Mean value thorem for $\alpha:[a,b]\to \mathbb R^2$ Let $\alpha:[a,b]\to \mathbb R^2$ continuous on $[a,b]$ and derivable on $(a,b)$. If $\alpha´(t)\neq 0 \forall t\in (a,b)$ prove that $\exists c\in (a,b)$ and $\exists \lambda\in \mathbb R$ such that $$\alpha(a)-\alpha(b)=\lambda\alpha´(c)$$
There has been many questions about the generalized mean value theorem, but I don´t consider this as a duplicate (and I hope it isn´t)
I have seen a counterexample given by $(\text{cos t},\text{sin t})$ in $[0,2\pi]$ but in this case this doesn´t work because $\lambda$ would just be $0$
I would really appreciate any ideas or hints
 A: It is simpler in terms of notation to use $\mathbb{C}$ instead of $\mathbb{R}^{2}$. Thus let $\alpha(t) = f(t) + ig(t)$ be a function from $[a, b]$ to $\mathbb{C}$ such that $\alpha$ is continuous on $[a, b]$ and differentiable on $(a, b)$. This simply means that $f, g$ are continuous on $[a, b]$ and differentiable on $(a, b)$ and $\alpha'(t) = f'(t) + ig'(t)$. Now we can see that $$\alpha(b) - \alpha(a) = f(b) - f(a) + i(g(b) - g(a))$$ By Cauchy's Mean Value Theorem we know that there is a $c \in (a, b)$ such that $$\{f(b) - f(a)\}g'(c) = f'(c)\{g(b) - g(a)\}$$ Note that $\alpha'(t) \neq 0$ so that one of the $f'(c), g'(c)$ is non-zero. Let's assume that $f'(c) \neq 0$ so that $g(b) - g(a) = (g'(c)/f'(c))(f(b) - f(a))$ and therefore $$\alpha(b) - \alpha(a) = f(b) - f(a) + i(g'(c)/f'(c))(f(b) - f(a))$$ and therefore $$\alpha(b) - \alpha(a) = \frac{f(b) - f(a)}{f'(c)}\cdot\{f'(c) + ig'(c)\} = \lambda \alpha'(c)$$ If $f'(c) = 0$ then we must have $g'(c) \neq 0$ and again the proof can be carried similarly by writing $f(b) - f(a) = (f'(c)/g'(c))(g(b) - g(a))$.
