Combination Theorem for Complex Derivatives/Sum Rule/Proof 2

Theorem
Let $D$ be an open subset of the set of complex numbers $\C$.

Let $f, g: D \to \C$ be complex-differentiable functions on $D$

Then $f + g$ is complex-differentiable in $D$, and its derivative $\paren {f + g}'$ is defined by:


 * $\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$

for all $z \in D$.

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

Let $z \in D$.

By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:


 * $\map f {z + h} = \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h}$


 * $\map g {z + h} = \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}$

where $\epsilon_f, \epsilon_g: \map {B_r} 0 \setminus \set 0 \to \C$ are complex functions that converge to $0$ as $h$ tends to $0$.

Then:

From Sum Rule for Limits of Complex Functions, it follows that $\ds \lim_{h \mathop \to 0} \map {\paren {\epsilon_f + \epsilon_g} } h = 0$.

By the Epsilon-Function Complex Differentiability Condition, it follows that:


 * $\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$