Definition:Continuous Complex Function

This page is about Continuous Mapping in the context of Complex Analysis. For other uses, see Continuous Mapping.

Definition

As the complex plane is a metric space, the same definition of continuity applies to complex functions as to metric spaces.

Let $A_1, A_2 \subseteq \C$ be subsets of the complex plane.

Let $f: A_1 \to A_2$ be a complex function from $A_1$ to $A_2$.

Let $a \in A_1$.

$f$ is continuous at (the point) $a$ if and only if:

The limit of $\map f z$ as $z \to a$ exists, and

$\ds \lim_{z \mathop \to a} \map f z = \map f a$

$f$ is continuous at (the point) $a$ if and only if:

$\forall \epsilon > 0: \exists \delta > 0: \forall z \in A_1: \cmod {z - a} < \delta \implies \cmod {\map f z - \map f a} < \epsilon$

Let $A_1$ be open in $\C$.

$f$ is continuous at (the point) $a$ if and only if:

$\forall \map {\NN_\epsilon} {\map f a}: \exists \map {\NN_\delta} a: f \sqbrk {\map {\NN_\delta} a} \subseteq \map {\NN_\epsilon} {\map f a}$

where $\map {\NN_\epsilon} a$ is the $\epsilon$-neighborhood of $a$ in $A_1$.

That is, for every $\epsilon$-neighborhood of $\map f a$ in $\C$, there exists a $\delta$-neighborhood of $a$ in $\C$ whose image is a subset of that $\epsilon$-neighborhood.

Let $A_1$ be open in $\C$.

$f$ is continuous if and only if:

for every set $U \subseteq \C$ which is open in $\C$, $f^{-1} \sqbrk{ U }$ is open in $\C$.