Combination Theorem for Complex Derivatives/Sum Rule
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 1
Let $z_0 \in D$ be a point in $D$.
Define $k : D \to \Z$ by $\map k {z_0} = \map f {z_0} + \map g {z_0}$.
Then:
\(\ds \map { k' } {z_0}\) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h\) | Definition of Derivative of Complex Function | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\paren {\map f {z_0 + h} + \map g {z_0 + h} } - \paren {\map f {z_0} +\map g {z_0} } } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} + \map g {z_0 + h} - \map f {z_0} - \map g {z_0} } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\paren {\map f {z_0 + h} - \map f {z_0} } + \paren {\map g {z_0 + h} - \map g {z_0} } } h\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \paren {\frac {\map f {z_0 + h} - \map f {z_0} } h + \frac {\map g {z_0 + h} - \map g {z_0} } h}\) | Complex Multiplication Distributes over Addition | |||||||||||
\(\ds \) | \(=\) | \(\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h + \lim_{h \mathop \to 0} \frac {\map g {z_0 + h} - \map g {z_0} } h\) | Sum Rule for Limits of Complex Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \map {f'} {z_0} + \map {g'} {z_0}\) | Definition of Derivative of Complex Function | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \forall z \in D: \, \) | \(\ds \map { \paren{f+g}' } z\) | \(=\) | \(\ds \map {f'} z + \map {g'} z\) | Definition of Derivative of Complex Function |
$\blacksquare$
Proof 2
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 $\S 1.1$