Is L'Hopitals rule applicable to complex functions?

I have a question about something I'm wondering about. I've read somewhere that L'Hopitals rule can also be applied to complex functions, when they are analytic. So if have for instance:

$$ \lim_{z \rightarrow 0} \frac{\log(1+z)}{z} \stackrel{?}{=} \lim_{z \rightarrow 0} \frac{1}{(1+z)} = 1 $$

Now i'm wondering if this is correct? Also if we take $|z|<1$, is it then correct?

Thanks,

Solution 1:

L'Hopital's rule is a local statement: it concerns the behavior of functions near a particular point. The global issues (multivaluedness, branch cuts) are irrelevant. For example, if you consider $\lim_{z\to 0}$, then it's automatic that only small values of $z$ are in play. Saying "take $|z|<1$" is redundant.

Generally, you have a point $a\in\mathbb C$ and some neighborhood of $a$ in which $f,g$ are holomorphic. If $f(a)=g(a)=0$, then $$\lim_{z\to a}\frac{f(z)}{z-a}=f'(a),\qquad \lim_{z\to a}\frac{g(z)}{z-a}=g'(a) \tag{1}$$ hence $$\lim_{z\to a}\frac{f(z)}{g(z)}= \lim_{z\to a}\frac{f(z)/(z-a)}{g(z)/(z-a)} =\frac{f'(a)}{ g'(a)}$$ Note that the above is a simple special case of the L'Hopital's rule, because we have (1). It's basically just the definition of derivative.