Let $D\subset \mathbb{C}$ be a domainband let $f,g: D\to \mathbb{C}$ be analytic. Let $z_0 \in D$ and suppose that $f(z_0)=g(z_0)=0$, suppose furthermore that $$\lim_{z\to z_0}\frac{f'(z)}{g'(z)}\in \mathbb{C}$$ exists. Prove that $$\lim_{z\to z_0}\frac{f(z)}{g(z)}$$ exists and calculate this limit.
This is what I've done: Since $z_0$ is a zero of both $f$ and $g$, we have that $$f(z)=(z-z_0)^{n}\tilde{f}(z), \hspace{0.5cm} \text{where} \hspace{0.5cm} \tilde{f}(z)\neq 0$$ $$g(z)=(z-z_0)^m\tilde{g}(z),\hspace{0.5cm} \text{where} \hspace{0.5cm} \tilde{g}(z)\neq 0.$$ Then we have that $f'=n(z-z_0)^{n-1}\tilde{f}+(z-z_0)^n\tilde{f}$ and similarly for $g$, so $$\lim_{z\to z_0}\frac{f'}{g'}=\frac{n(z-z_0)^{n-1}\tilde{f}}{m(z-z_0)^{m-1}\tilde{g}}$$ but since this exists, we have that $n\geq m\geq 1$, then $$\lim_{z\to z_0}\frac{f}{g}=(z-z_0)^{n-m}\frac{\tilde{f}}{\tilde{g}}$$.
grosso modo this is my idea. Any suggestion, please? looks like something is missing. I have the idea that the limit should be something like $n/m$ or $n-m$.
You have almost finished proving that $\lim \frac {f(z)} {g(z)}=\lim \frac {f'(z)} {g'(z)}$. Just consider the case $n>m$ and $n=m$ separately. In the first case both side are $0$. In the second case both are equal to $\frac {\overline f (z_0)} {\overline g (z_0)}$