### 3 Differentiating functions of a complex variable

The function $f\left(z\right)$ is said to be differentiable at $z={z}_{0}$ if

$\phantom{\rule{2em}{0ex}}\underset{\Delta z\to 0}{lim}\left\{\right\\frac{f\left({z}_{0}+\Delta z\right)-f\left({z}_{0}\right)}{\Delta z}\left\}\right\\phantom{\rule{1em}{0ex}}\text{exists.}$ Here $\Delta z=\Delta x+\text{i}\Delta y$ .

Apart from a change of notation this is precisely the same as the definition of the derivative of a function of a real variable. Not surprisingly then, the rules of differentiation used in functions of a real variable can be used to differentiate functions of a complex variable. The value of the limit is the derivative of $f\left(z\right)$ at $z={z}_{0}$ and is often denoted by $\frac{df}{dz}{|}_{z={z}_{0}}$ or by ${f}^{\prime }\left({z}_{0}\right)$ .

A point at which the derivative does not exist is called a singular point of the function.

A function $f\left(z\right)$ is said to be analytic at a point ${z}_{0}$ if it is differentiable throughout a neighbourhood of ${z}_{0}$ , however small. (A neighbourhood of ${z}_{0}$ is the region contained within some circle $\left|x-{z}_{0}\right|=r$ .)

For example, the function $f\left(z\right)=\frac{1}{{z}^{2}+1}$ has singular points where ${z}^{2}+1=0,$  i.e. at $z=±\text{i}$ .

For all other points the usual rules for differentiation apply and hence

So, for example, at $z=3\text{i},$ ${f}^{\prime }\left(z\right)=-\frac{6\text{i}}{{\left(-9+1\right)}^{2}}=-\frac{3}{32}\text{i}.$

##### Example 2

Find the singular point of the rational function $f\left(z\right)=\frac{z}{z+\text{i}}.$ Find ${f}^{\prime }\left(z\right)$ at other points and evaluate ${f}^{\prime }\left(2\text{i}\right)$ .

##### Solution

$z+\text{i}=0$ when $z=-\text{i}$ and this is the singular point: $f\left(-\text{i}\right)$ does not exist. Elsewhere, differentiating using the quotient rule:

${f}^{\prime }\left(z\right)=\frac{\left(z+\text{i}\right)\cdot 1-z\cdot 1}{{\left(z+\text{i}\right)}^{2}}=\frac{\text{i}}{{\left(z+\text{i}\right)}^{2}}.$ Thus at $z=2\text{i},$ we have ${f}^{\prime }\left(z\right)=\frac{\text{i}}{{\left(3\text{i}\right)}^{2}}=-\frac{1}{9}\text{i}$ .

The simple function $f\left(z\right)=\stackrel{̄}{z}=x-\text{i}y$ is not analytic anywhere in the complex plane. To see this consider looking at the derivative at an arbitrary point ${z}_{0}$ . We easily see that

$\begin{array}{rcll}R& =& \frac{f\left({z}_{0}+\Delta z\right)-f\left({z}_{0}\right)}{\Delta z}& \text{}\\ & =& \frac{\left({x}_{0}+\Delta x\right)-\text{i}\left({y}_{0}+\Delta y\right)-\left({x}_{0}-\text{i}{y}_{0}\right)}{\Delta x+\text{i}\Delta y}=\frac{\Delta x-\text{i}\phantom{\rule{1em}{0ex}}\Delta y}{\Delta x+\text{i}\phantom{\rule{1em}{0ex}}\Delta y}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}& \text{}\end{array}$

Hence $f\left(z\right)$ will fail to have a derivative at ${z}_{0}$ if we can show that this expression has no limit. To do this we consider looking at the limit of the function along two distinct paths.

Along a path parallel to the $x$ -axis:

$\phantom{\rule{2em}{0ex}}\Delta y=0$ so that $R=\frac{\Delta x}{\Delta x}=1$ , and this is the limit as $\Delta z=\Delta x\to 0$ .

Along a path parallel to the $y$ -axis:

$\phantom{\rule{2em}{0ex}}\Delta x=0$ so that $R=\frac{-\text{i}\phantom{\rule{1em}{0ex}}\Delta y}{\text{i}\Delta y}=-1$ , and this is the limit as $\Delta z=\Delta y\to 0.$

As these two values of $R$ are distinct, the limit of $\frac{f\left(z+\Delta z\right)-f\left(z\right)}{\Delta z}$ as $z\to {z}_{0}$ does not exist

and so $f\left(z\right)$ fails to be differentiable at any point. Hence it is not analytic anywhere.