Sum Rule for Complex Derivatives

Theorem

Let $\map f z, \map j z, \map k z$ be single-valued continuous complex functions in a domain $D \subseteq \C$, where $D$ is open.

Let $f$, $j$, and $k$ be complex-differentiable at all points in $D$.

Let $\map f z = \map j z + \map k z$.

Then:

$\forall z \in D: \map {f'} z = \map {j'} z + \map {k'} z$

Let $z_0 \in D$ be a point in $D$.

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

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds $

$=$

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

$\ds \leadsto \ \ $

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

$\ds \map {f'} z$

$=$

$\ds \map {j'} z + \map {k'} z$

$\blacksquare$