On the complex L'Hospital rule Let $f(z)$ and $g(z)$ be two complex functions (defined on a neighborhood of a point $z_0\in\mathbb{C}$) such that $f(z_0)=g(z_0)=0$ and $g'(z_0)\neq 0$. Is it true that $\lim_{z\rightarrow z_0}\frac{f(z)^2}{g(z)^2}=\frac{[f'(z)]^2}{[g'(z)]^2}$ ? If so, can we generalize this result to any integer power $n$ of  $\frac{f(z)}{g(z)}$ ? 
EDIT: Another related question would be as follows: Can we apply the L'Hospital rule consecutively ? That is, suppose that applying the L'Hospital rule each time the limit gives us $\frac{0}{0}$, so in this case can we continue to apply the L'Hospital rule (in a finite step) ? Do we need any restrictions on the functions $f$ and $g$ ?
 A: Yes, this is true if $f$ and $g$ are holomorphic at $z_0$. In this case $f$ and $g$ are analytic on an open neighborhood $U$ of $z_0$, so we have $$f(z) = \sum_{n=1}^\infty a_n(z-z_0)^n \text{ and } g(z) = \sum_{n=1}^\infty b_n(z-z_0)^n$$ for $|z-z_0|$ sufficiently small. Note that the $0$-th coefficient of each series is $0$ since $f(z_0) = g(z_0) = 0$.
Furthermore we have $b_1 = g'(z_0) \neq 0$. With this we obtain $$\lim_{z \to z_0} \frac{f(z)}{g(z)} = \lim_{z \to z_0} \frac{\sum_{n=1}^\infty a_n(z-z_0)^{n-1}}{\sum_{n=1}^\infty b_n(z-z_0)^{n-1}} = \frac{a_1}{b_1} = \frac{f'(z_0)}{g'(z_0)}.$$
Now we can use the continuity of $\mathbb{C} \to \mathbb{C}, z \mapsto z^n$ for $n \in \mathbb{N}_0$ to obtain $$\lim_{z \to z_0} \frac{f(z)^n}{g(z)^n} = \left(\lim_{z \to z_0} \frac{f(z)}{g(z)} \right)^n = \frac{f'(z_0)^n}{g'(z_0)^n}.$$
For $n < 0$ we can apply the same idea if $a_1 = f'(z_0) \neq 0$; otherwise the quotient $\frac{f(z)^n}{g(z)^n}$ will diverge to $\infty$ for $z \to z_0$.
EDIT: Yes, we can also apply L'Hospital multiple times as long as one of the functions is not equal to $0$ in an open ball around $z_0$: 
If we take $n_0 \in \mathbb{N_0}$ to be the minimal index $n$ such that $f^{(n)}(z_0) \neq 0$ or $g^{(n)}(z_0)  \neq 0$ (such an index exists by the above assumption), then we have $a_k = \frac{f^{(k)}(z_0)}{k!} = 0$ and $b_k = \frac{g^{(k)}(z_0)}{k!}  = 0$ for all $k < n_0$, so for $g^{(n_0)}(z_0) \neq 0$ we obtain $$\lim_{z \to z_0} \frac{f(z)}{g(z)} = \lim_{z \to z_0} \frac{\sum_{n=n_0}^\infty a_n(z-z_0)^{n-n_0}}{\sum_{n=n_0}^\infty b_n(z-z_0)^{n-n_0}} = \frac{a_{n_0}}{b_{n_0}} = \frac{\frac{f^{(n_0)}(z_0)}{n_0!}}{\frac{g^{(n_0)}(z_0)}{n_0!}} = \frac{f^{(n_0)}(z_0)}{g^{(n_0)}(z_0)}$$
and the first equation shows that the the quotient $\frac{f(z)}{g(z)}$ will diverge to $\infty$ for $z \to z_0$ if $g^{(n_0)}(z_0) = 0$ and $f^{(n_0)}(z_0) \neq 0$.
A: While Hôpital's rule works correctly for holomorphic functions it does not for arbitrary complexvalued functions. Consider the following example:
$$f(t):=t,\qquad g(t):=t\,e^{-i/t}\qquad(t>0)\ .$$
Then $\lim_{t\to0+}f(t)=\lim_{t\to0+}g(t)=0$. Furthermore
$$f'(t)=1,\qquad g'(t)=\left(1+{i\over t}\right)e^{-i/t}\ ,$$
so that $$\lim_{t\to0+}{f'(t)\over g'(t)}=\lim_{t\to0+}{t\over t+i}e^{i/t}=0\ .$$ 
But the limit
$$\lim_{t\to0+}{f(t)\over g(t)}=\lim_{t\to0+}e^{i/t}$$
does not exist!
