# Most natural intro to Complex Numbers [closed]

This is a soft question but I'm willing to ask.

There are few ways to introduce the field of complex numbers, but if You had the opportunity to write an elementary textbook, what would be the most intuitive introduction for a very sceptical audience?

Example of an "bad" answer: write down all the axioms and then quote Hilbert "existence in mathematics means freedom from contradiction".

## closed as primarily opinion-based by Thomas, user223391, hardmath, user147263, Jonas MeyerJul 17 '15 at 21:03

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• I didn't know that bad answer from Hilbert. I love it. – Git Gud Jul 17 '15 at 11:49
• Feynman (lectures on physics) has a nice introduction to the idea of negatives, rationals, irrationals, and then complexes, each as a necessary construction to express something that physicists need to talk about. It's not very rigorous, but it's compelling. – John Hughes Jul 17 '15 at 12:00
• That depends on what you are looking to explain and what what your audience is sceptical about. If an axiomatic definition doesn't satisfy you, then you could try the historical approach: complex numbers were born to give solutions to some equations, just like negative numbers and roots before them. – A.P. Jul 17 '15 at 12:01
• I would start my definition with the preliminary question "Why they are called COMPLEX numbers?" and skeptical people I feel would like all the history on the "imaginary" adjective. – Piquito Jul 17 '15 at 13:20
• Skeptical about what, exactly? – anomaly Jul 17 '15 at 16:08

$\mathbb{C}$ has very little to do with axioms. Just let $\mathbb{C} = \mathbb{R} \times \mathbb{R}$ and then define the appropriate operations on it. In particular:

$$(x,y)(x',y') = (xx'-yy',xy'+yx')$$

Then prove that arithmetic inside $\mathbb{C}$ behaves as expected.

Other possible approaches:

• let $\mathbb{C} = \mathbb{R}[i]/(i^2+1).$
• realize $\mathbb{C}$ as a set of $2 \times 2$ real matrices. (Courtesy of GEdgar).

Okay, but what if the audience is uber-skeptical?

1. If they don't accept that $\mathbb{R}$ exists, you can use an axiomatic set-theory (say, ZFC) to prove its existence. Similarly, if they don't accept that from $\mathbb{R}$ we can construct $\mathbb{R} \times \mathbb{R}$ with the aforementioned operations, then you may need to appeal to an axiomatic set-theory (or other foundation).

2. If they don't accept the usual axioms of set theory, applaud them for being very skeptical. Then challenge them to come up with their own ideas and their own formal system on which to secure mathematics. This is a surprisingly hard challenge, and is likely to increase their respect for existing foundations! Hopefully not so much that they stop pondering their own foundations, though. The rules of the game are:

• Most mathematical knowledge currently in existence should be "realizable" in their system. In particular, the things we currently know about differential equations and real analysis should be realizable within their system.
• The student should be able to describe - informally, at least - the intended semantics; furthermore, the axioms and inference methods of their formal system should be reasonably intuitive, given the intended semantics.
• Good answer. But I wonder if someone is "very skeptical", would he accept the real numbers? – GEdgar Jul 17 '15 at 13:04
• @GEdgar, certainly some constructivists reject the existence of $\mathbb{R}$. We could try bringing Dedekind-cuts into it, I suppose. Honestly, I think its probably better to just assume the existence of $\mathbb{R}$ at this level. – goblin Jul 17 '15 at 13:07
• A skeptical audience would point that a definition does not ensure the existence of the object. So $\mathbb{R}\times \mathbb{R}$ with the definition of the appropriate operations might not exist. – Ramiro Jul 17 '15 at 13:25
• @goblin The point remains. A skeptical audience would insist that defining an object and showing that the definition entails nice properties is not enough to prove the existence of the object. Yet another objection from a skeptical audience might be: "If complex numbers are just a notation to represent points in the plane or a quotient of a polynomial ring or a subset of 2x2 real matrices, then there is no NUMBER to solve equations like $x^2+1=0$ (and third and fourth degree equations)". – Ramiro Jul 17 '15 at 14:11
• Please note that the question is NOT to prove that the Complex Numbers exists. The questions is about how to introduce Complex Numbers to a very skeptical audience. There are many PERFECTLY correct ways to present $\mathbb{C}$ which will look either "too abstract" or "arbritrary" to a very skeptical audience. All the three different ways to define $\mathbb{C}$ that you mentioned in your answer are mathematically correct and elegant. – Ramiro Jul 17 '15 at 14:40

I don't like the introduction via $x^2+1=0$, because it's too easy for a skeptical person to say, why should that equation have a solution, anyway? I prefer to look at something like $x^3-6x-4=0$. Now, that has a real solution; the left side is negative for $x=2$ and positive for $x=3$, so there's a solution somewhere between 2 and 3. And you can express that solution as $$x=\root3\of{2+2\sqrt{-1}}+\root3\of{2-2\sqrt{-1}}$$ as you can check by cubing it and doing some algebra.

Historically, complex numbers came in as solutions of cubics, not quadratics.

• Nice. In my opinion, the best approach would use third and fourth degre equations. I would explore Cardano's formulas to show to the skeptical audience that the case you mentioned is not just a "coincidence" in one particular equation. – Ramiro Jul 17 '15 at 14:45
• Still someone can argue that $\sqrt{-1}$ doesn't make sense and you are expressing a valid solution with not valid numbers. They could say this is a fake solution and that there is a different ( and valid ) way to express this solution. – Integral Jul 17 '15 at 14:45
• So what is the analogue of 'negative for x=2 and positive for x=3, so there's a solution somewhere between 2 and 3' for 1st eq? – BCLC Jul 17 '15 at 16:26
• @BCLC, that's the point --- there isn't any real solution for $x^2+1=0$, and there isn't anything quite like the Intermediate Value Theorem for the complex plane (and even if there were, it would presuppose you had already accepted the complex plane as a thing, which is precisely what's at issue here). – Gerry Myerson Jul 17 '15 at 23:07
• @Integral, given an irreducible cubic with three real roots, there is no way to write down those roots in closed form in terms of the four arithmetic operations and square roots and cube roots on the coefficients, without introducing complex numbers. The proof of that statement isn't easy --- the only way I know goes through Galois Theory --- but it's a fact. – Gerry Myerson Jul 17 '15 at 23:11

I've had some fun and some success with middle school kids using this approach.

They know about the number line. They can understand that addition of a fixed quantity can be thought of as translation, right for positives and left for negatives. Note that translation by $a$ takes the number $0$ to $a$. They can understand a fraction like $a/2$ as what you need to translate by twice to translate by $a$.

Then we move on to multiplication by positive numbers. As a transformation, multiplying by a fixed $a$ is a change of scale. Note that scaling by $a$ takes the number $1$ to $a$. That leads to a definition of $\sqrt{a}$ as the number you have to scale by twice to scale by $a$.

Next we note that multiplication by $-1$ is a reflection about $0$ on the line. (Incidentally, that helps with "why is a negative times a negative a positive?") But what is half that reflection? What can you do twice to end up moving $x$ to $-x$? Just rotate the number line half a turn (counterclockwise is the right convention). Kids really like that idea. By analogy, that's a rescaling by the number that $1$ moves to - so we have $i$ in the right place in the newly introduced number plane. The additive structure of the rest of the number plane follows easily when interpreted as translation.

The full multiplicative structure is much harder - both historically and for now tired middle schoolers. You can do it with the distributive law, but the geometry is harder. You can't quite get to polar coordinates. But they will be delighted to find $\sqrt{i}$ as a rotation and then its real and imaginary parts since they know the length of the hypotenuse of an isosceles right triangle with side $1$ is $\sqrt{2}$.

If the audience is very skeptical, there will be no short way to introduce complex numbers. I think they would be suspicious of any "axiomatic" approach.

For such an audience, I would take a longer way, which happens to be historically relevant. I would present (and prove) the Cardano's algebraic formulas to solve third and fourth degree equations.

Then, I would show that those formulas produce correct REAL roots even when there are square roots of negative numbers involved in the calculation that cancel out. So this fact suggests that the square roots of negative numbers can be treated as numbers which really exist.

Then, I would remark that square roots of negative numbers, IF THEY EXIST, can be reduced to $a\sqrt{-1}$ where $a$ is a real number.

Then, I would algebraically deduce what would be the properties of such "imaginary" numbers when combined with real numbers.

Finally, I would come back and show that those "strange" numbers (or, we may call them "complex" numbers), resulting from the combination of real and "imaginary" numbers, work perfectly with Cardano's formulas and also with second degree equations.

It is a looong way to present complex numbers, but it corresponds (at least partially) to how historically the concept of complex numbers appeared.

• My issue with this is twofold. Firstly, in practice I have always found historical introduction to an idea painful to hear or read about. Please, just tell me the darn idea directly! Motivation is good, but all that history just kills me. Secondly, even after all that work, you still haven't really proved that $\mathbb{C}$ exists. You've just accumulated evidence that its plausible that $\mathbb{C}$ exists, and that we'd like $\mathbb{C}$ to exist. But this doesn't prove the existence of $\mathbb{C},$ nor does it prevent its existence from entailing an outright contradiction. – goblin Jul 17 '15 at 14:05
• @goblin The question is not to prove that the Complex Numbers exists. The questions is about how to introduce Complex Numbers to a very skeptical audience. There are many PERFECTLY correct ways to present $\mathbb{C}$ which will look either "too abstract" or "arbritrary" to a very skeptical audience. – Ramiro Jul 17 '15 at 14:32
• I disagree. The only way to introduce $\mathbb{C}$ to any audience - skeptical or not - is to construct it. Any other approach is asking for trouble. Also, "too abstract" and "too arbitrary" aren't legitimate critiques in my opinion. – goblin Jul 17 '15 at 14:48
• @goblin As a mathematician I agree with you: "too abstract" and "too arbitrary" aren't legitimate critiques. However the question is about how to introduce complex nunbers to a very skeptical audience, who does not know enough Mathematics (see in the question "an elementary textbook"). – Ramiro Jul 17 '15 at 15:00
• But surely you agree that we should only pander to the reader's legitimate critiques? – goblin Jul 17 '15 at 15:02

Just a soft answer - perhaps wrong.

To convince skeptic people...

# The "thing" $i$ is just a rotation over 90 degrees, so $i^2 = -1$

Note that $$x^2 - 2 \cdot \mathbf{1} = 0,$$

means that $x = \pm \sqrt{2}, \mathbf{1} = 1$, but we can also write $$x = \pm \left( \begin{array}{cc} \sqrt{2} & 0 \\ 0 & \sqrt{2}\end{array} \right), \mathbf{1} = \left( \begin{array}{cc} 1 & 0 \\ 0 & 1\end{array} \right).$$

As "four times a rotation over 90 degrees is $1$", we can consider $\mathbf{i}$ to be a rotation over 90 degrees. As "a rotation over 180 degrees is $-1$", we can say that $\mathbf{i}^2 = -1$.

We can now consider objects written as $$\mathbf{c}(x,y) = x + \mathbf{i} y,$$ keeping the rotation over 90 degrees in our mind.

While for real numbers, we cannot have $x^2 < 0$, for rotations $\mathbf{R}$, we CAN have $\mathbf{R}^2 = -1$.

Note that $$\mathbf{c}(x,y) = \left( \begin{array}{cc} x & -y \\ y & x\end{array} \right).$$

And note that $$x^2 + 2 \cdot \mathbf{1} = 0,$$ means that

$$\mathbf{x} = \pm \left( \begin{array}{cc} 0 & -\sqrt{2} \\ \sqrt{2} & 0\end{array} \right) = \pm \mathbf{i} \sqrt{2}.$$

• It is a nice natural approach. However, a skeptical audience would probably answer something like: "Then complex number do not exist, they are just a notation to represent a subset of $2\times 2$ real matrices. So there is NO NUMBER to solve equations like $x^2+1=0" (nor to use in the solution of third and fourth degree equations). – Ramiro Jul 17 '15 at 13:36 • @RamiroGuerreiro, true. It would depend on how questions are asked, and how numbers are defined... For what "mathematical object$x$" is$x^2+1=0$?... – johannesvalks Jul 17 '15 at 13:53 • @RamiroGuerreiro, that's not a skeptical objection per se. Its more like a philosophically-confused objection. – goblin Jul 17 '15 at 14:00 • @goblin, mathematics deals with objects and some objects we call numbers, some objects we call complex numbers, some objects we call matrices... – johannesvalks Jul 17 '15 at 14:02 • If your audience are skeptical, they will say that doesn't make sense to rotate real numbers. You just defined a nonsense operation to get your complex numbers. – Integral Jul 17 '15 at 15:00 Here are the main points I would consider: I. Motivation: Define a complex number, then state some reasons for studying complex numbers. It is assumed that the students know about real numbers. So tell them that equations like$x^2+4=0$has no solution in the real numbers; as a consequence introduce the concept of$i=\sqrt{-1}$, and then give them the solution to the equation above. 1. Algebra of Complex numbers: Addition and Subtraction of C.N, equations in C.N, Multiplication of complex numbers, Conjugate of a complex number, then division in C.N. 2. Representation, Forms and Applications: Talk about Visualisation (Argand's diagram), then talk about the three forms of a complex number (Standard: z=a+ib, Polar:$z=r(Cos\theta+iSin\theta)$and Euler:$z=re^{i\theta}$) and how to go from one to another. Finally, talk about DeMoivre's theorem et-cetera. C.N means Complex Numbers. • Right. But someone can say that$i=\sqrt{-1}$is nonsense or just a trick that makes impossible to possible without concrete essence. – Hulkster Jul 17 '15 at 12:19 • I can't tell you how to convince your students. I do Maths for fun! If you explain with an Argand diagram, and the person doesn't wanna accept...that's the person's cup of tea! – Chuks Jul 17 '15 at 12:21 • The Pythagoreans thought$\sqrt{2}$was a dangerous idea because it wasn't a rational number. There are still people (some finitists) who have problems with real numbers generally. If your criterion is "relates to the real world", lots of mathematics may be hard to motivate. If you agree with that guy Hilbert (and I do), then you follow the non-contradictions wherever they lead. That's part of why mathematics is fascinating. Oh, and complex numbers do have real-world applications in physics, for example. – user452 Jul 17 '15 at 12:29 For me one of the best examples comes from the Feynman Lectures. First he starts with positive integers and the operations you can use on those numbers. However gradually you run into problems that make it necessary to introduce other numbers to find solutions. Such as 2-5=x has no solution unless you introduce negative integers. Or 2/5=x doesn't work unless you introduce rational numbers. Eventually this results in x^2+1=0 to introduce the existence of imaginary numbers. Caltech has made these lectures available to read online for free and if you would like to see how Feynman explains it. http://www.feynmanlectures.caltech.edu/I_22.html suppose you have an equation like this:  x² = -1  now you want to solve it. how? x = sqrt(-1)  which can NOT be answered for$x \in \mathbb{R}$thus to obtain the solution, we introduce the Complex numbers, with their main axiom being: i = sqrt(-1) ...whenever there is something that can't be solved with the mathematics of the time, it grows into mathematics. e.g. Real numbers couldn't solve equations like this: x = 1/3 so in order to solve them, Rational numbers have been introduced. The same logic applies to Complex numbers, quod erat demonstrandum. • Your example is wrong!$\frac{1}{3}\in \mathbb{Q}$, but$\mathbb{Q}\subseteq \mathbb{R}$. – Chuks Jul 17 '15 at 12:28 • But if we have$0 \centerdot x = 3$? Can we grow some new "imaginary" math from that problem? (This is a joke but a sceptical person might pose that question). – Hulkster Jul 17 '15 at 12:30 • You could pose that, but division by zero does lead to contradictions.$\sqrt{-1}\$ does not. Teaching that difference would be an interesting thing! – user452 Jul 17 '15 at 12:33
• I Just realized this :/ What i wanted to say was basically, that e.g. sqrt(19) can't be shown as ratio of two numbers, thus it required the introduction of a new class of numbers: irrational numbers. im not native english, so please forgive me the mistake :) – Gewure Jul 17 '15 at 12:34
• @ Gewure, you are pardoned :) – Chuks Jul 17 '15 at 12:35