Question: Using Rudin's formulation of the L'Hospital Rule, show that it remains true if
a) $f:(a,b)\to \mathbb{R^k}$ is vector-valued and $g:(a,b)\to \mathbb{R}$
b) $f:(a,b)\to \mathbb{C}$ and $g:(a,b)\to \mathbb{C}$ are complex-valued functions and either $$\lim_{x\to a} \frac{Re (g'(x))}{Im (g'(x))}$$ or $$\lim_{x\to a} \frac{Im (g'(x))}{Re (g'(x))}$$ where $Im(f)$ and $Re(f)$ are the imaginary and real parts, respectively, of a complex-valued function.
Rudin's formulation - Theorem 5.13
Suppose $f$ and $g$ are real and differentiable in $(a,b)$, and $g'(x)\neq 0$ for all $x\in (a,b)$, where $-\infty\leq a<b\leq \infty$. Suppose $$\frac{f'(x)}{g'(x)} \to A \text{ as } x\to a.$$ If $f(x)\to 0$ and $g(x)\to 0$ as $x \to a,$ or if $g(x)=+\infty$ as $x\to a,$ then $$\frac{f(x)}{g(x)}\to A\text{ as }x\to a.$$
What I thought/did
For item a), I don't think that the proof Rudin offers for the usual real-valued case would need much change, some absolute values around the vectors would seem to me enough to do the trick. My only doubt is how/if one could change Theorem 5.9 to fit the new situation.
For item b), I'm not sure even where to start. I tried playing around with properties of complex numbers, but I couldn't find any promising path. And I don't think Rudin's proof could be tweaked/arguments reutilized as to fit this situation, apparently so different from the original one.