### 1 The Cauchy-Riemann equations

Remembering that $z=x+\text{i}y$ and $w=u+\text{i}v$ , we note that there is a very useful test to determine whether a function $w=f\left(z\right)$ is analytic at a point. This is provided by the Cauchy-Riemann equations. These state that $w=f\left(z\right)$ is differentiable at a point $z={z}_{0}$ if, and only if,

$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ at that point.

When these equations hold then it can be shown that the complex derivative may be determined by using either $\frac{df}{dz}=\frac{\partial f}{\partial x}$   or   $\frac{df}{dz}=-\text{i}\frac{\partial f}{\partial y}$ .

(The use of ‘if, and only if,’ means that if the equations are valid, then the function is differentiable and vice versa .)

If we consider $f\left(z\right)={z}^{2}={x}^{2}-{y}^{2}+2\text{i}xy$ then $u={x}^{2}-{y}^{2}$ and $v=2xy$ so that

$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=2x,\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=-2y,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial x}=2y,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial y}=2x.$

It should be clear that, for this example, the Cauchy-Riemann equations are always satisfied; therefore, the function is analytic everywhere. We find that

$\frac{df}{dz}=\frac{\partial f}{\partial x}=2x+2\text{i}y=2z$ or, equivalently, $\frac{df}{dz}=-\text{i}\frac{\partial f}{\partial y}=-\text{i}\left(-2y+2\text{i}x\right)=2z$

This is the result we would expect to get by simply differentiating $f\left(z\right)$ as if it was a real function. For analytic functions this will always be the case i.e. for an analytic function ${f}^{\prime }\left(z\right)$ can be found using the rules for differentiating real functions.

##### Example 3

Show that the function $f\left(z\right)={z}^{3}$ is analytic everwhere and hence obtain its derivative.

##### Solution

$\phantom{\rule{2em}{0ex}}w=f\left(z\right)={\left(x+\text{i}y\right)}^{3}={x}^{3}-3x{y}^{2}+\left(3{x}^{2}y-{y}^{3}\right)\text{i}$

Hence

$\phantom{\rule{2em}{0ex}}u={x}^{3}-3x{y}^{2}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}v=3{x}^{2}y-{y}^{3}.$

Then

$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=3{x}^{2}-3{y}^{2},\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial y}=-6xy,\phantom{\rule{2em}{0ex}}\frac{\partial v}{\partial x}=6xy,\phantom{\rule{2em}{0ex}}\frac{\partial v}{\partial y}=3{x}^{2}-3{y}^{2}.$

The Cauchy-Riemann equations are identically true and $f\left(z\right)$ is analytic everywhere.

Furthermore $\phantom{\rule{2em}{0ex}}\frac{df}{dz}=\frac{\partial f}{\partial x}=3{x}^{2}-3{y}^{2}+\left(6xy\right)\text{i}=3{\left(x+\text{i}y\right)}^{2}=3{z}^{2}$ as we would expect.

We can easily find functions which are not analytic anywhere and others which are only analytic in a restricted region of the complex plane. Consider again the function $f\left(z\right)=\stackrel{̄}{z}=x-\text{i}y$ . Here

$\phantom{\rule{2em}{0ex}}u=x$ so that $\frac{\partial u}{\partial x}=1,$ and $\frac{\partial u}{\partial y}=0;\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}v=-y$ so that $\frac{\partial v}{\partial x}=0,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial y}=-1.$

The Cauchy-Riemann equations are never satisfied so that $\stackrel{̄}{z}$ is not differentiable anywhere and so is not analytic anywhere.

By contrast if we consider the function $f\left(z\right)=\frac{1}{z}$ we find that

$\phantom{\rule{2em}{0ex}}u=\frac{x}{{x}^{2}+{y}^{2}};\phantom{\rule{1em}{0ex}}v=\frac{y}{{x}^{2}+{y}^{2}}.$

As can readily be shown, the Cauchy-Riemann equations are satisfied everywhere except for ${x}^{2}+{y}^{2}=0,$ i.e. $x=y=0$ (or, equivalently, $z=0$ .) At all other points ${f}^{\prime }\left(z\right)=-\frac{1}{{z}^{2}}$ . This function is analytic everywhere except at the single point $z=0$ .

Analyticity is a very powerful property of a function of a complex variable. Such functions tend to behave like functions of a real variable.

##### Example 4

Show that if $f\left(z\right)=z\stackrel{̄}{z}$ then ${f}^{\prime }\left(z\right)$ exists only at $z=0$ .

##### Solution

$f\left(z\right)={x}^{2}+{y}^{2}$ so that $u={x}^{2}+{y}^{2},\phantom{\rule{1em}{0ex}}v=0.$ $\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=2x,\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=2y,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial x}=0,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial y}=0.$

Hence the Cauchy-Riemann equations are satisfied only where $x=0$ and $y=0$ , i.e. where $z=0$ . Therefore this function is not analytic anywhere.

#### 1.1 Analytic functions and harmonic functions

Using the Cauchy-Riemann equations in a region of the $z$ -plane where $f\left(z\right)$ is analytic, gives

$\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\right\frac{\partial u}{\partial y}\left)\right=\frac{\partial }{\partial x}\left(\right-\frac{\partial v}{\partial x}\left)\right=-\frac{{\partial }^{2}v}{\partial {x}^{2}}$

and

$\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\right\frac{\partial u}{\partial x}\left)\right=\frac{\partial }{\partial y}\left(\right\frac{\partial v}{\partial y}\left)\right=\frac{{\partial }^{2}v}{\partial {y}^{2}}.$

If these differentiations are possible then $\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{{\partial }^{2}u}{\partial y\partial x}$ so that

$\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{{\partial }^{2}u}{\partial {y}^{2}}=0$ (Laplace’s equation)

In a similar way we find that

$\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}v}{\partial {x}^{2}}+\frac{{\partial }^{2}v}{\partial {y}^{2}}=0$ (Can you show this?)

When $f\left(z\right)$ is analytic the functions $u$ and $v$ are called conjugate harmonic functions .

Suppose $u=u\left(x,y\right)=xy$ then it is easy to verify that $u$ satisfies Laplace’s equation (try this). We now try to find the conjugate harmonic function $v=v\left(x,y\right).$

First, using the Cauchy-Riemann equations:

$\phantom{\rule{2em}{0ex}}\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=y\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-x.$

Integrating the first equation gives $v=\frac{1}{2}{y}^{2}+$ a function of $x$ . Integrating the second equation gives $v=-\frac{1}{2}{x}^{2}+$ a function of $y$ . Bearing in mind that an additive constant leaves no trace after differentiation, we pool the information above to obtain

Note that $f\left(z\right)=u+\text{i}v=xy+\frac{1}{2}\left({y}^{2}-{x}^{2}\right)\text{i}+D$ where $D$ is a constant (replacing $C\text{i}$ ).

We can write $f\left(z\right)=-\frac{1}{2}\text{i}{z}^{2}+D$ (as you can verify). This function is analytic everywhere.

Given the function $u={x}^{2}-x-{y}^{2}$

1. Show that $u$ is harmonic,
2. Find the conjugate harmonic function, $v$ .

$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=2x-1,\phantom{\rule{1em}{0ex}}\frac{{\partial }^{2}u}{\partial {x}^{2}}=2,\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=-2y,\phantom{\rule{1em}{0ex}}\frac{{\partial }^{2}u}{\partial {y}^{2}}=-2.$

Hence $\frac{{\partial }^{2}u}{\partial {x}^{2}}+\frac{{\partial }^{2}u}{\partial {y}^{2}}=0$ and $u$ is harmonic.

Integrating $\frac{\partial v}{\partial y}=2x-1$ gives $v=2xy-y+$ function of $x$ .

Integrating $\frac{\partial v}{\partial x}=+2y$ gives $v=2xy+$ function of $y$ .

Ignoring the duplication, $v=2xy-y+C,$ where $C$ is a constant.

Find $f\left(z\right)$ in terms of $z$ , where $f\left(z\right)=u+\text{i}v$ , where $u$ and $v$ are those found in the previous Task.

$f\left(z\right)=u+iv={x}^{2}-x-{y}^{2}+2xy\text{i}-\text{i}y+D,\phantom{\rule{1em}{0ex}}D\phantom{\rule{1em}{0ex}}\text{constant}.$

Now $\phantom{\rule{1em}{0ex}}{z}^{2}={x}^{2}-{y}^{2}+2\text{i}xy$ and $z=x+\text{i}y$ thus $\phantom{\rule{2em}{0ex}}f\left(z\right)={z}^{2}-z+D.$

##### Exercises
1. Find the singular point of the rational function $f\left(z\right)=\frac{z}{z-2\text{i}}$ . Find ${f}^{\prime }\left(z\right)$ at other points and evaluate ${f}^{\prime }\left(-\text{i}\right)$ .
2. Show that the function $f\left(z\right)={z}^{2}+z$ is analytic everywhere and hence obtain its derivative.
3. Show that the function $u={x}^{2}-{y}^{2}-2y$ is harmonic, find the conjugate harmonic function $v$ and hence find $f\left(z\right)=u+\text{i}v$ in terms of $z$ .
1. $f\left(z\right)$ is singular at $z=2\text{i}$ . Elsewhere

$\phantom{\rule{2em}{0ex}}{f}^{\prime }\left(z\right)=\frac{\left(z-2\text{i}\right).1-z.1}{{\left(z-2\text{i}\right)}^{2}}=\frac{-2\text{i}}{{\left(z-2\text{i}\right)}^{2}}$ $\phantom{\rule{2em}{0ex}}{f}^{\prime }\left(-\text{i}\right)=\frac{-2\text{i}}{{\left(-3\text{i}\right)}^{2}}=\frac{-2\text{i}}{-9}=\frac{2}{9}\text{i}$

2. $u={x}^{2}+x-{y}^{2}$   and    $v=2xy+y$

$\phantom{\rule{2em}{0ex}}\frac{\partial u}{\partial x}=2x+1,\phantom{\rule{1em}{0ex}}\frac{\partial u}{\partial y}=-2y,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial x}=2y,\phantom{\rule{1em}{0ex}}\frac{\partial v}{\partial y}=2x+1$

Here the Cauchy-Riemann equations are identically true and $f\left(z\right)$ is analytic everywhere.

$\phantom{\rule{2em}{0ex}}\frac{df}{dz}=\frac{\partial f}{\partial x}=2x+1+2y\text{i}=2z+1$

3. $\phantom{\rule{2em}{0ex}}\frac{{\partial }^{2}u}{\partial {x}^{2}}=2,\phantom{\rule{1em}{0ex}}\frac{{\partial }^{2}u}{\partial {y}^{2}}=-2$ therefore $u$ is harmonic.

$\phantom{\rule{2em}{0ex}}\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=2x$ therefore $v=2xy+$  function of $y$

$\phantom{\rule{2em}{0ex}}\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=2y+2$    therefore $v=2xy+2x+$  function of $x$

$\phantom{\rule{2em}{0ex}}\therefore \phantom{\rule{2em}{0ex}}v=2xy+2x+\phantom{\rule{1em}{0ex}}\text{constant}$

$\phantom{\rule{2em}{0ex}}f\left(z\right)={x}^{2}+2\text{i}xy-{y}^{2}+2x\text{i}-2y={z}^{2}+2\text{i}z$