Combination Theorem for Complex Derivatives/Multiple Rule

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

Let $w \in \C$.

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

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

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


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

where $\epsilon: B_r \left({0}\right) \setminus \left\{ {0}\right\} \to \C$ is a continuous functions with $\displaystyle \lim_{h \to 0} \epsilon \left({h}\right) = 0$.

Then:


 * $w f\left({z + h}\right) = w f \left({z}\right) + h \left({w f' \left({z}\right) + w \epsilon \left({h}\right) }\right)$

From Combination Theorem for Continuous Functions/Multiple Rule, it follows that $w \epsilon$ is a continuous function.

From Combination Theorem for Limits of Functions/Multiple Rule, it follows that $\displaystyle \lim_{h \to 0} w \epsilon \left({h}\right) = \left({ \lim_{h \to 0} w }\right) \left({ \lim_{h \to 0} \epsilon \left({h}\right) }\right) = 0$.

Then the Alternative Differentiability Condition shows that:


 * $\left({wf}\right)' \left({z}\right) = wf' \left({z}\right)$