# Continuity

A function $f$ is **continuous at a point** $z_0$ if the following
conditions are
satisfied:

- $\ds\lim_{z\rightarrow z_0}f(z)$ exists,
- $f(z_0)$ exists, and
- $\ds\lim_{z\rightarrow z_0}f(z) = f(z_0).$

Notice that **statement 3** actually contains
**statement 1** and **2**, since the
existence of the quantity on each side of the equation there is needed.

**Statement 3**
says that for each positive number $\epsilon,$ there is a positive number $\delta$ such
that

A function of a complex variable is said to be **continuous in a region** $R$ if it is
continuous at each point in $R.$

If a function $f$ is not continuous at a point $z_0$ then we say that $f$ is discontinuous at $z_0.$ For example, the function \[ f(z)=\frac{1}{1+z^2} \] is discontinuous at $z=i$ and $z=-i.$

**Example 1:**
We claim that the function $f(z)=\dfrac{1}{z}$ is continuous for every $z\neq 0.$

To prove this fix $z_0\in \C\setminus\{0\}.$ If $z\in \C\setminus\{0\},$ then \[ \abs{\frac{1}{z}-\frac{1}{z_0}} = \frac{\abs{z-z_0}}{\abs{z} \abs{z_0}}. \] If $\abs{z-z_0}\lt \dfrac{\abs{z_0}}{2},$ then

Thus, by choosing $\delta = \min\left\{ \dfrac{\abs{z_0}^2}{2}\epsilon, \dfrac{\abs{z_0}}{2} \right\}$ then we have that $$ \abs{z-z_0}\lt \delta \implies \left| \frac{1}{z}-\frac{1}{z_0}\right|\lt\epsilon. $$

**Remark:**
A *useful strategy* for showing that $f(z)$ is continuous at $z_0$ is
to obtain an estimate of the form

In the previous example we showed that a function is continuous at every point $z\in \C\setminus\{0\}.$ The following example shows how to determine continuity at a point using properties of limits.

**Example 2:**
Consider the function $f(z) = z^2 - iz+ 2.$
In order to show that $f$ is continuous at, say, $z_0=1-i,$
we must find $\ds \lim_{z\to z_0} f(z)$ and $f(z_0),$ then
check to see whether these two complex values are equal.

Using the properties of limits we have that

Furthermore, for $z_0=1-i$ we have

Since $ \ds\lim_{z\to z_0} f(z) = f(z_0),$ we conclude that $f(z) = z^2 - iz+ 2$ is continuous at $z_0 = 1-i.$

Suppose now that $f$ is continuous at $z$ and that $g$ is continuous at the point $f (z )$ in the $w$-plane. Since $g$ is continuous at $f (z ),$ for each $\epsilon\gt 0$ there is $\delta_2\gt 0 $ such that

Since $f$ is continuous at $z_0,$ this ensures that the neighborhood $\abs{z-z_0}\lt\delta_1$ can be made small enough such that the inequalities given in (\ref{composition}) hold. Therefore, the compostion $g\big(f(z)\big)$ is continuous.

Since $f(z_0)\neq 0$, we can take $\epsilon=\dfrac{|f(z_0)|}{2}.$ So we have

The algebraic properties of limits can also be restated in terms of continuity of complex functions. The proof of the following theorem is left as an exercise.

- $c\cdot f,$ with $c$ a complex constant,
- $f \pm g,$
- $f\cdot g,$ and
- $\dfrac{f}{g},$ if $g(z_0)\neq 0.$

**Exercise:**
Prove that polynomial functions are continuous
on the entire complex plane $\C.$