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So I'm a computer science graduate and I'm trying to understand imaginary numbers.

What I found is that it is often said: $i$ is the number defined such as $i^2=-1$, but this confuses me, as multiplication was defined as operation on real numbers, so how is it used now with imaginary or complex numbers?

My rough understanding is something along the lines: multiplication is an abstract operation,so we will just allow it for some abstract object we invent and call it $i$, and define $i*i$ as $-1$.

They say stuff like this is enlarging the real number system, but I don't fully understand this process.

I have studied calculus(single and multi-variable), linear algebra, probability and discrete mathematics in college. I just want to understand all these stuff of enlarging a number system, regarding multiplication as abstract operation performed on abstract objects, instead of thinking of it as repeated addition, etc..

I have searched the table of contents of many abstract algebra books, but failed to find what I want, may I just don't know what to search for and where.

Loai Ghoraba
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  • Does this help? https://cs.uwaterloo.ca/~alopez-o/math-faq/node18.html – ancient mathematician Nov 16 '23 at 21:05
  • You need, I reckon, to learn about polynomial rings, quotients of rings wrt an ideal. These words will be in the contents pages of abstract algebra textbooks (I hope). – ancient mathematician Nov 16 '23 at 21:06
  • This may answer your question, or at least help you to understand better the imaginary numbers: https://math.stackexchange.com/q/4805214/1215020 – César VB Nov 16 '23 at 21:06
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    Do you know what a vector space is? The complex numbers are a $2$-dimensional vector space with basis ${1,i}$. So you only need to define a multiplication on the basis. So do it by $1\cdot i=i\cdot 1=i$ and $i\cdot i=-1$. Now extend to all numbers $a+bi$ with $a,b$ real. This should dissolve your concern: "but this confuses me, as multiplication was defined as operation on real numbers". Yes, but now you have extended it. – Dietrich Burde Nov 16 '23 at 21:11
  • A small note: Since you have studied linear algebra, maybe you can borrow some intuition from algebra on matrices. You can view the complex number $a+bi$ as a matrix $$\begin{bmatrix}a&-b\b&a \end{bmatrix}$$ (more formally, there is an isomorphism mapping one object to the other). Addition, multiplication, and inversion of complex numbers corresponds exactly to addition, multiplication, and inversion on those matrices. If you are comfortable with extending those operations from real numbers to square matrices, this should give you - some - intuition about what's going on with complex numbers. – Christian E. Ramirez Nov 16 '23 at 21:15
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    You're right to be disturbed. The description you've been given is inaccurate. One can't generally, just make up some property and then say "and we'll invent a new number with this property and call it $i$.” I discussed this in a bit more detail here: https://math.stackexchange.com/a/4472292/25554 – MJD Nov 16 '23 at 22:29
  • Are you familiar with modular arithmetic (congruences)?, e.g. $\bmod 9!:\ 10\equiv 1\Rightarrow 10^n\equiv 1^n\equiv 1.,$ If so then similarly working with polynomials we can define $,\Bbb C = \Bbb R[i]/(i^2+1),$ or, equivalently, the ring of polynomials in indeterminate $,i,$ with real coef's, modulo $,i^2+1,,$ as Cauchy did. $\ \ $ – Bill Dubuque Nov 16 '23 at 22:43
  • @MJD Your comment is also inaccurate. In fact we can always adjoin a root of a polynomial $f(x)\in R[x]$ to a ring $R$ by constructing the quotient ring $,\hat R := R[x]/f(x)\cong R[x]\bmod f(x),,$ just as I did for $\Bbb C$ above. But generally this may not be an extension of the base ring $R,,$ i.e. the quotient may force some elements in $,R,$ to become equal in $,\hat R,,$ i.e. the natural image of $R,$ in $\hat R$ is not an injection (has kernel $K\neq 0).\ \ $ – Bill Dubuque Nov 16 '23 at 23:00
  • In the worst case all elements of $,R,$ may become equal in $,\hat R,,$ i.e. the kernel $(K=R),,$ i.e. $,R,$ could collapse to the zero ring, i.e. adjoining such a root is inconsistent. $\ \ $ – Bill Dubuque Nov 16 '23 at 23:06
  • Perhaps lost in the shuffle is the question of why $~i = \sqrt{-1},~$ and the field $~\Bbb{C}~$ were conjured in the first place. It is because [1] a way was found of folding Real Analysis into Complex Analysis, so that all of the results in Real Analysis were preserved under the Complex Analysis umbrella. [2] It simplified solving various math problems. For example, when looking for the 3 roots of $~(x-1) \times (x^2 + x + 1) = x^3 - 1 = 0,~$ since 2 of the 3 roots are not real, $~i = \sqrt{-1}~$ is needed here. ...see next comment – user2661923 Nov 17 '23 at 00:16
  • Then, once you eliminate the $~(x = 1)~$ root, the hard way of continuing is that $~\displaystyle x = \frac{1}{2} \left[ ~-1 \pm i\sqrt{3} ~\right].~$ The alternative, much easier, Complex Analysis approach is that the 3 roots are given by $$x = [ ~\cos(2n\pi/3) + i\sin(2n\pi/3) ~] ~: ~n \in {0,1,2}.$$ – user2661923 Nov 17 '23 at 00:20

2 Answers2

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The simplest way to construct complex numbers from real numbers, is to say a complex number is $a+ib$, where $a$ and $b$ are real numbers, and to express how addition and multiplication are done on these new numbers:

$(a+ib)+(c+id)=(a+b)+i(c+d)$
: addition is similar to the addition of vectors, if we consider $a+ib$ to be a 2-vector.

$(a+ib)\times(c+id) = (ac-bd)+i(ad+bc)$
: this one is less obvious, but boils down to the fact that $i^2=-1$.

So why choosing $i^2=-1$? Actually other rules have been used. But one of the main advantages of adding to the real numbers an element $i$ such that $i^2 =-1$, is that now polynomials have exactly as many roots as their degree (with potentially the same root more than once).

It may seem strange that just by adding 2 roots to $X^2+1$, one actually adds roots to all real polynomials of any degree, that lacked some roots, not just to the degree 2 polynomials. This can be intuitively understood by the fact that real polynomials can always be expressed as a product of degree 1 and degree 2 real polynomials. So by solving the degree 2, we actually solve all degrees.

Another way to understand $i$ is to remark that multiplication by $i$ is a $\pi/2$ rotation in the complex plane: $a+ib$ is vector $(a,b)$; multiplied by $i$ it becomes $ai-b$, i.e. vector $(-b, a)$, which is vector $(a,b)$ rotated by $\pi/2$. Which makes obvious that $i^2=-1$ : two $\pi/2$ rotations make a $\pi$ rotation.

By adding a $\pi/2$ rotation, we actually add any rotation $\theta$, because it can be expressed as $a+ib$, for $a = \cos \theta$ and $b = \sin \theta$. By developing $(\cos \theta + i \sin \theta) \times (\cos \phi + i \sin \phi)$, you'll see it magically makes $\cos(\theta+\phi)+i \sin(\theta+\phi)$. Having all rotations means $X^k=-1$ can be solved for any $k$, and explains (on an intuitive level) why all polynomials now have roots.

Perhaps you know that complex numbers can be further extended to quaternions (4 components instead of 2) and then octonions (8 components). But those further extensions, although useful (quaternions allow a simple expression of rotations in 3 dimensions), lose some important properties. For example, multiplication of quaternions is not commutative. So they are rarely used. Maths may create objects with any rules, but ony the useful ones are kept.

Jean-Armand Moroni
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It seems to me your problem is not with the imaginary numbers per se, but with the stipulation that $i$ is such an abstract object that $i^2=-1$.

And yes, as mentioned in the comments, you can just define $\mathbb C:=\mathbb R^2$ with peculiarly defined addition/multiplication, then set $i:=(0,1)$ and verify that it satisfies $i^2=-1$, but this is a different approach and it still does not make it clear why can we take some abstract $i$ out of thin air and impose $i^2=-1$ on it.

The proper way to explain that particular wording is based only on the addition/multiplication operations.

Let's start simple with $\mathbb N_0=\{0,1,2,\ldots\}$ with the addition (+) operation. As you have a CS background, you can imagine it in terms of OOP as a class with a member function "plus".

What is a negative one (and $\mathbb Z$), then? It's such an entity that sums with $1$ to zero. We can take an arbitrary object $\xi$ (denoted by $(-1)$) outside $\mathbb N_0$ and try to define a new set (class) $\mathbb N_0 \cup \{ \xi \}$ with a new overloaded (in OOP terms) function "plus", which acts just like $\mathbb N_0$'s "plus" on $\mathbb N_0$, but for $\xi + a$ returns the element previous to $a$.
You should now immediately notice that this "function" is undefined at $\xi+\xi$, because that should sum to 0 when added to 2, while no such element exists in the set (class) we're working - $\mathbb N_0\cup\{\xi\}$. To make it work, we have to add another artificial element from thin air - denoted $(-2)$ - to the set and define $(-1)+(-1):=-2$. But that still does not work, because now $(-2)+(-1)$ is undefined on $\mathbb N_0\cup\{-1,-2\}$. As you notice, we can go like this for arbitrarily long time and never get a properly defined "plus" function, which extends $\mathbb N_0$'s "plus" (i.e. acts like the normal plus on the $\mathbb N_0$ subset) and satisfies $(-1)+1=0$.

So what we want is an extension of $\mathbb N_0$ which supports a total function "plus" in the sense described above, but nothing more! There are a lot of these, but there's a unique -- up to isomorphism (of monoids, but let's not go there) -- "smallest" one, that we denote by $\mathbb Z$. I have no space to delve into details here, but I hope you can agree that the set you intuitively know as $\{\ldots,-2,-1,0,1,2,\ldots\}$ works here - it's "plus" function works like $\mathbb N_0$'s "plus", satisfies $(-1)+1=0$, and you cannot take any elements from $\mathbb Z$ without breaking these properties.

Now let's return to $i$. By analogy with above, here we have a set $\mathbb R$, but now the key operations include not only "plus", but "multiply" too.

Again, take an arbitray object outside $\mathbb R$ (formally, you can even take the set $\mathbb R$ itself!) and denote it by $i$. Construct the set $\mathbb R\cup\{i\}$ and try to define "plus" and "multiply" on it so that they coincide with the respective functions on $\mathbb R$, but also satisfy $i\cdot i=-1$.

You'll immediately notice that you need to add an element $2\cdot i$ to the set in an attempt to make the function "multiply" total, just like you had to add $(-1)+(-1)$ to $\mathbb N_0$ to get to $\mathbb Z$. Similarly, you'll have to add $-1$, because the multiplication function has to be defined on $i$ and $(-1)$, too.

The "smallest" set containing $\mathbb R$, on which "plus" and "multiply" are total functions restricting to the standard real addition and multiplication, with the additional condition that $i\cdot i=-1$, exists, and is denoted by $\mathbb C$.

To continue the OOP analogy, this is now only an "interface" - it just asserts that the object has methods satisfying some properties. To get an "implementation" of it, you have multiple options:

  1. Probably the easiest one for you is to use $\mathbb C:=\mathbb R \times \mathbb R$ and define $(+),(\cdot)$ in the familiar way.
  2. Another option (as mentioned in another answer) is to start with the polynomials $\mathbb R[x]=\{a_0+a_1x+\cdots+a_nx^n:n\in\mathbb N,a_i\in\mathbb R\}$ and consider any two of them equal if they differ by a factor of $x^2+1$.

But you didn't ask about the implementation, I feel, you asked about the interface.

Al.G.
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