The idea of complex numbers starts with a mathematical fiction that seems quite bizarre, but it leads to something remarkably powerful and useful, so it is worth getting grips with. You will already know that you can take the square root of any positive number, but whenever you square an ordinary number you get a positive result, so there is no way to calculate the square root of a negative number like $-1$. Well, suppose you *could* do it — let us see what happens if we *define* a symbol $i$ to be the square root of $-1$, so that

$$i^2=-1 . \tag{1}$$

(Some people call this $j$ rather than $i$, but I will always use $i$.)

Don’t worry about what this thing $i$ might *mean*: just bear with me for a bit as we play some mathematical games to see if it can do anything useful. The next step is to introduce combinations of ordinary numbers and $i$, called *complex numbers*: for example $5+3i$ or $2-2i$. The jargon is that $i$ is an “imaginary number”, as is $3i$ or $-2i$. These are the “imaginary parts” of the complex numbers $5+3i$ and $2-2i$, while the $5$ and the $2$ are the corresponding “real parts”.

As a first step to see if anything makes sense, let’s try multiplying these two complex numbers together. We can treat $i$ just as if it was a mathematical variable like $x$, and expand out the brackets as usual:

$$(5+3i)(2-2i) = 10 +6i -10i -6 i^2=10 -4i +6 = 16-4i \tag{2}$$

where we have used equation (1) to get rid of $i^2$. So the result of multiplying two complex numbers is another complex number — very reassuring.

Next, we should recall that any square root could actually have a $\pm$ sign in front of it: for example $\sqrt{4}=\pm2$. So $-i$ must also be a square root of $-1$. This suggests that the two complex numbers $x+iy$ and $x-iy$ might be closely related: we have just reversed the sign of $i$, to make the other choice for the square root of $-1$. These two complex numbers are called complex conjugates of each other. Now see what happens when we multiply the two together:

$$(x+iy)(x-iy)=x^2 +ixy -ixy -i^2 y^2 = x^2 + y^2. \tag{3}$$

This result has a simple geometrical interpretation, in a useful graphical representation of complex numbers called the “Argand diagram”. If we take the complex number $x+iy$, we can treat $x$ and $y$ as the Cartesian coordinates of a point, and plot it as in Fig. 1. The horizontal axis shows $x$, the real part of the complex number, so it is called the “real axis”. The vertical axis shows the imaginary part $y$, so it is called the “imaginary axis”. The complex conjugate of our complex number is also shown in the figure: it is the mirror-reflection of $z$ in the real axis.

But we don’t always use Cartesian coordinates — it is sometimes useful to use polar coordinates $r$ and $\theta$ such that

$$x=r \cos \theta, \mathrm{~~~} y=r \sin \theta. \tag{4}$$

This polar representation of our complex number is indeed useful. The radius $r$ is called the “magnitude” of the complex number, and the angle $\theta$ is called its “argument” or its “phase”. Now notice that we can reinterpret equation (3) as saying that a complex number multiplied by its complex conjugate gives the square of the magnitude. In symbols, if we denote the complex number by $z=x+iy$ and the complex conjugate by $z^*=x-iy$, then

$$z z^* =r^2 =|z|^2 \tag{5}$$

where $|z|$ is another common notation for the magnitude $r$ of the complex number.

We can use this result to complete our toolkit for elementary manipulations of complex numbers. We have already seen how to multiply two complex numbers — but what about dividing? So suppose we have complex numbers $a$ and $b$, and we want $a/b$. All we need to do it multiply on the top and bottom of the fraction by the complex conjugate of $b$:

$$\dfrac{a}{b}=\dfrac{a b^*}{b b^*}=\dfrac{a b^*}{|b|^2}. \tag{6}$$

The denominator is now an ordinary “real number” $|b|^2$, so we can see that the ratio $a/b$ is indeed just another complex number, which we could work out.

There is one more really important result involving complex numbers that we will need, and it relates to the combination $\cos \theta + i \sin \theta$ that appears in the polar expression

$$z=x+iy=r(\cos \theta +i \sin \theta) . \tag{7}$$

To get to this result we need a short digression on “Taylor series” or “power series”. It is often useful to express a function like cosine in terms of a sum of powers: is it possible to write

$$\cos \theta =\sum_{n=0}^\infty{a_n \theta^n} \tag{8}$$

where the coefficients $a_0$, $a_1$, $a_2,…$ are constants? Well, if we put $\theta =0$ in this equation, the left-hand side is $\cos(0)=1$, while the right-hand side is simply $a_0$ because all the other terms disappear. So we must have $a_0=1$. Now what happens if we differentiate equation (8)? We get

$$-\sin \theta = \sum_{n=1}^\infty{n a_n \theta^{n-1}} \tag{9}$$

and now if we set $\theta=0$ we find

$$a_1 = -\sin(0)=0. \tag{10}$$

Differentiating again and setting $\theta=0$ we find

$$a_2 = -\dfrac{\cos(0)}{2} = -\dfrac{1}{2}, \tag{11}$$

and we can continue this process to find the power series

$$\cos \theta = 1-\dfrac{\theta^2}{2}+\dfrac{\theta^4}{2 \times 3 \times 4}-\dfrac{\theta^6}{2 \times 3 \times 4 \times 5 \times 6} +…\tag{12}$$

If we go through the same process starting with the function $\sin \theta$, we find that the corresponding power series is

$$\sin \theta = \theta -\dfrac{\theta^3}{2 \times 3} + \dfrac{\theta^5}{2 \times 3 \times 4 \times 5}+… \tag{13}$$

Finally, we can repeat the process with the exponential function $e^x$ and find

$$e^x = 1+x+\dfrac{x^2}{2}+\dfrac{x^3}{2 \times 3}+\dfrac{x^4}{2 \times 3 \times 4}+… \tag{14}$$

Now comes the magic. If we substitute $x=i\theta$ into equation (14), we find

$$e^{i \theta} = 1+i \theta-\dfrac{\theta^2}{2}-i\dfrac{\theta^3}{2 \times 3}+\dfrac{\theta^4}{2 \times 3 \times 4}+… \tag{15}$$

Alternate terms are real and imaginary, and if we collect those real and imaginary terms together into two separate series, we recognise both of them from equations (12) and (13): we have the very striking result

$$e^{i \theta} = \cos \theta + i \sin \theta . \tag{16}$$

Notice that this allows us to rewrite equation (7) in the compact form

$$z=x+iy=r e^{i \theta}. \tag{17}$$

We will make extensive use of the result (16) in the development of the theory of vibration ( sections 2.2.2 and 2.2.7 are the first important examples), but to close this section I will just give one example of the power of the result to derive some useful results in trigonometry. Suppose we have two angles $A$ and $B$, and we multiply together the two results corresponding to equation (16):

$$e^{iA} e^{iB} = (\cos A +i \sin A)(\cos B +i \sin B). \tag{18}$$

Multiplying out the brackets and using familiar properties of the exponential function, this becomes

$$e^{i(A+B)} = \cos A \cos B -\sin A \sin B +i(\sin A \cos B+\cos A \sin B) \tag{19}$$

But

$$e^{i(A+B)} = \cos (A+B) + i \sin (A+B) \tag{20}$$

so equating the real parts and the imaginary parts of equation (19) we can deduce immediately that

$$\cos (A+B)= \cos A \cos B -\sin A \sin B \tag{21}$$

and

$$\sin (A+B) = \sin A \cos B+\cos A \sin B . \tag{22}$$

These results can, of course, be proved by trigonometry without using complex numbers, but it takes far, far longer than this simple calculation!