Combination Theorem for Complex Derivatives/Sum Rule

Theorem
Let $f, g: D \to \C$ be complex-differentiable functions, where $D$ is an open subset of the set of complex numbers.

Then $f + g$ is complex-differentiable in $D$, and its derivate $\left({f + g}\right)'$ is defined by:
 * $\left({f + g}\right)' \left({z}\right) = f' \left({z}\right) + g' \left({z}\right)$

Proof
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:

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

From Combination Theorem for Limits of Functions/Sum Rule, 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:


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