imaginary numbers

>>8829856
Consider the set of real linear polynomials in [math]x [/math] with addition and multiplication modulo [math](x^2 + 1) [/math]. This set along with those 2 operations are isomorphic (act like) 'complex numbers', and the 'imaginary numbers' are simply multiples of [math]x[/math].
Proof:
[math](a + bx) + (c + dx) = (a+c) + (b +d)x [/math] which acts like addition of two complex numbers [math]a + ib, c + id [/math]
Multiplication: [math](a + bx)(c + dx) = ac + (ad + bc)x + bdx^2 [/math] since multiplication acts modulo [math](x^2 + 1)[/math], we use division to reduce the quadratic to: [math](ac - bd) + (ad + bc)x [/math] which acts like the multiplication of complex numbers [math](a + ib)(c + id) [/math].

Not so imaginary now ay?

File: 1450158344230.png (661B, 473x454px) Image search: [Google]

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>>8829896
Damn, I wish I was smart enough to understand what you just said

>>8829896
OP smells underage, I perfectly understood this despite just having """college""" algebra knowledge desu.

Also imaginary numbers are the most interesting thing in non-university math, prove me wrong.

>>8829856
They literally aren't real

>>8829942
>OP smells underage, I perfectly understood this despite just having """college""" algebra knowledge desu.

That second poster isn't me, friendo!

>>8829945
I know, I'm talking to him because I don't think OP will understand it.

He gotta finish his high school math.

>>8829905
don't stroke anybody's ego on this board.

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Well imagine my shock

>>8829905
I had to make my post quick so I apologise if its a bit dense.
I assume you know what polynomials are, and what linear polynomials are. Consider just the collection of those linear polynomials, like: [math]1 + 2x, -4 + 2.5x [/math] and so on, polynomials just like this. Now, as you know, we can add these polynomials together, for example: [math](1 + 3x) + (5 - 2x) = (6 + 1x)[/math]. We can also multiply the polynomials, [math](1 + 3x)(5 - 2x) = 5 + 13x - 6x^2 [/math].

But, I'm not really interested in getting a quadratic again, I really just want to work with linear polynomials, since I understand them the best. So what I'll do is, every time I get a polynomial that has degree greater than 1 (a quadratic or above), I'll divide that polynomial by [math](x^2 + 1) [/math] (using long division or what have you), and the remainder that I get, that will be the result of multiplying [math](1 + 3x)(5 - 2x) = 5 + 13x - 6x^2 [/math].

Now, I want to divide [math] 5 + 13x - 6x^2 [/math] by [math](x^2 + 1)[/math], so we see that: [math]5 + 13x - 6x^2 = 5 + 13x - 6(x^2 + 1 - 1) = 5 + 13x - 6(x^2 + 1) + 6 = 11 + 13x - 6(x^2 + 1) [/math]. Clearly the remainder when we divide this by [math](x^2 + 1) [/math] is 11 + 13x.

So, we say that [math](1 + 3x)(5 - 2x) = 11 + 13x [/math]

Now, the point is that, addition and multiplication of the linear polynomials modulo [math](x^2 + 1)[/math] (which means exactly the process I outlined above), ACTS like the what people mean by complex numbers. That is:

[math](1 + 3i) + (5 - 2i) = (6 + i) [/math]
[math](1 + 3i)(5 - 2i) = 11 + 13i [/math].

As you can see there is therefore a correspondence between this collection of linear polynomials with the operations I gave it, and the complex numbers.

That means that these complex numbers can be constructed out of what we already have with real numbers. It is not simply a matter of, "let there be a square root of -1, just because!".

>>8829943
Technically nothing in math is real. Complex numbers are a tool to achieve results to operations where you would normally find a dead end. The results are plenty real though, quantum mechanics and a few other areas of physics wouldn't function without them, or at least it would be way more difficult.

>>8830038
You're completely misunderstanding his post, friend.

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>>8829994
WHUELL IMAHGUN MOY SHOWK

[math]
\\ z_1 = x_1+y_1i \; \; \; \; \; z_2 = x_2+y_2i
\\ z_1^* = x_1-y_1i \; \; \; \; \; z_2^* = x_2-y_2i
\\| z_1 | = \sqrt{x_1^2+y_1^2} \; \; \; \; \; | z_2 | = \sqrt{x_2^2+y_2^2}
\\ z_1+z_2 = x_1+x_2 + (y_1+y_2)i
\\ \left | z_1+z_2 \right |^2 = \left ( \sqrt{(x_1+x_2)^2+(y_1+y_2)^2} \right ) ^2 = (x_1+x_2)^2 +(y_1+y_2)^2
\\ z_1z_2^* = x_1x_2 -x_1y_2i +y_1ix_2 -y_1iy_2i = x_1x_2+y_1y_2 +(x_2y_1-x_1y_2)i
\\ z_1^*z_2 = x_1x_2 +x_1y_2i -y_1ix_2 -y_1iy_2i = x_1x_2+y_1y_2 +(x_1y_2-x_2y_1)i
\\ z_1z_2^* + z_1^*z_2 = 2(x_1x_2+y_1y_2) = \text{2Re}(z_1z_2^*) = \text{2Re}(z_1^*z_2)
\\ |z_1|^2+|z_2|^2 + z_1z_2^* + z_1^*z_2 = x_1^2+y_1^2 + x_2^2 + y_2^2 + 2(x_1x_2+y_1y_2)
\\ = (x_1^2 + 2x_1x_2 + x_2^2) + (y_1^2 + 2y_1y_2 + y_2^2) = (x_1+x_2)^2 +(y_1+y_2)^2
\\
[/math]