Theorem: Combination Theorem for Complex Derivatives/Quotient Rule

For all $z \in D$ with $\map g z \ne 0$:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$

Proof:
Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $\map {B_r} 0$.
Let $z \in D \setminus \set {x \in D: \map g z = 0}$.
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r_0 \in \R_{>0}$ such that for all $h \in \map {B_{r_0} } 0 \setminus \set 0$:
:$\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$
:$\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$
where $\epsilon_f, \epsilon_g: \map {B_{r_0} } 0 \setminus \set 0 \to \C$ are complex 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 $h \in \map {B_{r_1} } 0$, we have $\size {\map g {z + h} - \map g z} < \size {\map g z}$.
Then by Backwards Form of Triangle Inequality:
:$\map g {z + h} \ne 0$ for all $h \in \map {B_{r_1} } 0$
Put $r = \min \set {r_0, r_1}$.
Then for all $h \in B_r \setminus \set 0$:
{{begin-eqn}}
{{eqn | l = \dfrac {\map f {z + h} } {\map g {z + h} }
      | r = \dfrac {\map f z} {\map g z} + \dfrac {\map f {z + h} - \map f z} {\map g {z + h} } - \map f z \dfrac {\map g {z + h} - \map g z} {\map g z \map g {z + h} }
      | c = adding and subtracting the same quantity
}}
{{eqn | r = \dfrac {\map f z} {\map g z} + \dfrac {h \paren {\map {f'} z + \map {\epsilon_f} h} } {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } - \dfrac {\map f z h \paren {\map {g'} z \map {\epsilon_g} h} } {\map g z \paren {\map g z + h \paren {\map {g'} z + \map {\epsilon_g} h} } }
}}
{{eqn | r = \dfrac {\map f z} {\map g z} + h \dfrac {\map {f'} z \map g z + \map g z \map {\epsilon_f} h - \map f z \map {g'} z - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
      | c = with $\map {\epsilon_0} h = h \map {g'} z \paren {\map {g'} z + \map {\epsilon_g} h}$
}}
{{eqn | r = \dfrac {\map f z} {\map g z} + h \paren {\dfrac {\map {f'} z \map g z - \map f z \map {g'} z} { {\paren {\map g z}^2 } } + \map \epsilon h}
}}
{{end-eqn}}
where $\epsilon: \map {B_r} 0 \setminus \set 0 \to \C$ is defined by:
:$\map \epsilon h = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}$
From Combination Theorem for Limits of Complex Functions, it follows that:
{{begin-eqn}}
{{eqn | l = \lim_{h \mathop \to 0} \map \epsilon h 
      | r = \lim_{h \mathop \to 0} \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \lim_{h \mathop \to 0} \dfrac {\map g z \map {\epsilon_f} h - \map f z \map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
}}
{{eqn | r = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2} - \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2 } + \map g z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_f} h} {\paren {\map g z}^2 + \map {\epsilon_0} h} - \map f z \lim_{h \mathop \to 0} \dfrac {\map {\epsilon_g} h} {\paren {\map g z}^2 + \map {\epsilon_0} h}
}}
{{eqn | r = 0
}}
{{end-eqn}}
Then the Epsilon-Function Complex Differentiability Condition shows that:
:$\map {\paren {\dfrac f g}'} z = \dfrac {\map {f'} z \map g z - \map f z \map {g'} z} {\paren {\map g z}^2}$
{{qed}}