Complex-Differentiable Function is Continuous/Proof 2

Theorem

Let $f: D \to \C$ be a complex function, where $D \subseteq \C$ is an open set.

Let $f$ be complex-differentiable at $a \in D$.

Then $f$ is continuous at $a$.

Proof

For each $z \in D$:

\(\ds \lim_{w \mathop \to z} \map f w\)

\(=\)

\(\ds \map f z + \lim_{w \mathop \to z} \paren {\map f w - \map f z}\)

Sum Rule for Limits of Complex Functions

\(\ds \)

\(=\)

\(\ds \map f z + \lim_{w \mathop \to z} \paren {\frac {\map f w - \map f z} {w - z} \paren {w - z} }\)

\(\ds \)

\(=\)

\(\ds \map f z + \lim_{w \mathop \to z} \frac {\map f w - \map f z} {w - z} \lim_{w \mathop \to z} \paren {w - z}\)

Product Rule for Limits of Complex Functions

\(\ds \)

\(=\)

\(\ds \map f z + \map {f'} z \cdot 0\)

Definition of Complex-Differentiable Function

\(\ds \)

\(=\)

\(\ds \map f z\)

$\blacksquare$