Combination Theorem for Complex Derivatives/Sum Rule/Proof 1

Theorem

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

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:

$\ds \map { k' } {z_0}$

$=$

$\ds \lim_{h \mathop \to 0} \frac {\map k {z_0 + h} - \map k {z_0} } h$

$\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 $

$=$

$\ds \lim_{h \mathop \to 0} \paren {\frac {\map f {z_0 + h} - \map f {z_0} } h + \frac {\map g {z_0 + h} - \map g {z_0} } h}$

$\ds $

$=$

$\ds \lim_{h \mathop \to 0} \frac {\map f {z_0 + h} - \map f {z_0} } h + \lim_{h \mathop \to 0} \frac {\map g {z_0 + h} - \map g {z_0} } h$

$\ds $

$=$

$\ds \map {f'} {z_0} + \map {g'} {z_0}$

$\ds \leadsto \ \ $

$\ds \forall z \in D: \, $

$\ds \map { \paren{f+g}' } z$

$=$

$\ds \map {f'} z + \map {g'} z$

$\blacksquare$