Chronos, you got me all excited about all these things you can do in the complex world, which are not possible in the real world.
Just curious: What does an angle with a sin >1 look like on my trusty trig circle?
Can you take the log of a negative real number if you write it as the product of a positive number and i^2? what is the log of i? Gosh, did I tell you you got me all excited? 
That’s correct. In general, the n roots of a number will be the vertices of a regular n-gon centered at the origin.
One other thing I’ve always liked is that i[sup]i[/sup] is real. (It’s equal to exp(-pi/2), or 0.2089796.)
I’ve a soft spot for Euler’s “mathematical poem” myself: e[sup]i p [/sup] + 1 = 0. Five fundamental constants, addition, subtraction, exponentiation and equality all rolled into one tiny equation. Shame he couldn’t fit division in there as well!
Eh… for “subtraction” read “multiplication”, of course.
Btw, concerning the question about nth roots, there are n such roots of any number, real, imaginary or complex. For instance, the humble number 1 has three cube roots, two of which are complex. You should be getting the idea of how to evaluate these by now. They have some interesting properties when added, subtracted and multiplied 
In higher math, the sine as the ratio of sides of a triangle is not used. Rather, sin(x) = x + x[sup]3[/sup]/3! + x[sup]5[/sup]/5! + … A number whose sine is greater than 1 is simply a number whose series gets big a little faster than any real number’s.
As far as logarithms go…
log(c) is defined for any complex number c, including the negative reals. The problem has to do with the polar representations. (r, t) and (r, t + 2[symbol]p[/symbol]) represent the same point, but log((r, t)) and log((r, t + 2[symbol]p[/symbol])) are not equal. Things get ugly because of this.
The poor OP was just beginning to learn about imaginary numbers, folks! It’s like she asked for a cup of coffee and received instruction on how to make it starting with the bean! Forgetting polar coordinate and Euler (some pronounce “Oiler”) So, “Oiler vay”, Batman! Give the poor OP a break!
To me, imaginary numbers simply indicate there are cracks in the theory behind the very numbering system we use. (Of course, no system is perfect.) I say this because, as so many posts above are trying to show, those so-called imaginary numbers have very REAL application, in the engineering underworld! - Jinx 
If you want a number with a weird sine, use this formula:
sin(x + iy) = sin(x) cosh(y) + i cos(x) sinh(y)
So, the unwieldy pi/2 - i ln(sqrt(3) + 2) = 1.570796 - 1.316958 i has a sine of 2.
Hey! The reason I went into the complex plane is that OP mentioned it! “picturing the trigonometric circle where i is the segment sticking up in the air and -1 (i^2) is the segment going ‘straight and to the left’…where would i^1/2 be ?” And it’s a good thing, too. Complex roots are much simpler in terms of their polar form.
Well yeah, it’s scary- but it’s a GOOD kind of scary. It’s almost like finding religion when I look into such things…
Thanks again everybody for widening the ol’ horizon. I have to spend a few eons now digesting all that has been said here, especially the examples. Feel free to bring more on!
OK Malacandra, you are the cause of my current headache- I can visualize the 3 cubic roots of 1 on my trig circle, they have to be 1, e^iπ/3 and e^i2π/3.
Now, I can’t seem to figure out how to solve this from equations. Help, someone!
Starting from:
(x+iy)³=1
x³+3x²iy+3x(iy)²+(iy)³=1
x³+3ix²y-3xy²-iY³=1 which implies
x³-3xy²=1 and 3x²y-y³=0
I’m stuck. 
Not quite. They are exp(2[symbol]pi[/symbol]/3), exp(4[symbol]pi[/symbol]/3), and exp(6[symbol]pi[/symbol]/3), which is equal to 1. In general, the nth roots of one are given by exp(2k[symbol]pi[/symbol]/3), with 0 < k < n - 1.
Once you know the formula above, there’s no reason to try to solve the equations–they’re nasty.
Yes! I gotcha, Ultrafilter. In a momentary lapse of reason, I had the 180 degree angle on my circle equal π/2, not π (which halved all my roots). Thanks!
It’ll also help to know that:
exp(ix) = cos(x) + i sin(x)
In general, if the polar coordinates of any complex number are given by (r, theta), then the complex number can be represented as r exp(i theta).
Allow me to unstick you, then. Factorise your equations:
x(x[sup]2[/sup] - 3y[sup]2[/sup]) = 1
y(3x[sup]2[/sup] - y[sup]2[/sup]) = 0
The second equation has two solutions: y=0, or y[sup]2[/sup]=3x[sup]2[/sup]. (Actually the second of these implies two solutions, but we’ll come back to that). Going back to the first equation and substituting 3x[sup]2[/sup] for y[sup]2[/sup], we get:
-8x[sup]3[/sup] = 1 and so x = -1/2
Given that
y[sup]2[/sup]=3x[sup]2[/sup], there are two solutions:
y=sqrt(3)x and y = -sqrt(3)x, i.e. sqrt(3)/2 and -sqrt(3)/2
So this gives us the three cube roots of 1:
1
-1/2 + (sqrt(3)/2)i
-1/2 - (sqrt(3)/2)i
The complex roots are known as w and w[sup]2[/sup], and as I said, you can have fun adding and multiplying the roots together
From here on in the nth roots get hairy, and you’re better off with the “polar coordinates” approach.
Btw, this post was a sod to code!!
I know you meant multiplication instead of subtraction, but it’s relatively easy to divide zero by the square root of one minus v-squared over c-squared.
One point that should be made is that all this talk about logs of negative numbers and arc sines of numbers larger than 1 have to be taken with a grain of salt. Just as i (like any non-zero number) has two square roots, any number has infinitely many logarithms! The fact that e^2pii = 1 implies that you can add an arbutrary integer multiple of 2pii to any logarithm. Sure log (-1) = pii, but it also equals -pii and 3pii and -7pii and …
BTW, not only does every complex number have n nth roots, any polynomial equation of degree n with complex number coefficients has n solutions (although sometimes you get repeated roots, such as x^2 - 2x + 1 = 0 has 1 as a double root).
I don’t think it’s that big of a problem. It’s not unlike the arcsine function on the real numbers. Sure, there are an infinite number of solutions to sin(x) = 0.4, but we still define arcsin(0.4) as the “primary” value.
Now that is freaky - I never knew that. How can one rationalize (pun unavoidable) that? Take the unitary imginary number, raise it to the power of itself, and you get something that is not only real but some weird, probably transcendental thing. I can’t get my intuition around that. I know you can probably derive it logically from one of the identities mentioned in this thread, but I like to have an intuitive feel for things.
Even if you do principal values, you still have to worry about branch cuts.
i[sup]i[/sup] = exp(i * ln(i)) = exp(i/2 * ln(-1)) = exp(i/2 * i[symbol]p[/symbol]) = exp(-[symbol]p[/symbol]/2)
It’s hard to get an intuitive feel for complex exponentiation. Consider that I could’ve chosen any other branch of the logarithm and gotten a different answer.
You have to be careful there. I know that is your point, but the equality doesn’t necessarily hold. It’s the same reason that we don’t say -2 = Sqrt(4), because we know that 2 = Sqrt(4) and if that first were true equality, that would imply -2 = 2, since they’d both be equal to the same “thing”.