If $f$ and $g$ are holomorphic complex functions, is $f/g$ also holomorphic? What's the proof? $f$ and $g$ are both holomorphic functions on some domain $\Omega_1, \Omega_2$. Is $\frac f g$ going to be holomorphic on $\Omega_1\cap\Omega_2-\{z\mid g(z)=0\}$? What's the proof?
 A: *

*$g(z)\ne 0$ for $z\in \Omega_1\cap\Omega_2$.

*Complex function is holomorphic in $\Omega$ if there exist the derivative in each point of the domain. Then in usual manner one could prove that there exist derivative $\left(\frac{f}{g}\right)'(z)$ for all $z\in \Omega_1\cap\Omega_2$. Actually we should to check that 
$$
\left(\frac{f}{g}\right)'(z) = \frac{f'(z)g(z)-f(z)g'(z)}{g^2(z)}.
$$
The proof is the same as in real case (but before they prove product rule).
A: \begin{align*}
& \lim_{h\to0} \frac{\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}} h \\[10pt]
= {} & \lim_{h\to0} \frac{ f(x+h)g(x) - f(x)g(x+h) }{h g(x+h) g(x)}\\[10pt]
= {} & \lim_{h\to0} \frac{ \Big( f(x+h)g(x) - f(x)g(x) \Big) - \Big( f(x)g(x+h) - f(x)g(x)  \Big)}{hg(x)g(x+h)} \\[10pt]
= {} & \lim_{h\to0} \left( \frac 1 {g(x)g(x+h)} \left( g(x) \frac{f(x+h)-f(x)} h - f(x) \frac{g(x+h)-g(x)} h \right) \right) \\[15pt]
= {} & \frac{g(x)\left(\lim_{h\to0} \frac{f(x+h)-f(x)} h\right) - f(x) \left( \lim_{h\to0}\frac{g(x+h)-g(x)} h \right) }{\lim_{h\to0} g(x)g(x+h)} \tag 1 \\[15pt]
= {} & \frac{g(x)f'(x) - f(x) g'(x)}{g(x)^2}
\end{align*}
In $(1)$ we used an assumption that $f$ and $g$ are differentiable and a theorem that says differentiable functions are continuous.
