g(z) = z^n$. 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.