### 2 The limit of a function

The limit of $w=f\left(z\right)$ as $z\to {z}_{0}$ is a number $\ell$ such that $|f\left(z\right)-\ell |$ can be made as small as we wish by making $|z-{z}_{0}|$ sufficiently small. In some cases the limit is simply $f\left({z}_{0}\right)$ , as is the case for $w={z}^{2}-z$ . For example, the limit of this function as $z\to \text{i}$ is $f\left(\text{i}\right)={\text{i}}^{2}-\text{i}=-1-\text{i}$ .

There is a fundamental difference from functions of a real variable: there, we could approach a point on the curve $y=g\left(x\right)$ either from the left or from the right when considering limits of $g\left(x\right)$ at such points. With the function $f\left(z\right)$ we are allowed to approach the point $z={z}_{0}$ along any path in the $z$ -plane; we require merely that the distance $\left|z-{z}_{0}\right|$ decreases to zero.

Suppose that we want to find the limit of $f\left(z\right)={z}^{2}-z$ as $z\to 2+\text{i}$ along each of the paths (a), (b) and (c) indicated in Figure 1.

Figure 1

1. Along this path $z=x+\text{i}$ (for any $x$ ) and ${z}^{2}-z={x}^{2}+2x\text{i}-1-x-i$

That is: ${z}^{2}-z={x}^{2}-1-x+\left(2x-1\right)\text{i}$ .

As $z\to 2+\text{i},$ then $x\to 2$ so that the limit of ${z}^{2}-z$ is ${2}^{2}-1-2+\left(4-1\right)\text{i}=1+3\text{i}$ .

2. Here $z=2+y\text{i}$ (for any $y$ ) so that ${z}^{2}-z=4-{y}^{2}-2+\left(4y-y\right)\text{i}.$

As $z\to 2+\text{i},y\to 1$ so that the limit of ${z}^{2}-z$ is $4-1-2+\left(4-1\right)\text{i}=1+3\text{i}$ .

3. Here $z=k\left(2+\text{i}\right)$ where $k$ is a real number. Then

$\phantom{\rule{2em}{0ex}}{z}^{2}-z={k}^{2}\left(4+4i-1\right)-k\left(2+\text{i}\right)=3{k}^{2}-2k+\left(4{k}^{2}-k\right)\text{i}.$

As $z\to 2+\text{i},k\to 1$ so that the limit of ${z}^{2}-z$ is $3-2+\left(4-1\right)\text{i}=1+3\text{i}$ .

In each case the limit is the same.

Evaluate the limit of $f\left(z\right)={z}^{2}+z+1$ as $z\to 1+2\text{i}$ along the paths

1. parallel to the $x$ -axis coming from the right,
2. parallel to the $y$ -axis, coming from above,
3. the line joining the point $1+2\text{i}$ to the origin, coming from the origin.
1. Along this path $z=x+2\text{i}$ and ${z}^{2}+z+1={x}^{2}-4+x+1+\left(4x+2\right)\text{i}$ . As $z\to 1+2\text{i},x\to 1$

and ${z}^{2}+z+1\to -1+6\text{i}$ .

2. Along this path $z=1+y\text{i}$

and ${z}^{2}+z+1=1-{y}^{2}+1+1+\left(2y+y\right)\text{i}$ . As $z\to 1+2\text{i},y\to 2$ and ${z}^{2}+z+1\to -1+6\text{i}$ .

3. If $z=k\left(1+2\text{i}\right)$ then ${z}^{2}+z+1={k}^{2}+k+1-4{k}^{2}+\left(4{k}^{2}+2k\right)\text{i}$ . As $z\to 1+2\text{i},k\to 1$

and ${z}^{2}+z+1\to -1+6\text{i}$ .

Not all functions of a complex variable are as straightforward to analyse as the last two examples. Consider the function $f\left(z\right)=\frac{\stackrel{̄}{z}}{z}$ . Along the $x$ -axis moving towards the origin from the right

$\phantom{\rule{2em}{0ex}}z=x$ and $\stackrel{̄}{z}=x$ so that $f\left(z\right)=1$ which is the limit as $z\to 0$ along this path .

However, we can approach the origin along any path. If instead we approach the origin along the positive $y$ -axis $z=\text{i}y$ then

$\phantom{\rule{2em}{0ex}}\stackrel{̄}{z}=-\text{i}y$ and $f\left(z\right)=\frac{\stackrel{̄}{z}}{z}=-1$ , which is the limit as $z\to 0$ along this path .

Since these two limits are distinct then $\underset{z\to 0}{lim}\frac{\stackrel{̄}{z}}{z}$ does not exist .

We cannot assume that the limit of a function $f\left(z\right)$ as $z\to {z}_{0}$ is independent of the path chosen.

#### 2.1 Definition of continuity

The function $f\left(z\right)$ is continuous as $z\to {z}_{0}$ if the following two statements are true:

1. $f\left({z}_{0}\right)$ exists;
2. $\underset{z\to {z}_{0}}{lim}f\left(z\right)$ exists and is equal to $f\left({z}_{0}\right)$ .

As an example consider $f\left(z\right)=\frac{{z}^{2}+4}{{z}^{2}+9}$ . As $z\to \text{i},$ then $f\left(z\right)\to f\left(\text{i}\right)=\frac{{\text{i}}^{2}+4}{{\text{i}}^{2}+9}=\frac{3}{8}.$ Thus $f\left(z\right)$ is continuous at $z=\text{i}$ .

However, when ${z}^{2}+9=0$ then $z=±3\text{i}$ and neither $f\left(3\text{i}\right)$ nor $f\left(-3\text{i}\right)$ exists. Thus $\frac{{z}^{2}+4}{{z}^{2}+9}$ is discontinuous at $z=±3\text{i}$ . It is easily shown that these are the only points of discontinuity.

State where $f\left(z\right)=\frac{z}{{z}^{2}+4}$ is discontinuous. Find $\underset{z\to \text{i}}{lim}f\left(z\right)$ .

${z}^{2}+4=0$ where $z=±2i$ ; at these points $f\left(z\right)$ is discontinuous as $f\left(±2\text{i}\right)$ does not exist.

$\phantom{\rule{2em}{0ex}}\underset{z\to \text{i}}{lim}f\left(z\right)=f\left(\text{i}\right)=\frac{\text{i}}{{\text{i}}^{2}+4}=\frac{1}{3}\text{i}.$ It is easily shown that any polynomial in $z$ is continuous everywhere whilst any rational function is continuous everywhere except at the zeroes of the denominator.

##### Exercises
1. For which values of $z$ is $w=\frac{1}{z-\text{i}}$ defined? For these values obtain $u$ and $v$ and evaluate $w$ when $z=1-2\text{i}$ .
2. Find the limit of $f\left(z\right)={z}^{3}+z$ as $z\to \text{i}$ along the paths
1. parallel to the $x$ -axis coming from the right,
2. parallel to the $y$ -axis coming from above.
3. Where is $f\left(z\right)=\frac{z}{{z}^{2}+9}$ discontinuous?. Find the $\underset{z\to -\text{i}}{lim}f\left(z\right)$ .
1. $w$ is defined for all $z\ne \text{i}$ $\phantom{\rule{2em}{0ex}}w=\frac{1}{x+y\text{i}-\text{i}}=\frac{1}{x+\left(y-1\right)\text{i}}×\frac{x-\left(y-1\right)\text{i}}{x-\left(y-1\right)\text{i}}=\frac{x-\left(y-1\right)\text{i}}{{x}^{2}+{\left(y-1\right)}^{2}}.$

$\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}u=\frac{x}{{x}^{2}+{\left(y-1\right)}^{2}},\phantom{\rule{2em}{0ex}}v=\frac{-\left(y-1\right)}{{x}^{2}+{\left(y-1\right)}^{2}}$ .

When $z=1-2\text{i}$ ,   $x=1,\phantom{\rule{1em}{0ex}}y=-2$  so that $\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}u=\frac{1}{1+9}=\frac{1}{10},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}v=\frac{3}{10},\phantom{\rule{1em}{0ex}}z=\frac{1}{10}+\frac{3}{10}\text{i}$

1. $z=x+\text{i},\phantom{\rule{1em}{0ex}}{z}^{3}+z={x}^{3}+3{x}^{2}\text{i}-2x$ . As $z\to \text{i}$ ,    $x\to 0$   and   ${z}^{3}+z\to 0$
2. $z=y\text{i},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}{z}^{3}+z=-{y}^{3}\text{i}+y\text{i}$ . As   $z\to \text{i},\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}y\to 1$   and   ${z}^{3}+z\to -\text{i}+\text{i}=0$ .
2. $f\left(z\right)$ is discontinuous at $z=±3\text{i}$ . The limit is $f\left(-\text{i}\right)=\frac{-\text{i}}{-1+9}=-\frac{1}{8}\text{i}$ .