imaginary numbers

i^2=-1

what's else good to know

There's a technical sense in which that is 100% of all there is to know about i.

But maybe it's also helpful to know that i^3 = 1/i = i, and that absolutely any formula that is true for i is also true for -i.


Also, it doesn't come up as often, but sqrt(i) = (sqrt(2)/2) (1 +/- i).

Urgh.  I meant i^3 = 1/i = -i.

FAIL.

>>2
I think your second statement needs qualification. How about the trivial equation i + x = 0?

complex numbers are good to know.
probably not, but theyre another thing that involves the imaginary unit.

http://en.wikipedia.org/wiki/Imaginary_unit#Powers_of_i

These are a couple basic "identities"--by which I mean they're playing around with 'i to a power' to illustrate patterns. You can verify any of them for yourself pretty easily with some algebra and possibly a little scratching of the head. (Ask if you're having trouble.) Remember that:

BLAH ^ (-1) = 1/BLAH

If you're looking for what USES imaginary numbers have, most of electrical engineering would fall hard on its face without the concept of i (except EEs call it j instead of i for some reason I don't care to investigate). Scroll down a little on the link I gave to see Euler's beautiful formulas for a cursory glance at WHY that's the case. (Don't expect to fully appreciate or even be able to apply the equations unless you have a little background in stuff like that.)

Complex numbers are just vectors with two values in them. But because we can do arithmetic and calculus with these vectors we call them numbers.

>>4

No qualification needed.  If i + x = 0, then x = -i.
If -i + x = 0, then x = -(-i).

e^i*pi - 1 = 0

Shit, sorry:
e^i*pi + 1 = 0

>>10
Beatiful

>>1
http://en.wikipedia.org/wiki/Quaternion_group
http://en.wikipedia.org/wiki/Quaternion

>>6
So would QM and string theory.

>>7
You mean 2 dimensional.  A vector space either has 1 element (trivial), or an infinite number of elements.  Arithmetic and calculus are defined for vector spaces of arbitrary, finite dimension.

>>12
Fine. Complex numbers form a field. So we can call them numbers. Satisfied?

>>13
yes

>>12

Let P be a prime. Then Z / PZ is a field, and (Z/PZ)^n for n natural is a finite vector field.

>>12
In addition to >>15's point, >>7 said complex numbers were vectors with two elements (the implication being that the elements were real numbers), not "a vector space with two elements."

>>15
Good catch; I forgot about finite fields.  I assumed vector spaces over infinite fields.

>>16
mmm