Combination Theorem for Complex Derivatives/Sum Rule/Proof 2
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Theorem
- $\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$
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:
\(\ds \map {\paren {f + g} } {z + h}\) | \(=\) | \(\ds \map f z + h \paren {\map {f'} z + \map {\epsilon_f} h} + \map g z + h \paren {\map {g'} z + \map {\epsilon_g} h}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \map {\paren {f + g} } z + h \paren {\map {f'} z + \map {g'} z + \map {\paren {\epsilon_f + \epsilon_g} } h}\) |
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:
- $\map {\paren {f + g}'} z = \map {f'} z + \map {g'} z$
$\blacksquare$
Sources
- 2001: Christian Berg: Kompleks funktionsteori $\S 1.1$