How do I prove L'Hôpital's rule variation for complex functions

Question

I need to prove the following version of L'Hôpital's rule for complex functions:

Let $f$ and $g$ be analytical at $z_0$ such that $f(z_0)=g(z_0)=0$ and $g$ is not identically zero: so the following limits exist $($whether $\infty$ or finite$)$ and equal: $\displaystyle\lim_{z\to z_0}\frac{f(z)}{g(z)}=\lim_{z\to z_0}\frac{f'(z)}{g'(z)}$ .

I looked up online and found people using it but not a proof. I've found this, which I'm not sure why doesn't work here:

$$\lim_{z\rightarrow z_{0}}\frac{f(z)}{g(z)} =\lim_{z\rightarrow z_{0}}\frac{f(z)-f(z_{0})}{z-z_{0}}\cdot\frac{z-z_{0}}{g(z)-g(z_{0})} =\frac{f'(z)}{g'(z)}$$

My guess it's something with power series.

Thank you.

Answer 1

Assume that $z_0$ is a simple zero of both $f$ and $g$. Then, if$$f(z)=a_1(z-z_0)+a_2(z-z_0)^2+\cdots\text{ and }g(z)=b_1(z-z_0)+b_2(z-z_0)^2+\cdots,$$we have\begin{align}\lim_{z\to z_0}\frac{f(z)}{g(z)}&=\lim_{z\to z_0}\frac{a_1(z-z_0)+a_2(z-z_0)^2+\cdots}{b_1(z-z_0)+b_2(z-z_0)^2+\cdots}\\&=\lim_{z\to z_0}\frac{a_1+a_2(z-z_0)+\cdots}{b_1+b_2(z-z_0)+\cdots}\\&=\frac{a_1}{b_1}\\&=\lim_{z\to z_0}\frac{f'(z)}{g'(z)}.\end{align}Can you do the general case now?

Answer 2

I know this post is a bit old, but trying to do this same exercise and finding this post, I noticed some problems with the setting of the problem and the other answer given. I do this for anybody that stumbles onto the same exercise and can find this useful, and also this was the same as I would had wanted to find when I was searching for some help with the case where $g'(a)=0$.

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So the thing is that the function $g$ should have the property that $g'(a)\neq 0$, so that the limit $\underset{z\to a}{lim} \frac{f'(z)}{g'(z)}$ can exist in general, in the set $\mathbb{C}$ meaning it is finite.

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In the case of the "general case" that it is mentioned in the other answer, there is the following proposition from complex analysis:

Let $z_0\in \mathbb{C}$, $R>0$ and $f,g\in\mathcal{H}(D(z_0,R))$ (holomorphic functions on the disc with center $z_0$ and radius $R$).If $z_0$ is a zero of order $k$ of $f$ and of order $l$ of $g$, then $z_0$ is an isolated singularity of $h=f/g$ that is:
· removable if $k\geq l$
· a pole of order $l-k$ if $k<l$

We say that $z_0$ is a zero of order $k$ of $g$ if $0=g(z_0)=\cdots =g^{(k-1)}(z_0)$, but $g^{(k)}(z_0)\neq 0$.
So if $g'(a)=0$ (zero of order at least 2) then because of this proposition we will have that $f/g$ has a pole in $a$, so then $$\lim_{z\to a} \frac{f(z)}{g(z)}=\infty$$ I can provide proofs for the mentioned proposition and the fact that $a$ is a pole (we don't care about the order) of $f$ $\iff$ $\underset{z\to a}{lim}f(z)=\infty$.