# 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 $\left({f + g}\right)'$ is defined by:

$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$

for all $z \in D$.

## Proof

Denote the open ball of $0$ with radius $r \in \R_{>0}$ as $B_r \left({0}\right)$.

Let $z \in D$.

By the Alternative Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:

$f\left({z + h}\right) = f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right)$
$g\left({z + h}\right) = g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$

where $\epsilon_f, \epsilon_g: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ are continuous functions that converge to $0$ as $h$ tends to $0$.

Then:

 $\displaystyle \left({f + g}\right) \left({z + h}\right)$ $=$ $\displaystyle f \left({z}\right) + h \left({f' \left({z}\right) + \epsilon_f \left({h}\right) }\right) + g \left({z}\right) + h \left({g' \left({z}\right) + \epsilon_g \left({h}\right) }\right)$ $\displaystyle$ $=$ $\displaystyle \left({f + g}\right) \left({z}\right) + h \left({f' \left({z}\right) + g' \left({z}\right) + \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) }\right)$

From Sum Rule for Continuous Functions, it follows that $\epsilon_f + \epsilon_g$ is a continuous function.

From Sum Rule for Limits of Functions, it follows that $\displaystyle \lim_{h \to 0} \left({ \epsilon_f + \epsilon_g }\right) \left({h}\right) = 0$.

Then the Alternative Differentiability Condition shows that:

$\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$

$\blacksquare$