What is really the purpose of $i$?

We started talking about imaginary numbers again this year and asked this question in class, but nobody could really give a straight answer. So if anyone could tell me the real reason we have imaginary numbers that would be great! :)

Solution 1:

Historically, the imaginary and complex numbers were introduced for the resolution of the cubic equation: in the case of three real roots, you can't really do without an excursion in the complex numbers, even though the final results are real !

But a deeper reason is the Fundamental Theorem of Algebra, stating that every polynomial has at least one root: for it to be true, you must admit complex roots as well. Strange as it may sound, the simplicity of the theorem is so powerful that mathematicians prefer to accept complex numbers rather than dropping the fundamental theorem. This makes the set of complex numbers $\mathbb C$ a more natural domain than $\mathbb R$ for algebraic operations.

As you will later see, complex numbers allow numerous shortcuts in some computations and the theory of complex functions turns out to be rich and useful.

For the sake of illustration, the Euler's formula unifies the exponential and the trigonometric functions:

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

As an application, consider

$$(e^{ix})^3=(\cos x+i\sin x)^3=\cos^3x-3\cos x\sin^2x+i(3\cos^2x\sin x-\sin^3x)\\ =e^{i3x}=\cos 3x+i\sin 3x.$$

This establishes in a single go the two formulas

$$\cos 3x=\cos^3x-3\cos x\sin^2x,\\\sin3x=3\cos^2x\sin x-\sin^3x$$

which are more painful to get at by other means.

Solution 2:

The real numbers you are familiar with (positive, negative, and zero) are incomplete. They are part of a larger system, the complex numbers, and many matters that are mathematically strange or puzzling become clear when you view them with the full context. Many difficult problems become easy and many impossible problems become solvable.

Here is a short example.

For many reasons we spend a lot of time studying polynomial equations. For example, $x^2+x-6=0$ (this is an equation of the “2nd degree” because the highest power of $x$ is $x^2$) or $23x^5-12x^3+x+127 = 0$ (this is an equation of the “5th degree”). For equations of second degree and higher, there can be more than one solution. For example, $x^2+x-6=0$ is solved by both $x=2$ and by $x=-3$.

Sometimes, though, these equations appear to have no solution. For example, $x^2+4x+5=0$ has no solutions. One can ask how to tell when an equation has solutions, and how many it has. It's not hard, for example, to show that any equation of odd degree must have at least one solution. But it is hard to make much progress past that point.

If you look at the whole context, which is the complex numbers, the answer is rather simple. An equation of the $n$th degree always has a solution, and in fact it always has exactly $n$ solutions! For example, the equation $x^2+4x+5=0$ I mentioned earlier as having “no solutions” is solved by $-2+i$ and by $-2-i$.

The polynomials don't care that you think the complex numbers are strange. They are doing their own thing, and to them, the complex numbers are what is important and the real numbers are just a weird little province. If we want to deal with polynomials, we have to visit them in the place where they live. This is the complex numbers.

Here is a longer example.

There is a certain infinite series, $$1+x^2+x^4+x^6+\ldots$$ which (for reasons I don't expect you to appreciate right now) looks as though it should add up to $$\frac1{1-x^2}.$$ That is, you pick $x$, and then as you add up more and more terms of the series, the sum gets closer and closer to $\frac1{1-x^2}$. For instance, $$1 + \left(\frac13\right)^2 + \left(\frac13\right)^4 + \left(\frac13\right)^6 = 1.12483…$$ which is already quite close to $\frac1{1-\left(\frac13\right)^2} = \frac98 = 1.125$, and the following terms of the series make up the missing difference. You should try this yourself with $\frac12$.

So the series does add up the way we expect—if the absolute value of $x$ is less than $1$. But outside that range it does not work. (Try it for some $x$ bigger than $1$; it will be obvious that it is completely wrong.) Why not?

This is not just an idle question. This sort of infinite series is an important part of a method for solving differential equations, which are crucial to many problems of physics, architecture, and engineering. We need to understand when this method works and when it doesn't, so that our buildings don't fall down.

The reason that $\frac1{1-x^2}$ fails when $x$ is too large is subtle, but the basic cause is that $\frac1{1-x^2}$ is undefined at $-1$ and at $1$, because you would have to divide by zero. As you move past those undefined points something breaks.

Using the same methods, we can find an infinite series $$1-x^2+x^4-x^6+\ldots$$ that adds up to $$\frac1{1+x^2}.$$ This also works—but again, only if $x$ is between $-1$ and $1$. Why does it fail when $x$ is outside that range? The explanation of the previous paragraph doesn't make sense any more, because unlike $\frac1{1-x^2}$, the function $\frac1{1+x^2}$ is not undefined at $-1$ or at $1$. It is defined everywhere, so why doesn't the series work everywhere?

To understand what is really going on, you have to look at the complex numbers. If we think of the complex numbers as being arranged in a plane, it turns out that the series for $\frac1{1-x^2}$ works for any complex number inside a certain circle. The circle contains not just the real numbers whose absolute value is less than 1, but all the complex numbers whose absolute value is less than 1. The circle passes through $+1$ and $-1$, and the undefinedness of $\frac1{1-x^2}$ at those two points spoils the series everywhere outside this circle.

And now the failure for the series for $\frac1{1+x^2}$ makes sense: the series works anywhere inside that same circle. Why does it break on that circle when $\frac1{1+x^2}$ is perfectly well-defined at $1$ and at $-1$? Because the circle also goes through $i$ and $-i$, and at those points $\frac1{1+x^2}$ is not well-defined. It breaks in exactly the same way, for exactly the same reason!

The two functions that looked so different are actually almost exactly the same once you look at the entire picture. The complex numbers are the entire picture.

Solution 3:

The real reason? That sounds like you want a one-liner that you might not have heard before.

Take out the graph paper.

What happens if you keep applying $i$ to $1$? i.e, $\;i \times 1$, $\;i \times (i \times 1)$, $\;i \times (i \times (i \times 1))$, etc.

Apply $i$ to $1$ and you get $i$.
Apply $i$ to $i$ and you get $-1$.
Apply $i$ to $-1$ and you get $-i$.
Apply $i$ to $-i$ and you get $+1$.

So repeated application and you get to watch $1 \mapsto i \mapsto -1 \mapsto -i \mapsto 1 \dots$

Exercise: What happens if you keep applying $i$ to $1 + i$.

The real reason for $i$ is you get to rotate stuff by 90° using number multiplication.

There is nothing imaginary about rotations!

Solution 4:

The real purpose is be able to carry out mathematical operations in an unrestricted manner.

Why have negative numbers ? So you can subtract: $2-3=-1$

Why have fractional numbers ? So you can divide: $2\div 3=\frac{2}{2}$

Why have imaginary numbers ? So you can take square roots: $\sqrt{-1}=i$.

It turns out to be a minor miracle that with imaginary numbers ALL algebraic equations, eg, $x^7+4x^6+3x^5+76x^4+14x^3+12x^2+89x+100=0$ have (possibly imaginary) solutions.

The the range of applicability of the operations is greatly extended.

Solution 5:

@Yves Daoust gave great answer in the Algebric field and showed how powerful it can be so I won't add more in that regard.

Imaginary, or complex, numbers in physics.

In quantum physics there are systems of superposition, let's say you have 2 states, $A,B$ and let's say $a$ is the probability for $A$ to be the result $b$ is the probability of $B$ to be the result. To be more clear and follow how it is written in reality we will say $|A\rangle,|B\rangle$ instead of $A,B$.

Now the final position will be $|\psi\rangle$. One will think that $|\psi\rangle=a|A\rangle+b|B\rangle$ but in reality the probability is $a^2,b^2$, so we can also say it as say $|\psi\rangle=a|A\rangle-ib|B\rangle$(because $b^2=-(ib)^2$)

This is SO useful, think about it, this little, that look almost meaning, showing as connection between quantum physics and waves!

I won't get more into this and why it is so useful because it requires a little background in physics.

There are a lot of complex numbers in quantum physics, apparently quantum physics has almost nothing without complex numbers, for example Schrödinger equation: $i\hbar\dfrac{\partial}{\partial t}\Psi(\mathbf{r},t)=\hat H \Psi(\mathbf{r},t)$ where $\hbar$ is the reduced planck constant.

I can keep going, read about quantum physics if this is interest you