Combination Theorem for Complex Derivatives/Multiple Rule

Theorem
Let $D$ be an open subset of the set of complex numbers $\C$.

Let $f: D \to \C$ be a complex-differentiable function on $D$.

Let $w \in \C$.

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

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 $h \in B_r \left({0}\right) \setminus \left\{ {0}\right\}$:


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

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

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

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

Then the Alternative Differentiability Condition shows that:


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