In the following result
- $f$ is differentiable on a domain $D$ & i) $\Re f$ is constant or ii) $\Im f$ is constant or iii) $\arg f$ is constant or iv) $|f|$ is constant on $D\implies f$ is constant on D.
we see that whenever the image of $f$ (differentiable on a domain) lies on a line parallel to $x$-axis (or $y$-axis) or on the line $y=x,~f$ becomes constant on the domain. Can the result be generalized for any st. line as
Q.1: $f$ is differentiable on a domain $D$ and the image of $f$ lies on an arbitrary straight line. Does this imply $f$ is constant on $D?$
Also we know that
- $f$ is differentiable on a domain $D$ & $|f|$ is constant on $D\implies f$ is constant on D i.e. whenever the image of $f$ lies on a circle $f$ becomes constant on $D.$
Q.2: Can this be generalized for any conic on which the image of $f$ (differentiable on a domain) lies?
Q.3: How far can those observations be ultimately generalized?
Suppose $f^\prime(d)\neq 0$ for some $d\in D$. The inverse function theorem tells us that $f$ restricts to a diffeomorphism onto its image on some neighbourhood of $d$. Since dimension is a topological invariant, this means that $f(d)$ has a 2-dimensional neighbourhood in $\mathrm{im}(f)$.
If $\{d\in D|f^\prime(d)\neq 0\}$ is finite, then $f^\prime$ vanishes on an open set, hence $f$ is constant.
Thus, a non-constant continuously differentiable $f\colon D\to\mathbb{C}$ has infinitely many points $d$ such that $f(d)$ has a two dimensional neighbourhood in $\mathrm{im}(f)$.
This immediately disqualifies lines and conic sections as images of such $f$.