Derivative of Constant Multiple/Complex
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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 | |||||||||||
\(\ds \) | \(=\) | \(\ds c \map {D_z} {\map f z}\) |
$\blacksquare$