Complex-Differentiable Function is Continuous

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
Let $\map {N_r} 0$ denote the $r$-neighborhood of $0$ in $\C$.

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


 * $\map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$

where $\epsilon: \map {N_r} 0 \setminus \set 0 \to \C$ is a complex function with $\ds \lim_{h \mathop \to 0} \map \epsilon h = 0$.

We rewrite the of this equation to get:

By definition of continuous complex function, it follows that $f$ is continuous at $a$.

Proof 2
For each $z \in D$: