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# Complex Numbers

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## Introduction

Several researchers in the mid 1900's developed analytical solutions to linear elastic problems involving cracks by making use of Airy stress functions and complex numbers. So they are reviewed here. Special focus will be placed on the Cauchy-Riemann Equations [1], which are a set of equality relationships among various derivatives of functions involving complex numbers.

## Complex Numbers

The letter $$i$$ is used to represent the square root of -1 ($$i = \sqrt{-1}$$). And therefore

$\begin{eqnarray} i^2 & = & -1 \\ \\ i^3 & = & -1 \; * \; i & = & -i \\ \\ i^4 & = & -1 * -1 & = & 1 \end{eqnarray}$
A general complex number is written as $$x + i y$$ where $$x$$ and $$y$$ are themselves, two real numbers. For example, $$3 + i 4$$. The $$3$$ is called the real part and $$4$$ is the imaginary part because it is the coefficient of $$i$$ (even tho $$4$$ itself is a real number).

The letter $$z$$ is often used as a shortcut symbol for $$x + i y$$. So $$z = x + i y$$. And when one speaks of the real part of $$z$$, one is talking about $$x$$. Likewise, the imaginary part of $$z$$ is $$y$$. This is abbreviated as

$\text{Re}\,z = x \qquad \qquad \text{Im}\,z = y$
Recall that $$y$$ is a real valued quantity (like 4), so "$$\text{Im}\,z$$" is a real valued quantity too. It's only called the imaginary part because it is the coefficient of $$i$$.

Functions of complex numbers exist just like functions of real numbers. For example, $$z^2$$ is a function of the complex number $$z$$. $$Z(z)$$ is often used to represent a complex function (personally, I think it's confusing to have so many z's). So this would be written as

$Z(z) = z^2$
The real and imaginary parts are found by expanding the function out as follows.

$Z(z) \; = \; z^2 \; = \; (x + i y)(x + i y) \; = \; x^2 - y^2 + i 2 x y$
So

$\text{Re}\,Z = x^2 - y^2 \qquad \text{and} \qquad \text{Im}\,Z = 2 x y$
Derivatives of complex functions work just like those of real functions. For example

${d \, (z^2) \over dz} \; = \; 2z \; = \; 2x + i2y$
While it is clear that the derivative of $$z^2$$ is $$2z$$, it is not at all obvious when the functions are expressed in terms of $$x$$ and $$y$$. For example, compare the function $$Z(z) = x^2 - y^2 + i 2 x y$$ to its derivative $${dZ \over dz} = 2x + i2y$$. This is not at all intuitive, at least to me, even though it is correct. Finally, note that $$dZ \over dz$$ is equal to $$\partial Z \over \partial x$$. This is more than just coincidence, and it leads to the topic of Cauchy-Riemann equations, which are several important relationships between the derivatives of complex functions. The topic is central to Westergaard's solution for cracks.

## Cauchy-Riemann Equations

Let's go ahead and lay out what the Cauchy-Riemann relationships are first, using $$Z = z^2$$ as an example. Then we will come back and show why they are so.

The equations are relationships between the many different types of derivatives of complex functions. They are

$\begin{eqnarray} \text{Re} \, { dZ \over dz } & \; = \; & {\partial \, \text{Re}Z \over \partial x} & = & \;\;\;\;{\partial \, \text{Im}Z \over \partial y} \\ \\ \text{Im} \, { dZ \over dz } & \; = \; & {\partial \, \text{Im}Z \over \partial x} & = & - {\partial \, \text{Re}Z \over \partial y} \end{eqnarray}$
The equations lead to some interesting, if not intuitive, relationships among the derivatives. For example, just as any complex function, $$Z(z)$$, can be separated into real and imaginary parts like $$Z(z) = \text{Re}Z + i\,\text{Im}Z$$, so can its derivative. This looks like

${ dZ \over dz } = \text{Re} \, { dZ \over dz } + i \, \text{Im} \, { dZ \over dz }$
But thanks to the Cauchy-Riemann equations, the derivative can also be expressed as

$\begin{eqnarray} { dZ \over dz } & \; = \; & \text{Re} \, { dZ \over dz } \; + \; i \, \text{Im} \, { dZ \over dz } \\ \\ & \; = \; & {\partial \, \text{Re}Z \over \partial x} \; + \; i \, {\partial \, \text{Im}Z \over \partial x} \\ \\ & \; = \; & {\partial \, \text{Im}Z \over \partial y} \; - \; i \, {\partial \, \text{Re}Z \over \partial y} \\ \end{eqnarray}$

### Analytic Functions

Note that all this works only when $$Z = Z(z)$$. In such cases $$Z$$ is called analytic. The $$x$$'s and $$y$$'s are all contained within the $$z$$ variable and do not appear explicitly in the function. For example, $$Z(z) = z^2$$ is an analytic function, as well as $$Z(z) = \sin(z)$$ and even $$Z(z) = e^z \sin(z) / \sqrt{z}$$.

On the other hand, although $$Z = xz^2$$ is a perfectly good function, it does not fall within the category of analytic, and the Cauchy-Riemann equations will not apply because of the explicit presence of $$x$$ in the function.

### Cauchy-Riemann Example

Recall the $$z^2$$ function

$Z(z) \; = \; z^2 \; = \; (x + i y)(x + i y) \; = \; x^2 - y^2 + i 2 x y$
and its derivative.

${d \, (z^2) \over dz} \; = \; 2z \; = \; 2x + i2y$
The Cauchy-Riemann equations give the following results

$\begin{eqnarray} \text{Re} \, { dZ \over dz } & \; = \; & {\partial \, \text{Re}Z \over \partial x} & = & \;\;\;\;{\partial \, \text{Im}Z \over \partial y} & = & 2x \\ \\ \text{Im} \, { dZ \over dz } & \; = \; & {\partial \, \text{Im}Z \over \partial x} & = & - {\partial \, \text{Re}Z \over \partial y} & = & 2y \end{eqnarray}$

## Derivation of Cauchy-Riemann Equations

It is relatively easy to show that the Cauchy-Riemann equations must be true, and the proof involves differentials. Let's start with the simplist case of real functions of single variables, i.e., $$y = y(x)$$. For this function, a differential change is given by

$dy = {dy \over dx} dx$
So if the slope, $${dy \over dx}$$, is 5, and $$x$$ is changed by 2, ($$dx = 2$$), then it is easy to see that $$y$$ would change by 10 ($$dy = 5*2 = 10$$).

A higher dimensional real-number example for the case of $$z = z(x,y)$$ works as follows.

$dz = {\partial z \over \partial x} dx + {\partial z \over \partial y} dy$
Now going to complex numbers, $$Z = Z(z)$$, the differential change would be

$dZ = {dZ \over dz} dz$
But of course, $$z = x + i y$$, so the chain rule can be used to expand the differential as

$dZ = {dZ \over dz} {\partial z \over \partial x} dx + {dZ \over dz} {\partial z \over \partial y} dy$
And since $$z = x + i y$$, then $${\partial z \over \partial x} = 1$$ and $${\partial z \over \partial y} = i$$. So the above equation can be written as

$dZ = {dZ \over dz} dx + i {dZ \over dz} dy$
The next step is to partition $$dZ \over dz$$ into its real and imaginary parts

${dZ \over dz} = \text{Re} \, {dZ \over dz} + i \, \text{Im} \, {dZ \over dz}$
and substitute this into the above differential equation

$\begin{eqnarray} dZ & = & {dZ \over dz} dx + i {dZ \over dz} dy \\ \\ & = & \left( \text{Re} \, {dZ \over dz} + i \, \text{Im} \, {dZ \over dz} \right) dx + i \left( \text{Re} \, {dZ \over dz} + i \, \text{Im} \, {dZ \over dz} \right) dy \\ \\ & = & \text{Re} \, {dZ \over dz} dx + i \, \text{Im} \, {dZ \over dz} dx + i \, \text{Re} \, {dZ \over dz} dy - \text{Im} \, {dZ \over dz} dy \end{eqnarray}$
So there are four terms total: real and imaginary terms for $$dx$$, and real and imaginary terms for $$dy$$. Pay close attention to which terms are real versus imaginary. For example, $$\text{Im}{dZ\over dz} dy$$ is now a real term because no $$i$$ is present. Conversely, $$\text{Re}{dZ\over dz} dy$$ is now an imaginary term because it is the coefficient of $$i$$.

Now, we restart the process and will arrive at a different combination of four terms that must be equal to the four we already have. Start back over at $$Z(z) = \text{Re}\,Z + i\,\text{Im}\,Z$$ and recall that both $$\text{Re}\,Z$$ and $$\text{Im}\,Z$$ are in fact real functions of $$x$$ and $$y$$. Therefore, differentials of both the real and imaginary parts can be expressed directly in terms of $$dx$$ and $$dy$$ as

$\begin{eqnarray} dZ & = & {\partial \over \partial x} \left( \text{Re}\,Z + i\,\text{Im}\,Z \right) dx + {\partial \over \partial y} \left( \text{Re}\,Z + i\,\text{Im}\,Z \right) dy \\ \\ & = & {\partial \, \text{Re}\,Z \over \partial x} dx + i \, {\partial \, \text{Im}\,Z \over \partial x} dx + {\partial \, \text{Re}\,Z \over \partial y} dy + i \, {\partial \, \text{Im}\,Z \over \partial y} dy \end{eqnarray}$
This is the second expression we needed with four terms for $$dZ$$ in terms of $$dx$$ and $$dy$$. And it must be that these four terms are equal to the first four. For example, the real term involving $$dx$$ requires the following equality

$\text{Re}\, {dZ \over dz} = {\partial \, \text{Re}\,Z \over \partial x}$
while the real term involving $$dy$$ requires

$\text{Im}\, {dZ \over dz} = - {\partial \, \text{Re}\,Z \over \partial y}$
and so on. Indeed, the entire set of equality relationships is

$\text{Re}\, {dZ \over dz} = {\partial \, \text{Re}\,Z \over \partial x} \qquad \qquad \text{Re}\, {dZ \over dz} = {\partial \, \text{Im}\,Z \over \partial y}$  $\text{Im}\, {dZ \over dz} = {\partial \, \text{Im}\,Z \over \partial x} \qquad \qquad \text{Im}\, {dZ \over dz} = - {\partial \, \text{Re}\,Z \over \partial y}$
which can be written more compactly as

$\begin{eqnarray} \text{Re} \, { dZ \over dz } & \; = \; & {\partial \, \text{Re}Z \over \partial x} & = & \;\;\;\;{\partial \, \text{Im}Z \over \partial y} \\ \\ \text{Im} \, { dZ \over dz } & \; = \; & {\partial \, \text{Im}Z \over \partial x} & = & - {\partial \, \text{Re}Z \over \partial y} \end{eqnarray}$

### Real and Imaginary Parts

It has been convenient to use the function $$Z(z) = z^2$$ because it is so easily separated into its real and imaginary parts.

$Z(z) \; = \; z^2 \; = \; (x + i y)(x + i y) \; = \; x^2 - y^2 + i 2 x y$
But it is not always this easy, although it is always possible. For example, the function $$Z(z) = \sin(z)$$ is not easily separated. Doing so requires several identities as follows.

$\begin{eqnarray} Z(z) & = & \sin(z) \\ \\ & = & \sin(x + i y) \\ \\ & = & \sin(x) \cos(iy) + \cos(x) \sin(iy) \\ \\ & = & \sin(x) \cosh(y) + i \cos(x) \sinh(y) \end{eqnarray}$
When helpful mathematical identities are not available to aid the partitioning of a complex function into its real and imaginary parts, then the method of last resort is a Taylor series expansion of the function.

$\begin{eqnarray} Z(z) & = & \sin(z) \\ \\ & = & z - {z^3 \over 6} + {z^5 \over 120} \, + \, ... \\ \\ & = & (x+iy) - {(x+iy)^3 \over 6} + {(x+iy)^5 \over 120} \, + \, ... \\ \\ & = & (x - {1\over 6}x^3 + {1\over 2}xy^2 \, + \, ...) + i \, (y - {1\over 2}x^2 y + {1\over 6}y^3 \, + \, ...) \end{eqnarray}$