3 Differentiating functions of a complex variable

The function $f\left(z\right)$ is said to be differentiable at $z=z_0$ if

$\underset{\Delta z\to 0}{lim}\left\{\right.\frac{f\left(z_0+\Delta z\right)-f\left(z_0\right)}{\Delta z}\left\}\right.\text{exists.}$ 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 $z=\pm \text{i}$ .

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

$f^{\prime }\left(z\right)=-\frac{2z}{{\left(z^2+1\right)}^2}\left(\text{quotient rule}\right)$

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 $f\left(z\right)=\bar{z}=x-\text{i}y$ 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

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:

$\Delta y=0$ 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:

$\Delta x=0$ so that $R=\frac{-\text{i}\Delta y}{\text{i}\Delta y}=-1$ , 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.