# 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 1

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 $h \in \map {N_r} 0 \setminus \set 0$:

$(1): \quad \map f {a + h} = \map f a + h \paren {\map {f'} a + \map \epsilon h}$

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 right hand side of $(1)$ to get:

 $\ds \map f {a + h}$ $=$ $\ds \map f a + h \map {f'} a + h \map \epsilon h$ $\ds \leadsto \ \$ $\ds \map f z$ $=$ $\ds \map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a}$ substituting $z = a + h$ for $z \in \map {N_r} a$ $\ds \leadsto \ \$ $\ds \lim_{z \mathop \to a} \map f z$ $=$ $\ds \lim_{z \mathop \to a} \paren {\map f a + \paren {z - a} \map {f'} a + \paren {z - a} \map \epsilon {z - a} }$ Take the limit on both sides $\ds$ $=$ $\ds \map f a + \lim_{z \mathop \to a} \paren {z - a} \map {f'} a + \lim_{z \mathop \to a} \paren {\paren {z - a} \map \epsilon {z - a} }$ Combination Theorem for Limits of Complex Functions $\ds$ $=$ $\ds \map f a$ as $\ds \lim_{z \mathop \to a} \paren {z - a} = 0$, and $\ds \lim_{z \mathop \to a} \map \epsilon {z - a} = 0$

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

$\blacksquare$

## Proof 2

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 $\ds$ $=$ $\ds \map f z$

$\blacksquare$