Using de l'Hopital for complex functions? I was wondering if de l'Hopital's rule also applies to complex functions.
Some background information:
This question came up as I was trying to calculate $\displaystyle \lim_{z \to 1} {z^n - 1 \over z -1}$. If we let $f(z) = z^n - 1$ and $g(z) = z-1$ then $\lim_{z \to 1} f(z) = \lim_{z \to 1} g(z) = 0$ and both functions are differentiable on any open interval (=line segment in $\mathbb C$) with endpoint $1$ and $\lim_{z \to 1} f'(z) / g'(z) = n$. 
These thoughts created two questions for me:

Question (1): As per title: Can de l'Hopital's rule be applied to complex maps and if needed, what additional assumptions are needed?

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Question (2): If de l'Hopital's rule cannot be applied, what common method is used to evaluate a limit of a fraction of complex
  maps like in this example?

 A: For your first question check this post Is L'Hopitals rule applicable to complex functions?
For the second question, another way to find limits, which is more general, is to consider the Taylor expansion of the functions areound that point, and do any cancellations. For example, let's say you want o find $\lim_{z\to 0} \sin z/z$. You can write
$$\frac{\sin z}{z}= \frac{z-z^3/3!+ z^5/5! - \dots}{z}= 1-z^2/3!+z^4/5!-\dots$$
and the latter converges to $1$ as $z\to 0$. Another example is $\lim_{z\to 0} \frac{e^z-1-z}{z^2}$. We write
$$\frac{e^z-1-z}{z^2}= \frac{(1+z+z^2/2!+z^3/3!+\dots)-1-z}{z^2}$$
$$=\frac{z^2/2+z^3/3!+\dots}{z^2}= \frac{1}{2}+ z/3!+\dots$$
and this converges to $1/2$ as $z\to 0$. Using this method you avoid applying lots of times L'Hospital's rule (for example here you would need to apply it twice), and also you only need to know the first few Taylor coefficients, rather than all of them. 
In your case you can write
$$\frac{z^n-1}{z-1}= \frac{(z-1)(1+z+\dots+z^{n-1})}{z-1}=1+z+\dots+z^{n-1}$$
and this clearly converges to $n$ as $z\to 1$.
