Imaginary and complex numbers. Perhaps they are incorrectly named, and Gauss even called for the ‘imaginary’ numbers to be renamed lateral numbers. However, most young mathematicians are not concerned with their naming but instead might be asking questions such as:

- Do they even have a meaning?
- Do they relate to any other area of mathematics?

These are justified questions. One might start with their discovery. Though complex numbers are commonly taught to students through considering the roots of the function $f(x)=x^2+1$, their origin is in fact from investigating cubic curves in the 16th century.

We start just before the 16th century. Cubic equations (equations of the form $ax^3+bx^2+cx+d=0$) had been reduced to simpler cubics in the form $x^3+px+q=0$. These cubics without an $x^2$ term were known as depressed cubics and this made solving cubic equations much easier.

We now skip ahead to the 16th century. An Italian professor, Scipione del Ferro, at Bologna University had found a formula to solve the depressed cubics with $p$ positive and $q$ negative. On his death bed in 1526, he confided his proof and formula to his student Antonio Fiore. Fiore was at the university Bologna, where they regularly held mathematical competitions. Fiore now had the formula that mathematicians had been searching for and with the proof in hand, challenged a self-taught Niccolò Tartaglia. In an amazing feat, Tartaglia was able to derive the formula for the cubic before the competition and won against Fiore. The formula (without proof) was again passed on from a Tartaglia to Gerolamo Cardano. Cardano simply worked backwards and reconstructed the proof.

One problem was that part of the formula could have led to taking the root of a negative number and Cardano was the first to really consider it as a possibility. However, he never included this in any of his writing. On the other hand, Rafael Bombelli, in his writing ‘l’Algebra’ explicitly made note of this. It took time for complex numbers to be accepted but eventually, their usefulness outweighed the difficulty some mathematicians had in understanding them.

The cubic history of the complex numbers is fascinating. However, looking at quadratic functions might be easier. Let us reconsider the curve $f(x)=x^2+1$. According to an algebraic principle named the fundamental theorem of algebra, every polynomial with degree $n$ must have exactly $n$ roots. Therefore, since this curve has degree 2, it should have exactly two roots. An attempt to find the roots of $f(x)$ might look like this:

$$x^2+1=0 $$

$$x^2=-1 \, \, \, \therefore \, \, \, \text{“No solutions”}$$

This doesn’t make sense though. The fundamental theorem of algebra states that there should be 2 roots, yet this answer suggests there are none. To correct the mistake, the phrase “No solutions” should read “No real solutions”. Taking the square root yields the result that $x=\pm \sqrt{-1}$. This produces exactly two roots for the function. $\sqrt{-1}$ was special and was given its own symbol, $\mathrm{i}$, and become the imaginary unit.

## Creating beauty

Mathematicians’ belief in the existence of $\mathrm{i}$ as a plausible mathematical concept was vital for what is considered by some to be the most beautiful equations in all of mathematics. The first was discovered by Abraham De Moivre. We will slowly work towards it by first noticing that:

$$(\cos \theta+\mathrm{i}\sin \theta)^2 =\cos^2\theta – \sin^2\theta +(2\cos \theta \sin \theta)\mathrm{i}=\cos2\theta+\mathrm{i}\sin2\theta \, \, \text{(1)}.$$

Maybe we can start to see a pattern and we might even conjecture that:

$$(\cos\theta+\mathrm{i}\sin\theta)^n= \cos(n\theta)+\mathrm{i}\sin(n\theta).$$

Let $P(n)$ be the mathematical statement that equation (1) is true for all real values of $\theta$. We will prove this by induction on $n$ starting from the `base case’, $n=1$. We clearly have:

$$(\cos\theta+\mathrm{i}\sin\theta)^1=\cos1\theta+\mathrm{i}\sin1\theta$$

so $P(1)$ is true. Now, for the inductive step, we assume the `Inductive Hypothesis’ that $P(k)$ is true for some positive integer $k$. That means:

$$(\cos\theta+\mathrm{i}\sin\theta)^k=\cos k\theta+\mathrm{i}\sin k\theta.$$

We will use this to show that $P(k + 1)$ is correct.:

\begin{align*}

(\cos\theta+\mathrm{i}\sin\theta)^{k+1}&=(\cos\theta+\mathrm{i}\sin\theta)^k (\cos\theta+\mathrm{i}\sin\theta)\\

&=(\cos k\theta+\mathrm{i}\sin k\theta)(\cos\theta+\mathrm{i}\sin\theta)\\

&=\cos k\theta\cos\theta+\mathrm{i}\cos k\theta\sin\theta-\sin k\theta\sin\theta+\mathrm{i}\sin k\theta\cos\theta\\

&=\cos k\theta\cos\theta-\sin k\theta\sin\theta+\mathrm{i}(\sin k\theta\cos\theta+\cos k\theta\sin\theta)\\

&=\cos (k\theta+\theta)+\mathrm{i}\sin(k\theta+\theta)\\

&=\cos((k+1)\theta)+\mathrm{i}\sin((k+1)\theta)

\end{align*}

We conclude that, if $P(k)$ is true, then $P(k+1)$ is true. Therefore, since $P(1)$ is true $P(n)$ is true for all positive integers $n$ by the principle of mathematical induction.

The second formula I want to mention follows from the first and is perhaps even more aesthetically pleasing. Although discovered by Roger Coates, the equation is named Euler’s identity due to how much Euler was able to manipulate and further this identity. It is given as;

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

This proof we are going to give is one of three proofs that Euler gave to try and prove this beauty. To get to this mathematical gem, we start by considering the expansion of $e^x$:

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$

Next, consider the case where the index is an imaginary number, say $\mathrm{i}x$:

\begin{align*}e^{\mathrm{i}x}&=1+\frac{\mathrm{i}x}{1!}+\frac{(\mathrm{i}x)^2}{2!}+\frac{(\mathrm{i}x)^3}{3!}+\frac{(\mathrm{i}x)^4}{4!}+\frac{(\mathrm{i}x)^5}{5!}+\frac{(\mathrm{i}x)^6}{6!}+\frac{(\mathrm{i}x)^7}{7!}+\frac{(\mathrm{i}x)^8}{8!}+\cdots \\

&=1+\mathrm{i}\frac{x}{1!}-\frac{x^2}{2!}-\mathrm{i}\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{i}\frac{x^5}{5!}-\frac{x^6}{6!}-\mathrm{i}\frac{x^7}{7!}+\frac{x^8}{8!}+\cdots

\end{align*}

This can be separated into real parts and imaginary parts:

$$e^{\mathrm{i}x}=(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}+ \cdots ) + \mathrm{i}(\frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!} +\cdots)$$

If you are familiar with the Maclaurin series expansion, you will quickly realise that the infinite sums in parenthesis are the expansions for $\cos x$ and $\sin x$ respectively. Ergo:

$$e^{\mathrm{i}x}=\cos x+\mathrm{i}\sin x$$

However, to show that $x$ here is an angle measured in radians, the formula is most commonly seen as:

$e^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta$,

and therefore $e^{i\pi} = -1$. Wow! The elegance of Euler’s identity is that it yields more wonderful results. For example, we can continue exploring by taking $\theta$ to be the sum of two angles. Let these angles be A and B, where A and B are measured in radians. This produces the following identities which we used in the proof of De Moivre’s formula:

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

but also:

$$(\cos A+\mathrm{i}\sin A)(\cos B+\mathrm{i}\sin B) = \cos A\cos B-\sin A\sin B+\mathrm{i}(\cos A\sin B+\sin A\cos B).$$

Equating real and imaginary parts gives:

$$\cos A\cos B-\sin A\sin B= \cos(A+B)$$

$$\cos A\sin B+\sin A\cos B= \sin(A+B)$$

In the same way we could replace $\theta$ with $A-B$ to derive expressions for $\cos(A-B)$ and for $\sin(A-B)$. In addition, all the other trigonometric identities used in A-level mathematics can be derived just from these identities.

Many students might be aware of the geometric proof of these identities and might still be struggling to memorise these sometimes-confusing identities. However, I hope that after seeing this you might gain a deeper understanding of your learning, truly enjoying the hidden beauty of mathematics instead of rote learning mathematics. Perhaps the elegance of using Euler’s formula is that it highlights something quite interesting: there is a clear link between algebra and geometry, namely complex numbers. Furthermore, it provides evidence that mathematics is a collection of unified and collected ideas.

## Why do they matter?

The discovery of complex numbers was extremely important in algebra. To fully appreciate this, we must consider our number system, something that many students are never explicitly taught. Initially, humans made use of natural numbers $\{1,2,3…\}$ to count naturally occurring objects- for example, 20 cows or 3 apples. Integers were then discovered to understand concepts such as debt. Integers extended the number of system and the natural numbers were a subset of the integers as shown by the image below. The next extension of the numbers system was the discovery of the rational numbers, which are defined as the ratio of two integers. The number system was yet again extended through the introduction of the real numbers, a set which contained the rational and irrational numbers. Mathematicians thought that the number system was complete, and that algebra was now understood. However, the discovery of the complex numbers exposed the incompleteness of the number system. In fact, the complex numbers were the final extension of the number system and are what is known as the algebraic closure of the number system. They completed traditional algebra.

## More fun!

Let’s reconsider the function curve $f(x)=x^2+1$ we were working with at the beginning. When we attempted to find the roots of this function, we arrived at the equation $x^2=-1$. Now consider the case $z^n=\pm1$, where $n$ is a real number. The first equation, $z^n=1$, is a classic equation in complex analysis. Its solutions are named the roots of unity as they are the roots of one (unity). They are very interesting, but I was more intrigued with a variation of the roots of unity. Replacing 1 with $i$ to get:

$$z^n=\mathrm{i}.$$

I came across this idea whilst considering if I could take the square root of $\sqrt{-1}$. Since:

$$e^{\mathrm{i}\theta}=\cos\theta+\mathrm{i}\sin\theta,$$

we could find a value of $\theta$ such that $\cos\theta=0$ and $\sin\theta=1$ to obtain an expression for $\mathrm{i}$. For example, $\theta=\pi/2$. Note: there are infinitely many values of $\theta$ that satisfy these requirements since $\sin\theta$ and $\cos\theta$ are periodic functions. However, working with values in the range $-\pi < \theta \leq \pi$ is easier. So we have $e^{\mathrm{i} \pi/2}=\cos \pi/2+\mathrm{i}\sin \pi/2$, or:

$$e^{\mathrm{i} \pi/2}=\mathrm{i}.$$

As a result, taking roots becomes simpler. For example:

$$\sqrt{\mathrm{i}}=e^{\mathrm{i}\pi/4}$$

$$e^{\mathrm{i} \pi/4}=\cos \frac{\pi}{4}+\mathrm{i}\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}+\mathrm{i}\frac{\sqrt{2}}{2}.$$

Think about the shape that all the roots of $\mathrm{i}$ would make on the complex plane. Try and plot these points on some graphing software such as Desmos. Could you explain why they form this shape by considering Euler’s identity and your knowledge of trigonometry.

We can continue exploring. What about taking the $\mathrm{i}$-th root of $\mathrm{i}$? Recall that $e^{\mathrm{i}\pi/2}=\mathrm{i}$, so maybe:

$$\sqrt[\mathrm{i}]{\mathrm{i}}= (e^{\mathrm{i} \pi/2})^{1/\mathrm{i}}$$

and

$$\sqrt[\mathrm{i}]{\mathrm{i}}=e^{\pi/2}.$$

Since $e$ and $\pi$ are real numbers, we can deduce that $\sqrt[\mathrm{i}]{\mathrm{i}}$ is a real number. This is a stunning result but there is a slight problem. In the same way that the principal value of $\sin^{-1}(1/2)$ is $\pi/6$ rads, $e^{\pi/2}$ can be considered to be the principal value of $\sqrt[\mathrm{i}]{\mathrm{i}}$. There are actually infinitely many values we could assign to the expression $\sqrt[i]{i}$.

In conclusion, complex numbers are something of beauty. They are just as real as real numbers, just as tangible and just as necessary. As I said before, complex numbers show that mathematics is not a collection of separated ideas but instead consists of linked and connected fields. In this case, we have seen the intrinsic link between complex analysis, geometry and algebra.

## A challenge

But the fun doesn’t end there. As you might have seen, investigating complex numbers really is fascinating. Complex numbers can even be applied to situations that don’t seem to require them. We leave it is a challenge to the reader to use complex numbers to evaluate the following integral:

$$\int_{0}^{1}\frac{\sin(\log x)}{\log x}dx.$$

A solution will be uploaded to this blog in the near future.

*A solution has now been uploaded! See this article.*