Derivative of Constant Multiple/Complex

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 $c \in \C$ be a constant.

Then:

$\forall z \in D : \map {D_z} {c \map f z} = c \map {D_z} {\map f z}$

Proof

\(\ds \map {D_z} {c \map f z}\)

\(=\)

\(\ds c \map {D_z} {\map f z} + \map f z \map {D_z} c\)

Product Rule for Complex Derivatives

\(\ds \)

\(=\)

\(\ds c \map {D_z} {\map f z} + 0\)

Complex Derivative of Constant

\(\ds \)

\(=\)

\(\ds c \map {D_z} {\map f z}\)

$\blacksquare$