# Continuity of Complex Functions

Recall from the Limits of Complex Functions page that if $A \subseteq \mathbb{C}$, $z_0$ is an accumulation point of $\mathbb{C}$, and $f : A \to \mathbb{C}$ then the limit of $f$ as $z$ approaches $z_0$ equals $A$ denoted $\displaystyle{\lim_{z \to z_0} f(z) = A}$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $z \in A$ and $\mid z - z_0 \mid < \delta$ then $\mid f(z) - A \mid < \epsilon$.

With the definition of a limit of a complex function defined we can now also define the concept of continuity of a function $f$ at a point $z_0$.

Definition: Let $A \subseteq \mathbb{C}$ and let $f : A \to \mathbb{C}$. Then $f$ is said to be Continuous at $z_0 \in A$ if $\displaystyle{\lim_{z \to z_0} f(z) = f(z_0)}$. If $f$ is continuous for every point $z_0 \in A$ then $f$ is said to be Continuous on $A$. If $f$ is not continuous at $z_0$ then $f$ is said to be Discontinuous at $z_0$. |

Many of the basic functions that we come across will be continuous functions. For example, all polynomials are continuous on $\mathbb{C}$, i.e., $f(z) = a_0 +a_1z^1 + a_2z^2 + ... + a_nz^n$ is continuous at every $z_0 \in \mathbb{C}$.

For a more complicated example, consider the following function:

(1)This is a rational function. Notice that the numerator of this function is simply a polynomial and is continuous at every $z_0 \in \mathbb{C}$. Problems arise when the denominator equals $0$ though for which $f$ becomes undefined. Notice that $1 + z^2 = 0$ if and only if $z = \pm i$. So $f$ is actually discontinuous at $z = \pm i$ since $f(\pm i)$ is not even defined!

We will now state some of the familiar properties regarding continuous functions.

Lemma 1: Let $f : A \to \mathbb{C}$. Then $f$ is continuous at $z_0$ if and only if for every $\epsilon > 0$ there exists a $\delta > 0$ such that $f(B(z_0, \delta)) \subseteq B(f(z_0), \epsilon)$. |

**Proof:**We have that $f$ is continuous at $z_0$ if and only if $\displaystyle{\lim_{z \to z_0} f(z) = f(z_0)}$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $|z - z_0| < \delta$ then $|f(z) - f(z_0)| < \epsilon$. Equivalently, $f$ is continuous at $z_0$ if and only if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $z \in B(z_0, \delta)$ then $f(z) \in B(f(z_0), \epsilon)$, that is:

Lemma 1 is a useful classification for continuous functions. It states that a function $f$ is continuous at $z_0$ provided that for any $\epsilon > 0$ we can find a $\delta > 0$ (possibly depending on $\epsilon$) for which whenever we are $\delta$ close to $z_0$ we can guarantee that $f$ is $\epsilon$ close to $f(z_0)$.

Theorem 2: Let $f$ and $g$ be continuous functions at $z_0 \in \mathbb{C}$. Then:a) $f + g$ is continuous at $z_0$.b) $f - g$ is continuous at $z_0$.c) $fg$ is continuous at $z_0$.d) $\displaystyle{\frac{f}{g}}$ is continuous at $z_0$ provided that $g(z_0) \neq 0$ |

*The proofs of (a) - (d) are omitted but follow from the limit laws for complex functions.*