On the mean value theorem in $\mathbb R^2$ Consider the following claim:
If $A$ is a (complex) unital Banach algebra and $f: \mathbb R \to A$ is differentiable with $f' = 0$ then $f$ is constant. 
The proof uses that for $\tau \in A^\ast$: $(\tau\circ f)' = 0$ and therefore $\tau \circ f$ is constant.
The latter is apparently a basic fact but not basic enough for me as I do not understand it. I appreciate any insight, here are my doubts:
Note that $\tau \circ f$ is a map from $\mathbb R$ to $\mathbb C$. It was not obvious to me that the mean value theorem generalizes to complex valued maps so I did some checking on Wikipedia and, apparently, the MVT does not generalize to $\mathbb R^n$. Unfortunately, it does not mention whether $n=2$ is special.

Is there a mean value theorem for differentiable functions $[a,b]\to
 \mathbb R^2$?

Disclaimer: After trying to prove it I believe the answer to be no but I could not find suitable $a,b$ and $f$ for a counterexample. Since $ \mathbb C$ and $\mathbb R^2$ are homeomorphic it should suffice to discuss the question in $\mathbb R^2$. 
 A: 
Is there a mean value theorem for differentiable functions $[a,b]\to
 \mathbb R^2$?

Yes, there are several theorems that are called mean value theorem for that situation (and the more general of maps to $\mathbb{R}^n$).
One mean value theorem is
Let $f \colon [a,b] \to \mathbb{R}^n$ a differentiable function. If $\lVert f'(t)\rVert \leqslant M$ for all $t\in [a,b]$, then $\lVert f(b) - f(a)\rVert \leqslant M\cdot\lvert b-a\rvert$.
That version already implies that a function with $f' \equiv 0$ is constant on an interval, since $M = 0$ is then an admissible choice.
Another mean value theorem is
Let $f\colon [a,b]\to \mathbb{R}^n$ a differentiable function. Then there are $c_1,\dotsc,c_n \in (a,b)$ with
$$f(b) - f(a) = \begin{pmatrix}f_1'(c_1)\\\vdots\\ f_n'(c_n)\end{pmatrix}\cdot (b-a).$$
This is evidently just the mean value theorem for $g\colon [a,b]\to\mathbb{R}$ applied to all components. This also implies that a function with $f' \equiv 0$ is constant on intervals.
The $c_i$ in the above generally have to be different points, the classical example is $f(t) = (\cos t, \sin t)$, with $f(0) = f(2\pi)$, but $f'(t) \neq 0$ for all $t$, so there is no $c \in (0,2\pi)$ with $f(2\pi)-f(0) = f'(c)\cdot 2\pi$.

The proof uses that for $\tau\in A^\ast$: $(\tau\circ f)′=0$ and therefore $\tau\circ f$ is constant.

Apply real and imaginary part, $(\operatorname{Re} \tau\circ f)' = \operatorname{Re} (\tau\circ f)' = 0$, and analogously for the imaginary part. Thus real and imaginary parts of $\tau\circ f$ are constant, hence $\tau\circ f$ is constant (hence $f$ is constant).
