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Here is an example taken from an OER Physics course created at the University of Manchester.

Complex Numbers

1.1 Basic Properties

1.1.1 Introduction

Up to now you have seen 4 major sets of numbers:

• The set $\mathbb{N}$ of natural, $0,1,2,...$(non-negative integers). The sum and product of two natural numbers are natural numbers.
• The set $\mathbb{Z}$ of all integers, $0,\pm 1,\pm 2...$. This closes under addition, subtraction and multiplication.
• The set $\mathbb{Q}$ of all rational numbers (fractions) $p∕q$ like $3∕5$, closes under division as well.
• To this we add the irrational numbers, like $\sqrt{2},3^{1∕3}$
• And obtain $ℝ$ the set of real numbers includes $1,2.34,\pi ,4∕5,e=e^1$ etc

There is a lot of subtle mathematics associated with them. Are there more rational numbers than integers? More reals than rationals? We can also try to solve equations. In physics we usually mean “find a real number that solves the equation”.

Example 1.1:

Find the zero of the function (polynomial) $p\left(x\right)=x^2-1$.

Solution:

$$\begin{array}{rcll}p\left(x\right)=x^2-1=0⇝x=\pm \sqrt{1}=\pm 1.& & & \text{(1.1)}\end{array}$$

There are equations we can’t solve in this way, e.g. the roots or zeroes of the polynomial $q\left(x\right)=x^2+1$, are not real, i.e.

$$\begin{array}{rcll}q\left(x\right)=x^2+1=0⇝x=\pm \sqrt{-1}=????.& & & \text{(1.2)}\end{array}$$

The square root of $-1$ is not defined within the real numbers. There is no real zero of $q\left(x\right)$(look at the curve $q\left(x\right)=x^2+1$). We define new numbers (complex numbers) so that we can solve any equation of the kind of ( 1.2 ).

The symbol $i$ is called ‘complex unit’ with the property

$$\begin{array}{rcll}{i}^2=-1.& & & \text{(1.3)}\end{array}$$

Question: What is ${\left(-i\right)}^2$?
Using $i$, we can solve

$$\begin{array}{rcll}q\left(x\right)=x^2+1=0⇝x=\pm i.& & & \text{(1.4)}\end{array}$$

In $z=x+iy$, $x=\text{Re}\left(z\right)$ is called real part and $y=\text{Im}\left(z\right)$ is called imaginary part of the complex number $z$. Complex numbers are elements of a set called $ℂ$.

So far this may seem a bit artificial. However, we now can solve arbitrary quadratic equations, i.e., find solutions in $ℂ$:

Example 1.2:

Solve $z^2-2z+5=0$.

Solution:

$$\begin{array}{rcll}{z}^2-2z+5& =& 0& \\ & ⇝& z_{1∕2}=1\pm \sqrt{\frac{{\left(-2\right)}^2}{4}-5}=1\pm \sqrt{-4}=1\pm \sqrt{4}\sqrt{-1}=1\pm 2i.& \end{array}$$

In fact, within the complex numbers one can always find the root of a quadratic equation (and in fact all the roots of an arbitrary polynomial, i.e. solve equations like $z^{12}+4z^4+17=0$ etc., although for this last example there is no general formula as in the quadratic case.)

Actually, complex numbers first arose in the 15th century in the solution of cubic equations of the form $z^3+bz+c=0$. The general solution of such equations are

$$\begin{array}{rcll}z& =& -\left(\frac{2^{\frac{1}{3}}b}{{\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}\right)+\frac{{\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}{32^{\frac{1}{3}}},& \\ z& =& \frac{\left(1+i\sqrt{3}\right)b}{2^{\frac{2}{3}}{\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}-\frac{\left(1-i\sqrt{3}\right){\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}{62^{\frac{1}{3}}},& \\ z& =& \frac{\left(1-i\sqrt{3}\right)b}{2^{\frac{2}{3}}{\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}-\frac{\left(1+i\sqrt{3}\right){\left(-27c+\sqrt{108b^3+729c^2}\right)}^{\frac{1}{3}}}{62^{\frac{1}{3}}},& \text{(1.5)}\end{array}$$

and therefore inolve the complex unit $i$, even when the resulting solutions are real!

1.1.2 Basic Definitions

If we look at a pair of two complex numbers

$$\begin{array}{rcll}{z}_1=x_1+i{y}_1,z_2=x_2+i{y}_2,& & & \text{(1.6)}\end{array}$$

we can define all the standard algebraic manipulations.

Equality:

$$\begin{array}{rcll}{z}_1=z_2\iff x_1=x_2\text{and}{y}_1=y_2.& & & \text{(1.7)}\end{array}$$

Addition:

$$\begin{array}{rcll}{z}_1\pm z_2=\left(x_1\pm x_2\right)+i\left(y_1\pm y_2\right).& & & \text{(1.8)}\end{array}$$

Multiplication:

$$\begin{array}{rcll}{z}_1{z}_2& =& \left(x_1+i{y}_1\right)\left(x_2+i{y}_2\right)=x_1{x}_2+i\left(x_1{y}_2+x_2{y}_1\right)+i^2{y}_1{y}_2& \\ & =& \left(x_1{x}_2-y_1{y}_2\right)+i\left(x_1{y}_2+x_2{y}_1\right).& \text{(1.9)}\end{array}$$

Zero:

$$\begin{array}{rcll}z=x+iy=0\iff x=0\text{and}y=0.& & & \text{(1.10)}\end{array}$$

Other:

$$\begin{array}{rcll}z& =& x+iy⇝iz=i\left(x+iy\right)=ix-y=-y+ix& \\ \left(-i\right)i& =& -\left(i^2\right)=-\left(-1\right)=1& \\ i^3& =& i\left(i^2\right)=-i& \\ i^4& =& \left(i^2\right)\left(i^2\right)=1& \\ \frac{1}{i}& =& \frac{i}{i^2}=\frac{i}{-1}=-i.& \text{(1.11)}\end{array}$$

(Sometimes one writes $z^{\ast }$ instead of $\bar{z}$.) We have $$\begin{array}{rcll}\text{Re}\left(z\right)=\text{Re}\left(\bar{z}\right),\text{Im}\left(z\right)=-\text{Im}\left(\bar{z}\right).& & & \text{(1.12)}\end{array}$$

From this definition, we find

$$\begin{array}{rcll}\text{Re}\left(z\right)=x=\frac{z+\bar{z}}{2},\text{Im}\left(z\right)=y=\frac{z-\bar{z}}{2i}.& & & \text{(1.13)}\end{array}$$

(see next workshop for some exercises).

Division: we use a trick by calculating $z\bar{z}$:

$$\begin{array}{rcll}z\bar{z}=\left(x+iy\right)\left(x-iy\right)=x^2+y^2.& & & \text{(1.14)}\end{array}$$

(Check this !). For $z\ne 0$,

$$\begin{array}{rcll}\frac{1}{z}=\frac{\bar{z}}{z\bar{z}}=\frac{x-iy}{x^2+y^2}=\frac{x}{x^2+y^2}-i\frac{y}{x^2+y^2}.& & & \text{(1.15)}\end{array}$$

Other more complicated examples were dealt with in class and the workshop.