# How aliens would find complex numbers

1. Feb 13, 2009

### Gerenuk

I was trying to think how to introduce complex numbers in a more natural way. I find defining $\mathrm{i}=\sqrt{-1}$ just to not get stuck in maths and then be surprised by the power of complex numbers unsatisfactory. There are probably other ways, but they are abstract, too? Here is some visual way, so that even simple-minded aliens would get this (at least it would be visual if I could draw pictures here; its possible to draw all this reasoning here very graphical).

Aliens have trees $\Psi$. Many of them mean $\Psi\Psi\Psi\Psi$. To shorten notation the aliens introduce natural number algebra multiplication of object to write $4\Psi$. They notice that sometimes parts are important so the introduce positive rational numbers. An by considering algorithms they define rational number outcomes of equations.

But the alien notice that not all trees are equal, but have some age and cycle through state every year. So with the use of the positive number system they introduce the aging operator written as $\{t\}\Psi$, which means age by time $t$.

The age is periodic and additive
$$\{t+1\}=\{t\}$$
$$\{a\}\,\{b\}=\{a+b\}$$
Of course the age can be combined with the multiplication operator.

Now for some reasons I don't know the aliens assume that the sum of trees with different age is equal to a single age again
$$A\{a\}+B\{b\}=C\{c\}$$

----- to be continued in next post (physicsforum has problems) ---------

Last edited: Feb 13, 2009
2. Feb 13, 2009

### Gerenuk

Now they define some handy combinations
$$c(a)=\frac{1}{2}(\{a\}+\{-a\})$$
$$s(a)=\frac{1}{2}\,\{-1/4\}(\{a\}-\{-a\})$$
as these combination will allow them to deduce results from real algebra only.

With their known rules they try squaring these combinations and find
$$c(a)^2=\frac{1}{2}(c(2a)+1)$$
$$s(a)^2=1-c(a)^2$$
$$\{1/2\}+\{-1/2\}=\{1/2\}(\{0\}+\{-1\})=2\{1/2\}$$
With these rules and square-rooting the can calculate an arbitrary $c(2^{-n})$ and $s(2^{-n})$
(for some reason they always chose the positive square root; note that here they have to assume that the result will be a number of the form $T\{t\}$ to deduce the square root in calculations!)

$$c(a+b)=c(a)c(b)-s(a)s(b)$$
$$s(a+b)=s(a)c(b)+c(a)s(b)$$
they can split an arbitrary number $t$ into a sum of $2^{-n}$ and thus calculate $c(t)$ or $s(t)$

It is also easy to check that
$$\{a\}=c(a)\{0\}+s(a)\{1/4\}$$
This is useful for an intermediate form when adding numbers. The result of the addition will be $A\{0\}+B\{1/4\}$ and this should be equated to the resulting number $C\{c\}=Cc(c)\{0\}+Cs(c)\{1/4\}$. So
$$\frac{A}{B}=\frac{c(c)}{s(c)}$$
gives c and
$$A^2+B^2=C^2$$ gives C.

The aliens can now sum arbitrary age operators!

------ TBC --------------

Last edited: Feb 13, 2009
3. Feb 13, 2009

### Gerenuk

Now they go on to infinitesimals.

They define the resulting number $\pi$
$$\lim\frac{s(\varepsilon)}{\varepsilon}=2\pi$$
$$\lim\frac{c(\varepsilon)}{\varepsilon}=1$$

This way they find
$$\frac{\mathrm{d}}{\mathrm{d}t}\{t\}=\lim \{t\}\frac{\{\varepsilon\}-1}{\varepsilon}=2\pi\{1/4\}\{t\}$$
and hence by solving the differential equation
$$t=\frac{\{-1/4\}}{2\pi}\int_1^{\{t\}}\frac{\mathrm{d}x}{x}$$
and also
$$\{t\}=1+\int_0^t 2\pi\{1/4\}\{x\}\mathrm{d}x=1+\int_0^t 2\pi\{1/4\}\left(1+\int_0^t 2\pi\{1/4\}\{x\}\mathrm{d}x\right)\mathrm{d}x=\ldots=\sum_{k=0}^{\infty} \frac{(2\pi\{1/4\}t)^k}{k!}$$
They could find series expansions for c(a) and s(a) now.

Now they got a lot of useful results.

Last edited: Feb 13, 2009
4. Feb 14, 2009

### lurflurf

So you have dressed up the simple fact that exp(i*x) is a homomorphism from (R/2piR,+) to (C||z|=1,*)?
That is from addition of reals such that we have a~b is (a-b)/(2pi) is an integer
to multiplication of complex numbers with modulus 1.

5. Feb 14, 2009

### Gerenuk

Actually that's the point that I dressed it up. I could draw all this with nice natural pictures and teach it to a little child. It wouldn't even know it does complex algebra.

And I also derive the actual rules for calculations, which is the main part. The homomorphism is a very small part in that argumentation and comes from the logic of aging.

6. Feb 14, 2009

### arildno

What's so un-alien about Caspar Wessel's original geometric argument concerning how to define multiplication and addition of line segments?

Or Hamilton's definition of the complex numbers?

Neither uses the "square root of -1" as a starting point..

7. Feb 14, 2009

### confinement

The aliens would naturally become familiar with the real numbers, since they are the unique complete totally ordered field.

Furthermore, it would be natural for them to seek an extension field of the reals in which every polynomial with real coefficients will have a solution (since it is possible to prove the existence of such an extension field for the case of polynomials over an arbitrary field). Upon constructing it (which could be done in a general way by forming the quotient space of the ring of polynomials over the reals with a maximal principle ideal generated by an irreducible polynomial, for example generated by $x^2 + 1$ (human's choice)).

Then they would find that this extension field, the complex numbers, is algebraicly closed i.e. every polynomial with complex coefficients has a complex root. Then they would discover that every differentiable function in a complex variable is analytic i.e. has infinitely many derivatives and is equal to its taylor series. Then they would find that complex values are part of the description of the universe e.g. quantum physics ...

8. Feb 14, 2009

### Office_Shredder

Staff Emeritus
Maybe they don't really care about complete totally ordered fields? Their eyes could be mis-formed so they only see a countably dense subset of everything in front of them, so just stop at developing Q.

9. Feb 15, 2009

### Gerenuk

Don't know these one's. Internet reference?

10. Feb 15, 2009

### Gerenuk

And then, just as we do, they would wonder why the purely mathematical approach led to something that is a necessary structure for real world quantum mechanics.

The problem is that the mathematical method loses the connection to imaginable physics.

With my approach they might say "Hey, maybe that means in QM the phase determines time in a cycle of states." I haven't worked this idea out yet though....

11. Mar 4, 2009

### Gerenuk

Thanks for the link. It looks interesting at first glance and I'll go through it.

However, I never said complex numbers are mysterious. They are simple and in fact my explantion from above makes the definition slightly more complex. But these definitions are more related to the real world.

12. Mar 4, 2009

### Tac-Tics

You don't need to mention square roots at all in defining complex numbers. You could simply define an algebra over vectors in a plane. Vectors can be added as usual or multiplied together through a special rule. Multiplying two vectors together results in a third whose length is equal to the product of the lengths and whos angle is the sum of the angles.

Then, you basically work backwards and arrive at their useful representation as algebraic objects.

13. Mar 4, 2009

### Gerenuk

That misses the point of my representation. I do not wish to define some new mathematical objects. I'm sure there are plenty of way to do that.

I have a picture where everything has a simple real world representation.
Defining that lengths multiply and angles add in 2D has no real world picture.

14. Nov 14, 2009

### Xezlec

I know this is an old thread, but I've been trying pretty hard to figure out complex numbers for a long time. I don't understand them in the least, and this prevents me from being able to do much of anything with them competently. Unfortunately, this explanation loses me at "Now they define some handy combinations". You say this is a real-world picture, but I'm not seeing it. Guess I'm dumber than a child.

One thing that had occurred to me was that what I'm really trying to understand is what it means to raise an arbitrary operator to an imaginary power. I mean, I know what a whole number is because I can apply an operator recursively to something n times. I know what a negative number is because I can apply the inverse of that operator n times. I know what a rational number is because I can break an operator into parts such that applying some fractional piece of that operator n times gives me the same effect as applying my original operator. But I have no idea what it looks like to apply an operator i times. If I knew that, I would understand complex numbers.

https://www.physicsforums.com/showthread.php?t=316327&highlight=complex+power+operator" touches on the subject. I've found stuff online about fractional powers of operators and how it's connected to imaginary powers of operators. It seems like an amazing subject that I would love to know more about, but everything I've found on the subject has been much too advanced and jargon-heavy for me to follow. I wonder if anyone has a more accessible explanation.

And, lest anyone wonder why it's so important to me to understand this, I'm just sick and tired of seeing things like "oh, the wavelength is imaginary" and having no freaking clue what that means (yes, I know what it means in the case of wavelength. It's just an example). It seems like any quantity can be imaginary, and what that does to the situation seems completely random. It seems the only rule to making a quantity imaginary is "pull some random unrelated concept out of the sky and measure that instead of the concept that quantity usually measures".

Last edited by a moderator: Apr 24, 2017
15. Nov 14, 2009

### Gerenuk

It's just for that to see you need to have seen some weird mathematical thinking, which deal with really abstract concepts. Like group theory or abstract algebra could be the key words here.

The best definition I know is that you would take Taylor series of the exponential to define imaginary exponents. In that sense an imaginary power is just a mathematical definition. Once you know all rules for them by heart, you might feel that you understand imaginary exponents.

It's all about knowing the rules. Can you imagine what fractional Fourier transform means (even though here it's only real number)? Probably once one knows all the rules one would understand.

If I were to teach complex number to engineers, I might say follow the usual rules but introduce a new special number i. And all I can say about i is that you should leave it as it is. However, if you see a $i^2$ then you can replace it by -1. This is enough for mathematicians, but I probably should also teach some rules for exponentiation which aren't easily derived!
(actually not all rules of real algebra work for complex numbers. logarithms are multivalued and the numbers cannot be ordered by < and >).

That doesn't have to do anything with a "real" wavenumber. It only states that if you solve the mathematical problem, you should replace trigonometric functions with exponentials. Now instead of stating the complicating procedure how to replaces sin and cos with exp, it's easy to say "ah, just plug in an imaginary number into sin(x) and this will give you effectively exponentials, provided you do all the maths according to the abstract rules of imaginary numbers".

16. Nov 14, 2009

### DrGreg

Non-integer powers of a linear operator can be defined via the Spectral Theorem, in those circumstances when it is valid.

For example, over a finite-dimensional vector space, if the operator can be diagonalised, you apply the powers to each of the eigenvalues in the diagonal matrix.

As another (infinite-dimensional) example, in Fourier analysis, the differential operator (d/dt) in the time-domain corresponds to multiplication in the frequency domain, so you can apply your power to the multiplier function.

17. Nov 14, 2009

### Xezlec

Meh. I just find that "knowing the rules" isn't sufficient. It's enough to solve a problem when you're told exactly what to do, but it isn't enough to figure out what to do in the first place. For that, you need intuition, and a list of rules doesn't give me that.

I flunked electromagnetic engineering the first time I took it, because I was just desperately trying to slog my way through problems based on a set of arcane and unintuitive rules. Most of the time, I just had no idea how to even start to attack a problem. The second time through, I made a point of scouring the net for visualizations of each of the concepts in vector calculus that were used, and then I aced it easily, because I could picture everything.

18. Nov 14, 2009

### Xezlec

Thanks! I'm going to look into this some more.

19. Nov 15, 2009

### Tac-Tics

Complex numbers are simple. If you're having a hard time understanding them, you're doing something wrong.

20. Nov 15, 2009

### Xezlec

Feel free to help me figure out what.