Suppose throughout that $E$ is a complex normed vector space.
Question: For which $E$ does it hold that if $D\subset\Bbb C$ is a domain, $f:D\to E$ is holomorphic and $||f(z)||$ is constant then $f$ is constant?
For reference below I'm going to call such spaces cnc spaces (for "constant norm implies constant").
(Why the several-complex-variables tag: If $E$ is not cnc then in some sense the boundary of the unit ball contains analytic disks. So I wonder if maybe at least in the finite-dimensional case this has something to do with strict pseudo-convexity of the unit ball. I know nothing about scv...)
Context: The positive answers to this question show that $\Bbb C^2$ with the euclidean norm is a cnc space. In fact one of the answers shows that any Hilbert space is cnc. (Replace the pair of power series in that answer by a single power series with $E$-valued coefficients...)
I conjectured that any Banach space is cnc. But no:
Example: $\Bbb C^2$ with the $\ell_\infty$ norm $||z||_\infty=\max(|z_1|,|z_2|)$ is not cnc. Consider the function $f(z)=(1,z)$ in the unit disk.
My work so far: My failed proof that every Banach space is cnc leads naturally to a condition that does imply cnc. Having no idea whether it's equivalent to one of the standard [adverb]-convex conditions, I'm going to invent another silly name and call it qc, for "quite convex":
Definition $E$ is qc if for every $\Lambda\in E^*$ with $\Lambda\ne0$ there exists exactly one $x\in E$ with $||x||=1$ and $\Lambda x=||\Lambda||$.
So for example any Hilbert space is qc, as is $L^p(\mu)$ for $\sigma$-finite $\mu$ and $1<p<\infty$.
Easy Theorem. If $E$ is qc then $E$ is cnc.
Proof: Say $D$ is connected, $f:D\to E$ is holomorphic and $||f(z)||=1$ for all $z$. Fix $p\in D$. Hahn-Banach shows there exists $\Lambda\in E^*$ with $||\Lambda||=1$ and $\Lambda f(p)=1$. Now since $|\Lambda f(z)|\le 1$ for every $z$, MMT shows that $\Lambda\circ f$ is constant.
So for every $z\in D$ we have $\Lambda f(z)=\Lambda f(p)=||f(z)||=||f(p)||=||\Lambda||=1$; the definition of qc was contrived precisely so that this should imply $f(z)=f(p)$.
It doesn't seem obviously impossible that assuming $E$ is not qc one could construct an example analogous to the $f(z)=(1,z)$ above to show that $E$ is not cnc... Edit: No, the answer below shows that $L^1(\mu)$ is cnc, while it certainly is not qt.