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,

$\frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}\text{and}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x}$ 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

$\frac{\partial u}{\partial x}=2x,\frac{\partial u}{\partial y}=-2y,\frac{\partial v}{\partial x}=2y,\frac{\partial v}{\partial y}=2x.$

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

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

Hence

$u=x^3-3x{y}^2\text{and}v=3x^{2}y-y^3.$

Then

$\frac{\partial u}{\partial x}=3x^2-3y^2,\frac{\partial u}{\partial y}=-6xy,\frac{\partial v}{\partial x}=6xy,\frac{\partial v}{\partial y}=3x^2-3y^2.$

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

Furthermore $\frac{df}{dz}=\frac{\partial f}{\partial x}=3x^2-3y^2+\left(6xy\right)\text{i}=3{\left(x+\text{i}y\right)}^2=3z^2$ 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 $f\left(z\right)=\bar{z}=x-\text{i}y$ . Here

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

The Cauchy-Riemann equations are never satisfied so that $\bar{z}$ 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

$u=\frac{x}{x^2+y^2};v=\frac{y}{x^2+y^2}.$

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 $f\left(z\right)=z\bar{z}$ then $f^{\prime }\left(z\right)$ exists only at $z=0$ .

Solution

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

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

$\frac{{\partial }^{2}u}{\partial x\partial y}=\frac{\partial }{\partial x}\left(\right.\frac{\partial u}{\partial y}\left)\right.=\frac{\partial }{\partial x}\left(\right.-\frac{\partial v}{\partial x}\left)\right.=-\frac{{\partial }^{2}v}{\partial x^2}$

and

$\frac{{\partial }^{2}u}{\partial y\partial x}=\frac{\partial }{\partial y}\left(\right.\frac{\partial u}{\partial x}\left)\right.=\frac{\partial }{\partial y}\left(\right.\frac{\partial v}{\partial y}\left)\right.=\frac{{\partial }^{2}v}{\partial y^2}.$

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

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

In a similar way we find that

$\frac{{\partial }^{2}v}{\partial x^2}+\frac{{\partial }^{2}v}{\partial y^2}=0$ (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:

$\frac{\partial v}{\partial y}=\frac{\partial u}{\partial x}=y\text{and}\frac{\partial v}{\partial x}=-\frac{\partial u}{\partial y}=-x.$

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

$v=\frac{1}{2}\left(y^2-x^2\right)+C\text{where}C\text{is a constant}$

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.

Task!

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

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

Answer

Answer

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.

Answer

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 $z$ .

Answer