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> We've arrived back at the axiom that all numbers have two square roots of opposite signs.

While I like the geometric content of the rest of your post, I have to quibble on two points:

    * that a number has two (usually distinct) square roots is a theorem, not an axiom.  That is, it's not 'built in' (unlike, say, commutativity of addition), but rather can be proven on the basis of 'built-in' properties.

    * while one may technically say that i and -i have 'opposite signs', it's a good idea not to talk about the sign of a complex number at all (except possibly in the sense that sgn(z) = z/|z|).  It's better to say that the two square roots are opposite; or, if one wants to be more precise, that they are additive inverses.

In higher maths, commutativity of addition can be proved from the Peano axioms without taking it as an axiom. It's a common homework problem.

Do kids not learn this in 8th grade any more? I'm seriously not trying to be snarky, I just can't think of a way to write that question that sounds unsnarky. I thought everyone learned about the polar representation of complex numbers.

You need to differentiate between "are told" and "learn". Personally I was told a lot of things in my high school math classes, I learned very few of them.

(Author of the above article)

I learned the polar representation, yep, but didn't connect as i as an "rotation operation" that took you from one dimension to the other. For all I knew, having i be 2-dimensional was like saying "We can represent condiments by having an axis for ketchup and and an axis for mustard".

It was 2 random quantities represented as an (x,y) pair, without the notion that they were deeply connected.

No, they don't. If they do learn it, it's certainly not in 8th grade.

My town's school system was the second in the entire state, and they taught Algebra I (ie, factoring/FOIL) in 8th grade. I went to a well-respected private school, and they taught complex numbers in 10th grade, and polar coordinates only in the honors section of 10th/11th grade classes. For those, they taught matrix multiplication and determinants for 2x2 matrices only (ie, not 3x3 or up). This means that any real linear algebra was completely out of the picture.

This was well more than enough to score a very good score on the SAT II Math IIC exam (is this still even around?)

I was fortunate that I was able to teach myself outside of my school's curriculum (and my school also let me place out of certain classes), but overall, the standards are far below what you might hope.

On the last question, I find it really interesting to note that the nth root of i will have n solutions evenly distributed around the unit circle. (As will any unit vector.) So for the cube root of i, we'll also have -0.5√3 + 0.5i and -i as solutions, which are located at 5π/12 = 30° + 120° and 8π/12 = 30° + 2 * 120°. (On another note -i * -i * -i = -1 * -i = i. It's so simple and yet not obvious unless you do a lot of complex math...which I do not.)

Read about the roots of unity (solutions to a^n = 1) to see the same concept in action. http://en.wikipedia.org/wiki/Root_of_unity

It all comes out of the properties of rotation. We can first show that there are n nth roots of unity. It comes down to a^n = 1 -> a^n * a^n = 1 -> (a * a)^n = 1, which effectively shows that any multiple of the angle represented by a will also be a solution. If you use the full fledged exponent formulas, it's easy to see that there will be only n of these. Then if b^n = i (our case), b^n * a^n = i * 1 -> (b * a)^n = i. So adding any of the nth roots of unity to b will get a new solution for the nth root of i. Since one of the roots of unity is always 1, aka an angle of 0, we will only get n-1 new solutions from this. Hence there are n solutions. :)

EDIT: New here, didn't know the markup cues would interfere with my equations.

One more question; what is the "i"th root of i? Please try to use the angle metaphor :)

To answer the question, e^(it) =cos(t)+ i * sin(t). That means that when t=pi/2, cos(t)=0 and sin(t)=1, so i = e^(i * pi/2)

To get the 'i'th root we divide the exponent by i, so i^(1/i) = e^((i * pi/2)/i) which is e^(pi/2).

As for the angle metaphor, that works for multiplying by complex numbers. The usual "understanding" of exponentiation is repeated multiplication, but taking a power of "i" can't be understood as multiplying together "i" lots of the number. Similarly the "i"th root can't be understood like that.

But you knew that.

Sure! You might want to check out http://betterexplained.com/articles/intuitive-understanding-...

I like this question because it really works your intuition.

The basics: x^y means "grow at x, for y units of time". I see "2^3" as "grow at 2x for 3 units of time".

Having a base of i means your "growth" is a rotation at 90 degrees, no scaling. So i^(1/2) means a 45 degree rotation, i^3 means a 270 rotation, etc.

Raising this to the i power (or 1/i power, which is -i) means the growth that was originally purely rotational is now rotated. So instead of growing at i, you are growing at (i * 1/i = 1). So, we should expect a positive real number, greater than 1 since our growth is positive.

How long do we actually grow for? Well, the base of "i" is really e^(i* pi/2), which means "Start at 1.0 and rotate continuously for pi/2 seconds". We've now modified this to "Start at 1.0 and grow at 1.0 for pi/2 seconds", which is e^(pi/2).

So the answer is i^(1/i) = e^(pi/2) ~ 4.8

It's a bit tough with text-only, read the above article for more diagrams.

Note that 2575.97 also has the property that raised to the ith power gives i. There are infinitely many such numbers.

or 0.008983291.. Are there countably infinite solutions or otherwise? See infinity discussion few days ago at HN http://news.ycombinator.com/item?id=4526049

PS: Just curious, i am not a mathematician.

It is countably infinite. Let's try to find them all.

A relatively simple way to understand this is that i^i = e^(i * log(i)) for every possible log of i. So all we need to do is understand what values log(i) could have (there are actually many), and then we can work it out. But log(z) just undoes e^z, so we need to understand e^z.

Now let's work backwards. If z = x + y i with x and y real, then x tells us the absolute value of e^z and y tells us the angle. The absolute value of i is 1, so any possible solution to log(i) has real part 0. The angle that we want to wind up with is 90 degrees, or pi/2. Therefore y can be ..., -3.5 pi, -1.5 pi, .5 pi, 2.5 pi, 4.5 pi, ... .

Therefore log(i) has to be one of 1.5 pi i, -.5 pi i, -2.5 pi i, -4.5 pi i, ... .

Now i^i is e^(i log(i)) so it can be any of ..., e^(3.5 pi), e^(1.5 pi), e^(-.5 pi), e^(-2.5 pi), e^(-4.5 pi), ... .

Unless I've made a trivial calculation error, that is the whole list.

simple explanation of PI: http://en.wikipedia.org/wiki/File:Pi-unrolled-720.gif

This is of course much trickier because to define z^w for complex numbers requires that one choose a complex logarithmic function. There are infinitely many choices for this. Everyone has decided to use the principle logarithm but it could be consistently defined using any branch cut.

e^(pi/2)

This is is because exponentiation to an imaginary power can also be thought of as rotation. You have presumably heard e^(pii) = -1. That's because it's a half-rotation away from 1. Half of that half rotation would be i, and is sqrt(e^(pi i)), which is e^(pi/2*i), which is equivalent to (e^(pi/2))^i, and the ith root of that is obviously (e^(pi/2)).

Thanks for posting that :). My gut-check about whether I've intuitively understood a topic is whether I can intuit new results. I'd wager 95% of people who "learned" imaginary numbers couldn't work out the cube root of i in their head like that (I was in this camp all through high school and college, which made me realize I didn't really get them).