Question

What are imaginary numbers? Well, wonder no more. This is the most solid explanation I could find online, and it's really good.

http://math.stackexchange.com/questions/199676/what-are-imaginary-numbers

Answer 1

yooo you're in luck.. I'm taking a course in complex variables.

Answer 2

uhmm I don't have a scanner at home, but msg me your email, I'll scan my notes tomorrow and email them to you.

Answer 3

What are imaginary numbers?

Not real......

Answer 4

basically an imaginary number's an ordered pair

Answer 5

its when you try to take the square root of a negative number

Answer 6

There is only one imaginary number, i, defined by i^2 = -1

Answer 7

not really... a+bi are all imaginary.. a is the real part, b is the imaginary part.

Answer 8

b is real

Answer 9

you guys are confusing this student, lol

Answer 10

Stir,stir......

Answer 11

if a=0, the number is called pure imaginary.. Idk otherwise my professor was lying to me. No we're not, everyone knows that i is imaginary.

Answer 12

i is appended to the real number system by fiat......

Answer 13

lol

Answer 14

which is defined by:
(a,b)(c,d) = (ac-bd, ad+bc)

Answer 15

this conversation is really good for this student, keep it up guys

Answer 16

i meant i is an ordered pair, (0,1)

Answer 17

Then you make up an imaginary plane to go with this imagijnary number.....

Answer 18

The link I provided in the OP is a solid construction of how we (us normal humans, and bahrom7893) conjectured the Platonic existence of imaginary numbers. For further reading, check out "Mathematics: Its Content, Methods and Meaning" by A.D. Aleksandrov, et al.

Answer 19

Then the student says "Does it work in 3D?"

Answer 20

lol,

Answer 21

i^2 = i*i = (0,1)(0,1) = (0*0-1*1,0*1+1*0)=(-1,0)... that's where i^2 = -1 came from.

Answer 22

I was like WOOOOOOOOOOOOOOWW

Answer 23

The real wow comes from http://en.wikipedia.org/wiki/Euler%27s_identity .

Answer 24

yea that too.. i need to read its proof

Answer 25

it seems the questions stater already knew much info but wanted to see how we explain complex numbers, lol

Answer 26

the sqrt of -1 came from messing about with negative logs....

Answer 27

it seems that you don't know badrefs lol

Answer 28

never heard of that one estudier..

Answer 29

That's because I'm on very intermittently. I can't get to know everyone around here.

Answer 30

Yeah, logarithms provided the construction of imaginaries. Let me pull one up.

Answer 31

my professor lied to me.... :/

Answer 32

Not necessarily. I learned something else in math. In physics we constructed Platonic imaginaries through logs.

Answer 33

Because in physics we were more concerned with the "existence" of things.

Answer 34

@bahrom7893 - that's cos complex analysis is for pure mathematicians (they have to justify their existence, y'see:-)

Answer 35

That explains it. I can't find the reference right now, sorry. :<

Answer 36

Let me see if I can dig it up.....

Answer 37

Save us @estudier !

Answer 38

This might be it

http://www.maa.org/editorial/euler/How%20Euler%20Did%20It%2046%20e%20pi%20and%20i.pdf

Answer 39

"...They were perplexed because they had equally convincing (and flawed) arguments to "prove" that ln(-x) = ln(x)..."

Answer 40

"They" being Bernoulli and Euler

Answer 41

Anyway, the point is ln(-1) = pi*i

Answer 42

I'll try to post something as to how you can get a quantity that evaluates to -1 without all the hoopla.....

Answer 43

A really amusing question from Math Underflow http://math.stackexchange.com/questions/202172/why-is-i-0-498015668-0-154949828i

What does \(i!\) evaluate to?

Answer 44

Take a pair of vectors uv with the normal rules for multiplication etc and so write

uv = 1/2(uv+vu) + 1/2(uv-vu)

So that first term is basically u dot v and we'll call the second one u (wedge) v.

uv = u.v + u wedge v

vu = u.v - u wedge v

Multiply these two

uvvu = (u.v9^2 -(u wedge v)^2

Since vv = |v|^2 -> (u wedge v)^2 =-|u|^2|v|^2sin^2 theta

So whatever u wedge v is, it's square is a negative scalar