Corollary

For a vector function $f$ of a complex variable, defined and continuous on an open set $A \subset \mathbf{C}$, and with values in a normed space over $\mathbf{C}$, to be constant, it suffices that it have zero derivative at every point of $A$.

Indeed, let $a$ be an arbitrary point of $A$; the set $B$ of points $z$ of $A$ where $f(z) = f(a)$ is closed because $f$ is continuous; it is also open, as is shown by applying prop. 4 (with $m = 0$) to a convex open neighbourhood, contained in $A$, of an arbitrary point of $B$; so is identical to $A$.