Is there one? What is it?

'tis Ergot's question, but he's lazy, so eh?

i is an imaginary unit, a complex number whose square equals -1.

i, of course, then leads us into entirely new fields of mathematics altogether...

[quote author=Explicit[nK] link=topic=14053.msg143724#msg143724 date=1138577638]
i is an imaginary unit, a complex number whose square equals -1.
[/quote]

Then i is basically a logic bomb, because there is no way to get x*x to equal -1. =/

Yes there is: i * i
Didn't you pay attention to what we just said?

Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p

[quote author=Joe link=topic=14053.msg143839#msg143839 date=1138672186]
Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p
[/quote]

It shouldn't hurt your head seeing as to how it's pretty straight forward.

[quote author=Joe link=topic=14053.msg143839#msg143839 date=1138672186]
Well, yeah, i * i, but that just hurts my head. Yes, I fully understood what you said, it just hurts my head thinking about it =p
[/quote]You don't stick your tongue out when your head hurts.  You squint your eyes and clench your jaw.  You are obviously just trying to toy with us.

i = sqrt -1
i[sup]2[/sup] = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.

With the introduction of the imaginary number i, it can be proven that every polynomial of degree n can be factored as a product of n polynomials of degree 1. It's called the Fundamental Theorem of Algebra.

http://mathworld.wolfram.com/FundamentalTheoremofAlgebra.html
http://mathworld.wolfram.com/PolynomialFactorization.html

This is a very important result. It means that, by extending numbers onto 2 axes, algebra is "complete" (purely algebraic equations, such as x^2 + 1 = 0, are solvable).

[quote author=Shout link=topic=14053.msg143865#msg143865 date=1138686043]
i = sqrt -1
i[sup]2[/sup] = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics.

[/quote]Assuming that x is being limited to real numbers, this is true.  However, if complex numbers are allowed, then your statement goes to shit.  Other than that, way to tell Joe off :0

[quote author=Shout link=topic=14053.msg143865#msg143865 date=1138686043]
i = sqrt -1
i[sup]2[/sup] = -1

Now lets say you had something like f(x) = sqrt(x). The only things you can put into f(x) are positive numers and 0. This is called useful mathematics
[/quote]

Obviously you have no idea how useful imaginary numbers can be, especially in applied mathematics and Real Analysis.

Also, here's something to think about:

(-1)^(1/2) * (-1)^(1/2) = (-1*-1)^1/2 = 1^(1/2) = 1


As far as "imaginary" numbers not existing, and being hard to think about, they "exist" just as much as negative numbers do.  You just haven't thought about them that much.

Now prove or disprove that R^n can be related to R^(n-1) in the following fashion:
There exists x in R^(n-1) and 1-1 f(x) s.t. f(x)=r is not a member of R^(n-1) (and is not infinite!).
Furthermore f^(-1)(r)=x

Example:
Find a 1-1 f(x) that could generate an imaginary number relative to Z (call it j).

Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.

[quote author=rabbit link=topic=14053.msg145073#msg145073 date=1139360870]
Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.
[/quote]

Um.. ln(-(0!)) != -ln(0!).  ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi

...  + 2pi*k*i  ;)

[quote author=dxoigmn link=topic=14053.msg145085#msg145085 date=1139363774]
[quote author=rabbit link=topic=14053.msg145073#msg145073 date=1139360870]
Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.
[/quote]

Um.. ln(-(0!)) != -ln(0!).  ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi
[/quote]HAH!  I forgot logs don't like negatives...which explains why I confused myself...

[quote author=rabbit link=topic=14053.msg145207#msg145207 date=1139454333]
[quote author=dxoigmn link=topic=14053.msg145085#msg145085 date=1139363774]
[quote author=rabbit link=topic=14053.msg145073#msg145073 date=1139360870]
Hm.

e^(pi)i = -1
e^(pi)i = i^2
0! = 1 = -(i^2)
e^(pi)i = -(0!)
e^(pi)Sqrt(-(0!)) = -(0!)
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
(pi)Sqrt(-(0!)) = -(0)
(pi)i = 0

Somehow I find that wrong... o well.
[/quote]

Um.. ln(-(0!)) != -ln(0!).  ln(-(0!)) = ln(|-(0!)|) + i*arg(-(0!)) = ln(-(0!)) + i*arg(-(0!)) = i*pi
[/quote]HAH!  I forgot logs don't like negatives...which explains why I confused myself...
[/quote]

Don't like negatives? 

[quote author=rabbit link=topic=14053.msg145073#msg145073 date=1139360870]
ln(e^(pi)Sqrt(-(0!))) = ln(-(0!))
(pi)Sqrt(-(0!)) = -ln(0!)
[/quote]
This transformation is the wrong one.
1) ln(e^pi X) != pi * X, more like pi + ln(X).
2) ln(-X) != -ln(X), more like ln(X) + ln(-1).

And the correct value of ln(-1) is pi*i, of course.

[quote author=Rule link=topic=14053.msg145210#msg145210 date=1139456977]
Don't like negatives? 
[/quote]I've always been taught that ln(t) is undefined for negative values of t, is this wrong?

z = e^(w)

w = u + iv

z = e^(u)e^(iv)

x + iy = e^(u)e^(iv)

re^(itheta) = e^(u)e^(iv)

r = e^(u)  --->  u = log r
Also,  v = arg (z)

Therefore  log (x+iy) = log(z) =  log (r) + iarg(z) = log(|z|) + iarg(z)

Therefore log(-1) = log(1) + ipi + 2*pi*k*i, where k is any integer.

The principal branch of log (-1) = log(1) + ipi

Although my comment about the "principal branch" may seem trivial, "branch chasing" of multi-valued functions becomes a very important topic in analysis.

Although, we like to define logarithms unambiguously by imposing the restriction that 0 <= Im[log(z)] < 2*pi.

Edit: Im[], not arg() ;)

[quote author=Yoni link=topic=14053.msg145568#msg145568 date=1139677898]
Although, we like to define logarithms unambiguously by imposing the restriction that 0 <= Im[log(z)] < 2*pi.

Edit: Im[], not arg() ;)
[/quote]

I think the most often used branch of log(z) is the "Principal Branch,"
Log(z) = log(|z|) + iArg(z),  where Arg(z) (capital 'a') takes the argument of z
from the set (-pi, pi].  I believe this is also a standard in computer science.  One of the reasons we like the argument in (-pi, pi] is because it allows us to define 1-1 analytic inverse trigonometric functions, that take the values we're used to in real analysis. 

But there are problems with just using this branch of log(z); this principal branch has a branch cut along the negative real axis (it isn't analytic there).  If we want to find the derivative of a complex function at z = x < 0, we have to choose another branch of log. 

Also, for various applications, we may want a branch of a function like
(z^2-1)^1/2  analytic outside of the circle |z-i/2| = 1, etc.  An immediate application of this sort of thing is "Keyhole Contour Integration": if you want to find say "Integral(0, 1) x^(alpha-1) * (1-x)^(-alpha) dx"  0< alpha < 1, you'd start by finding a branch of z^(alpha-1)(z-1)^(-alpha) with a cut on the real axis from 0 to 1.

Here are some fairly well written articles on these topics:

Multivalued Functions
Complex Integration
Integration of Multi-valued Functions
Keyhole integration ("residue at infinity")

Not really quite as related, but you might find this interesting after reading the notes titled "Complex Integration":
Tricks for summing series