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VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: #1procrastinator on March 05, 2012, 04:40:58 pm

Title: Formal definition of complex numbers
Post by: #1procrastinator on March 05, 2012, 04:40:58 pm: I was reading a book on complex numbers where they derived i^2 = -1 from defining the complex numbers as ordered pairs and the addition and multiplication operations as:

(a,b) + (c,d) = (a+c, b+d)
(a,b) * (c,d) = (ac-bd, bc+ad)

The addition one is somewhat intuitive, but how did they choose the multiplication one? Did they already know i^2 = -1? Which came first? The book is 'Complex Numbers A-Z' by the way...
Title: Re: Formal definition of complex numbers
Post by: Lasercookie on March 05, 2012, 04:51:39 pm: I'm not really too sure, but I drew the points on a graph and I'm guessing that the vectors here are from the origin to that point.
i.e. (a,b) is a vector starting at Origin and ending at (a,b)

$(a,b) * (c,d)$
So, that's the same as: $(a + bi)(c + di)$ ?

so then expanding out and rearranging into x + yi form: $(ac - bd) + (bc + ad)i$ . Which you can convert to that other notation: $(ac-bd, bc+ad)$
Title: Re: Formal definition of complex numbers
Post by: TrueTears on March 05, 2012, 04:53:30 pm: Ahhh complex numbers A-Z by titu and andrica right? that's a really good book, read it quite a while ago, it's just an application of the definition, we define i^2=-1 and then (a,b) is just notation for a+bi, and the result follows from laseredd's working.
Title: Re: Formal definition of complex numbers
Post by: kamil9876 on March 05, 2012, 07:38:21 pm: Quote
we define i^2=-1 and then (a,b) is just notation for a+bi, and the result follows from laseredd's working.

Nope judging from what the OP said, they are doing it the other way around.
Quote from: #1procrastinator on March 05, 2012, 04:40:58 pm
I was reading a book on complex numbers where they derived i^2 = -1 from defining the complex numbers as ordered pairs and the addition and multiplication operations as:

(a,b) + (c,d) = (a+c, b+d)
(a,b) * (c,d) = (ac-bd, bc+ad)

In that case most likely they define $i=(0,1)$ and $-1$ as $(-1,0)$ (in fact any real number $x$ would be identifying it with $(x,0)$ ). So it's all about pluging $a=c=0$ and $b=d=1$ into the second property any hoping you get $(-1,0)$ .

Of course the motivation for the definition comes from the natural way you learn in high school, as the second post shows. There are other approaches to defining the complex numbers that get the best of both worlds: rigorous and intuitive (using Quotient Rings)

as an exercise you may as well show that (a,0)*(b,0)=(ab,0) which shows that the x-axis does indeed behave like the real numbers (i.e in fancy language they are 'isomorphic')
Title: Re: Formal definition of complex numbers
Post by: #1procrastinator on March 09, 2012, 01:37:49 pm: Yeah they introduced the addition and multiplication of ordered pairs first then defined i = (0,1) . So then the way you multiply ordered pairs is because of i = sqrt(-1)?

EDIT:

some noob questions

Consider the set R x {0}, together with the addition and multiplication operations defined on R^2. The function

f: R -> R x {0}, f(x) = (x, 0)

is bijective and moreover,

(x, 0) + (y, 0)

The reader will not fail to notice that the algebraic operations defined on R x {0} are similar to the operations on R; therefore we can identify the ordered pair (x, 0) with the number ex for all x E R. Hence we can use, by the above bijection f, the notation (x, 0) = x

What are the algebraic operations on R? Algebra rules?
Title: Re: Formal definition of complex numbers
Post by: kamil9876 on March 09, 2012, 09:10:34 pm: I guess that exercise is touching on the point I made earlier. That the x-axis, i.e the subset $\{(x,0)| x\in \mathbb{R} \}$ can be shown to be "isomorphic" to the real numbers with their usual operations, i.e addition and multiplication are done in the usual way e.g $(7,0)+(6,0)=(13,0)$ and $(7,0)*(6,0)=(42,0)$ .
Title: Re: Formal definition of complex numbers
Post by: #1procrastinator on March 14, 2012, 06:05:26 pm: ^ thanks