identifying a point in the complex plane why is it x+iy and

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>>9059032
Because their addition is necessary for their arithmetic. Multiply a complex by another complex, and the addition comes into play. Actually, anything that isn't multiplication/division by a pure real or addition/subtraction will make that addition sign important.

Sage for stupid question. Next time, take it to >>>/sci/sqt

>What is vector addition?

>>9059041

> Because their addition is necessary for their arithmetic

that's a shit explanation. you can say that addition is just as necessary for arithmetic with integers or real numbers, yet the cartesian coordinates are pairs and not sums: (x,y) not (x+y)

>>9059057
if you write e1 = (1,0) and e2 = (0,1) then your (x,y) is x*e1 + y*e2. that's the point. the usual real basis over C is {1, i} which correspond to (1,0) and (0,1)

>>9059032
They want to pretend it's a number not a vector, so that a complex number exponential looks like a real number exponent.

It is just convention - When defining the complex plane via Cartesian Product of the Real Numbers with + and * operations most books use the notation (x,y) for the complex numbers, and after they stablished the properties of + and * they start using the notation x + iy for convention and facility...

Specially after you have the fact that:
(x,y) = x(1,0) + y(0,1) and (1,0) is equivalent to the real number 1 and (0,1) satisfies (0,1)*(0,1) = -1, this (0,1) is 'alike' to sqrt(-1) = i.

Thus we represent (x,y) = x + iy, but using (x,y) is not wrong, just remember the right way to do the product and you'll be fine.

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>>9059057
Multiply (2, 3) by (4, 2) without using that addition sign.

>>9059086
There is a more pernicious reason why.

There actually is no such thing as "square root of -1", because complex numbers are a whole different field than real numbers, and in this context "-1" is actually exp (pi*i).

you can do it the other way too.

Have you ever done physics with unit vectors? Same principle. The hatted vectors get carried like an algebraic quantity. Just a reminder for which component that quantity represents.

>>9059100
Yeah, yeah, there are all that branch stuff, that is why I used the world "alike" and not "is"

i2 = -1, as if it were sqrt(-1), I was careful not to state that sqrt(-1) = i, which is only well defined after without the sqrt domain well stablished

>>9059094
>
hit the nail on the head

>>9059032
it's just a notation. complex numbers (=elements of the complex plane) ARE just pairs of real numbers. you make the following convention: complex numbers in the form of (x,0) will be denoted just by x and complex numbers in the form of (0,y) will be denoted iy. and since (x,y) = (x,0) + (0,y), you obtain (x,y) = x + iy. so it's just a different notation for the usual cartesian plane. the reason for this is that it makes multiplication look more natural.

>>9059131
You are so wrong. Complex numbers are INHERENTLY the addition of a real and imaginary number.

You can't say that they are just unconnected pairs of numbers. Case and point: take i squared. This is just (0,1) by (0,1), but it equals (-1,0). You CANNOT get this result with vectors/pairs-of-numbers without abandoning symmetry.

When you interpret it correctly as addition, and you multiply, any "i squares" will get directly subtracted from the real part, as it should.

>>9059032
>what are versors

>>9059639
>You CANNOT get this result with vectors/pairs-of-numbers without abandoning symmetry.
Of course you can. It's all about how you define multiplication, dummy. Example:

[math] (a,b)(c,d) = (ac - bd, ad + bc) [/math]

>>9059094
See >>9059659

Fucking brainlets...

>>9059100
>I have never taken a course in abstract algebra
It shows.

>>9059100
>because complex numbers are a whole different field than real numbers

Complex numbers are actually a commutative unitary algebra over the real numbers.

Anyways, you should know that in actual mathematics people no longer define i as the square root of -1. The reason is that this is problematic. You are doing an operation on a type of number (real numbers) which is invalid, and then you are claiming this invalid construction is now a new object.

It is good intuition, but bad formulation. Complex numbers are then first constructed from [math] \mathbb{R}^2 [/math] with vector addition and then embedded with the multiplication in >>9059659.

Now this construction is rigorous and then indeed this becomes an algebra after you have proven the necessary conditions.

>>9059659
Now I want you to use your big boy brain and tell me if that's symmetrical.

Hint: plug in (1,0) * (1,0), you get (1,0)

Plug in (0,1) * (0,1) and you will not get (0,1). You literally cannot define multiplication without abandoning symmetry. You can only regain symmetry when you treat the number as an addition of two other numbers.

(bi + a)(bi + a) = (a + bi)(a + bi)

When complex numbers were invented, people did not immediately see the geometric interpretation of adding an orthogonal dimension to the real number line. Back then the application of complex numbers was purely algebraic, so the notation made sense.

>>9059867
>(bi + a)(bi + a) = (a + bi)(a + bi)

(a,b)(a,b) = [(a,0) + (0,b)] [(a,0) + (0,b)] = [(0,b) + (a,0)] [(0,b) + (a,0)]

Why would you think you would lose commutativity? It literally makes no fucking sense to think that.

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>>9059659
>>9059669
That definition follows directly from FOILing out the addition notation. There is no redefinition of multiplication involved there, and using comma notation just obfuscates the fact that it's still using the rules of addition notation in order to get a result.

>(a+bi)(c+di) = ac + adi + bci + bdi2 = (ac - bd) + (ad + bc)i

>>9060117
Look, people only use the vector notation to formalize complex numbers. To turn them into rigorous math.

When doing computation people still use a+bi because that is the intuition. But to make it rigorous that "i" has to come from somewhere and just saying [math] i = \sqrt{-1} [/math] is not rigorous.

>>9059032
ITT: People who don't know what a basis is

>>9059032
There is a categorical justification for this. The pair notation is usually reserved for the direct product [math]\mathbb{R}\oplus\mathbb{R}[/math].
In the category of rings for example, this is explicitly not the same as the complex numbers because the product structure differs.
The underlying set is the same, but writing the elements of both rings with the same notation would be confusing.

I wrote >>9059639

>>9059639
>>9059867
ARE YOU FUCKING RETARDED?
>You literally cannot define multiplication without abandoning symmetry
what the fuck does that mean? [math]\mathbb{C}[/math] = the vector space [math]\mathbb{R}^2[/math] endowed with an extra operation. this operation is DEFINED by the formula [math](a,b)(c,d) := (ac-bd,ad+bc)[/math]. it just so happens that with this operation, [math]\mathbb{C}[/math] become a field and an algebra over the reals, so we call this operation "multiplcation". this is the standard definition, because it's the most elementary, yet rigorous. yes, there are other ways how to define [math]\mathbb{C}[/math] (muh field extension) but they all yield cannonically isomorphic results. and now, please tell me where exactly am I "so wrong".

>>9060360

I wrote >>9059131 my bad

>>9059896
this