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:

$\forall z \in D : D_z \left({c f \left({z}\right)}\right) = c D_z \left({f \left({z}\right)}\right)$


Proof

\(\displaystyle D_z \left({c f \left({z}\right)}\right)\) \(=\) \(\displaystyle c D_z \left({f \left({z}\right)}\right) + f \left({z}\right) D_z \left({c}\right)\) Product Rule for Complex Derivatives
\(\displaystyle \) \(=\) \(\displaystyle c D_z \left({f \left({z}\right)}\right) + 0\) Complex Derivative of Constant
\(\displaystyle \) \(=\) \(\displaystyle c D_z \left({f \left({z}\right)}\right)\)

$\blacksquare$