Proof of L'Hospital's rule in the form of$\frac{\infty }{ \infty }$ for complex functions Let $f$ and $g$ be complex functions that are continuous and differentiable in a neighborhood of $z_0$ and $g'(z)$ does not equal zero. If $\lim_{z\to\ z_0}f(z)=\lim_{z\to\ z_0}g(z)=\infty$ and the limit $\lim_{z\to\ z_0} \frac{f(z) }{ g(z) }$ exists prove:
$\lim_{z\to\ z_0} \frac{f(z) }{ g(z) }=\lim_{z\to\ z_0} \frac{f'(z) }{ g'(z) }$
I've seen many proofs for real functions for L'Hospital's rule in the form of $\frac{\infty  }{ \infty }$ but I am yet to find any formal proof for complex functions, does this rule apply to complex functions in the first place?
 A: In fact, L'Hopital is pretty much a non-issue here.
Thm: Assume both $f(z),g(z)\to \infty$ at $z_0.$ Then there exists $L\in (\mathbb C \cup\{\infty\})$ such that
$$\tag 1 \lim_{z\to z_0}\frac{f(z)}{g(z)} = \lim_{z\to z_0}\frac{f'(z)}{g'(z)}=L.$$
Proof: Since $f(z),g(z)\to \infty$ as $z\to z_0,$ these functions have poles of order $M,N$ at $z_0$ respectively, where $M,N$ are positive integers. Thus we can write
$$f(z)=\sum_{k=-M}^{\infty}a_k(z-z_0)^k,\,g(z)=\sum_{k=-N}^{\infty}b_k(z-z_0)^k,$$
for $z\in D(z_0,r)\setminus \{z_0\}$ for some $r>0,$ where $a_{-M},b_{-N}$ are nonzero numbers in $\mathbb C.$
Suppose $M=N.$ Then
$$\frac{f(z)}{g(z)} = \frac{\sum_{k=-M}^{\infty}a_k(z-z_0)^k}{\sum_{k=-M}^{\infty}b_k(z-z_0^k)} = \frac{\sum_{k=-M}^{\infty}a_k(z-z_0)^{k+M}}{\sum_{k=-M}^{\infty}b_k(z-z_0)^{k+M}}.$$
This is the quotient of two analytic functions in $D(z_0,r),$ whose nonzero values at $z_0$ are $a_{-M},b_{-M}$ respectively. Clearly the limit of this is $a_{-M}/b_{-M}.$
The quotient of derivatives in this case is
$$\frac{f'(z)}{g'(z)}=\frac{\sum_{k=-M}^{\infty}ka_k(z-z_0)^{k-1}}{\sum_{k=-M}^{\infty}kb_k(z-z_0)^{k-1}}.$$
By the same reasoning as above, the limit of this is $(-Ma_{-M})/(-Mb_{-M}) = a_{-M}/b_{-M}.$ We have thus proved the theorem for the case $M=N.$
In the case $M<N,$ $f(z)$ is on the order of $(z-z_0)^{-M}$ and $g(z)$ is on the order of $(z-z_0)^{-N}$ as $z\to z_0.$ It follows that $f(z)/g(z)\to 0.$ The same is true for the quotient of derivatives. If $M>N,$ then both $f(z)/g(z),f'(z)/g'(z)\to \infty.$ The proof of the theorem is complete.
