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

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

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

\begin{eqnarray}\label{continuouscondition} \left|f (z) - f(z_0)\right| \lt \epsilon \quad \text{whenever}\quad 0 \lt |z - z_0| \lt \delta \end{eqnarray}

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

\[ \abs{z}= \abs{z_0+z-z_0} \geq \abs{z_0}-\abs{z-z_0} \gt \frac{\abs{z}}{2} \]
and therefore
\[ \abs{\frac{1}{z}-\frac{1}{z_0}} \lt \frac{2}{\abs{z_0}^2}\abs{z-z_0}. \]

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

$$|f(z)-f(z_0)|\leq C|z-z_0|\;\; \text{for }\, z \;\;\text{near}\;\; z_0.$$
This guarantees that $|f(z)-f(z_0)|\lt\epsilon $ whenever $|z-z_0|\lt\dfrac{\epsilon}{C},$ so that we can take $\delta=\dfrac{\epsilon}{C}$ in the formal definition of limit.

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

\[ \lim_{z\to z_0} f(z) = \lim_{z\to 1-i } \left(z^2 - iz+ 2\right) = \left(1-i\right)^2-i(1-i)+ 2 = 1-3i. \]

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

\[ f(z_0) = f(1-i) = \left(1-i\right)^2-i(1-i)+ 2 = 1-3i. \]

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

A composition of continuous functions is itself continuous.
Let $w = f (z)$ be a function that is defined for all $z$ in a neighborhood $|z - z_0 | \lt \delta_1$ of a point $z_0,$ and let $W=g(w)$ be a function whose domain of definition contains the image of that neighborhood under $f.$ Then the composition $W=g\big(f(z)\big)$ is defined for all $z$ in the neighborhood $|z - z_0 | \lt \delta_1.$

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

\begin{eqnarray}\label{composition} \abs{g\big(f(z)\big)- g\big(f(z_0)\big)}\lt \epsilon \;\;\text{whenever}\;\; \abs{f(z)- f(z_0)}\lt \delta_2. \end{eqnarray}
Geometric interpretation of composition of functions.

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.

If a function $f$ is continuous and nonzero at a point $z_0,$ then $f(z)\neq 0$ throughout some neighborhood of that point.
Suppose that $f$ is continuous and $f(z_0)\neq 0.$ Then for every $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that $|z-z_0|\lt\delta$ implies

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

$$|f(z)-f(z_0)|\lt\frac{\abs{f(z_0)}}{2}\quad \text{whenever}\quad |z-z_0|\lt\delta.$$
Now, if there is a point $z$ in the neighborhood $|z - z_0| \lt\delta$ at which $f(z)=0,$ we have that $$|f(z_0)|\lt\frac{f(z_0)}{2},$$ which is a contradiction. Hence the theorem is proved.

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.

Suppose that $f$ and $g$ are continous at $z_0,$ then the following are continuous functions at $z_0:$
  1. $c\cdot f,$ with $c$ a complex constant,
  2. $f \pm g,$
  3. $f\cdot g,$ and
  4. $\dfrac{f}{g},$ if $g(z_0)\neq 0.$

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

Complex Differentiation