Combination Theorem for Complex Derivatives/Quotient Rule

Theorem
Let $f, g: D \to \C$ be complex-differentiable functions, where $D$ is an open subset of the set of complex numbers.

Let $\dfrac f g$ denote the quotient of the functions $f$ and $g$.

Then $\dfrac f g$ is complex-differentiable in $D \setminus \left\{ {x \in D }\, \mid \,{g \left({z}\right) = 0}\right\}$. For all $z \in D$ with $g \left({z}\right) \ne 0$:
 * $\left({\dfrac f g}\right)' \left({z}\right) = \dfrac{ f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{ \left({g \left({z}\right) }\right)^2 }$

where $\left({\dfrac f g}\right)'$ denotes the derivative of $\dfrac f g$.

Proof
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $B_r \left({0}\right)$.

Let $z \in D \setminus \left\{ {x \in D }\, \mid \,{g \left({z}\right) = 0}\right\}$.

By the Alternative Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in B_{r_0} \left({0}\right) \setminus \left\{ {0}\right\}$:


 * $f\left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
 * $g\left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$

where $\epsilon_f, \epsilon_g: B_{r_0} \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$.

From Complex-Differentiable Function is Continuous, it follows that $g$ is continuous at $z$.

Then there exists $r_1 \in \R_{>0}$ such that for all $h \in B_{r_1} \left({0}\right)$, we have $\left\vert{g \left({z + h}\right) - g \left({z}\right) }\right\vert < \left\vert{g \left({z}\right) }\right\vert$.

Then $g \left({z + h}\right) \ne 0$ for all $h \in B_{r_1} \left({0}\right)$ by Reverse Triangle Inequality.

Put $r = \min \left({r_0, r_1}\right)$.

Then for all $h \in B_r \setminus \left\{ {0}\right\}$:

where $\epsilon: B_r \left({0}\right) \setminus 0 \to \C$ is defined by:


 * $\epsilon \left({h}\right) = \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) } - \dfrac{f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{\left({g \left({z}\right) }\right)^2 } + \dfrac{ g \left({z}\right) \epsilon_f \left({h}\right) - f \left({z}\right) \epsilon_g \left({h}\right) }{\left({g \left({z}\right) }\right)^2 + \epsilon_0 \left({h}\right) }$

From Combination Theorem for Continuous Functions, it follows that $\epsilon$ is continuous.

From Combination Theorem for Limits of Functions, it follows that:

Then the Alternative Differentiability Condition shows that:


 * $\left({\dfrac f g}\right)' \left({z}\right) = \dfrac{ f' \left({z}\right) g \left({z}\right) - f \left({z}\right) g' \left({z}\right) }{ \left({g \left({z}\right) }\right)^2 }$