Complex-Differentiable Function is Continuous

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

Suppose that $f$ is complex-differentiable at $z \in D$.

Then $f$ is continuous at $z$.

Proof
By the Epsilon-Function Complex Differentiability Condition, it follows that there exists $r \in \R_{>0}$ such that for all $h \in \map {B_r} 0 \setminus \set 0$:


 * $\ds \lim_{h \mathop \to 0} \map f {z + h} = \map f z + h \paren {\map f z + \map \epsilon h}$

It follows from definition of limit that for all $\epsilon \in \R_{>0}$, there exists $\delta \in \R_{>0}$ with this property:

If $\cmod {h - 0} < \delta$, then $\cmod {\map f {z + h} - \map f z} < \epsilon$.

Put $z' = z + h$, so $\cmod {z' - z} = \cmod h$.

It follows that:

If $\cmod {z' - z} < \delta$, then $\cmod {\map f {z'} - \map f z} < \epsilon$.

Then $f$ is continuous at $z$ by definition.