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Complex numbers: reinventing the wheel?

  1. Dec 28, 2013 #1

    bobie

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    I am studying complex numbers and, hard as I try, I cannot see great difference between them and the conjugate numbers known and used since 500 B.C. (http://en.wikipedia.org/wiki/Quadratic_formula#Historical_development) to solve a quadratic equation
    [tex]p/2 \pm \sqrt(p/2 ^2\pm q) [/tex]
    where the sum of the conjugates gives p and the multiplication q.

    It seems to me that they just factorized these "conjugate, triangular or pythagorean" numbers (doing separate operations) and used them to find the root in
    [itex]x^3 -px = q[/itex] when p is < 0,
    setting √p/3 as the hypothenuse, x/2 and √(p/3-x/2 ^2) as legs of the triangle and parts of the conjugate numbers
    u , v = [tex]x/2\pm \sqrt{p/3-x/2 ^2}[/tex]
    the (separate) sum of the conjugates u (x/2+x/2) + v (+√ -√=0) gives x and the (cross,separate) multiplication gives (x/2 ^2 + √.. ^2) the squared hypothenuse p/3, satisfying cardano formula u * v = p/3:
    (x/2 + √...) * (x/2 - √...) = (x/2 ^2 + p/3 -x/2 ^2)

    Why call them complex or imaginary numbers as the negative square root is taken as an absolute value? can someone , please, point out the differences apart from the cross multiplication?

    Thanks
     
    Last edited: Dec 28, 2013
  2. jcsd
  3. Dec 28, 2013 #2

    stevendaryl

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    I'm not exactly sure what you're claiming. In ordinary real number arithmetic, the equation

    [itex]x^2 + 1 = 0[/itex]

    has no solution. If you try to come up with two numbers [itex]\alpha_+[/itex] and [itex]\alpha_-[/itex] such that

    [itex]\alpha_+ + \alpha_- = 0[/itex]
    [itex]\alpha_+ \times \alpha_- = -1[/itex]

    there are no such numbers. Complex analysis just adds a new number, [itex]i[/itex] to the mix, and declares:

    [itex]\alpha_+ = +i[/itex]
    [itex]\alpha_- = -i[/itex]

    is the solution.
     
  4. Dec 28, 2013 #3

    mathman

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    Minor error: [itex]\alpha_+ \times \alpha_- = 1[/itex]
     
  5. Dec 28, 2013 #4

    stevendaryl

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    Right.
     
  6. Dec 29, 2013 #5

    bobie

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    Your example has the same problem as mine, please consider x=6 and p=30:
    [tex]x^3-px=q → (2a)^3 -3*2a*b= 2c → (2*3)^3 - 3*(2*3)*10 = 2*18 [/tex]
    if you try to apply the regular cardano formula: [itex]c\pm \sqrt{c^2-b^3} →18 \pm \sqrt{18^2-10^3}[/itex]
    this equation apparently has no solution, exactly like yours, since 234-1000 =-676: there is no such squareroot and, furthermore, there are no two real numbers u, v whose product is p/3 (= b =10) but it has a solution with a different approach: the square and cube of the hypotenuse of a right triangle .
    Cardano et alii changed the formula: [itex]c \pm \sqrt{b^3 - c^2} → 1000-324 = 676 → \sqrt{676} = 26[/itex]
    then set 26 (√b^3-c^2) and 18 (= c) as legs of a right triangle with hypotenuse √b^3 and used that to find the legs (a and [itex]\sqrt{b-a^2}[/itex]) of the triangle with hypotenuse √b and λ'= λ/3 , getting direcly x with a new formula. The procedure of complex numbers is a sort of justification, derivation and can be skipped altogether.

    Did Cardano discover that property of the right triangle or he just applied it without realizing it?
     
    Last edited: Dec 29, 2013
  7. Dec 29, 2013 #6

    Nugatory

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    Well, we have to call them something, and "useful and fascinating abstraction formed from ordered pairs of real numbers and obeying the following definitions of addition and multiplication ..." would be a bit unwieldy.

    As for why we generally prefer the ##a+bi## formalism to the conjugate formalism.... Try expressing something like Schrodinger's equation in the conjugate formalism.
     
  8. Dec 29, 2013 #7

    bobie

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    Conjugate numbers is probably more appropriate, but surely not imaginary and no negative square root, you seem to agree. Just a technique to square the hypotenuse considering the legs a,b of a triangle as conjugate numbers u,v:
    a+b, a-b, so that u*v = the hypotenuse (a^2+b^2) c, and u*u = the legs of the squared triangle (a^2-b^2 ; 2ab). You surely can express Schroedinger and all without i, why not?

    Do you happen to know if it was Cardano who discovered that? He does not seem to realize what he is actually doing. Have you ever heard outside the scope of complex numbers that you in order to double the angle of a right triangle you must set [itex] b'= \sqrt{b^2-a^2}, a' = 2ab[/itex] and doing so you get [itex]c' = c^2[/itex]? Is there any trace of that in Euclid?

    Thanks
     
    Last edited: Dec 29, 2013
  9. Dec 29, 2013 #8

    stevendaryl

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    I still don't know what you are claiming. In the case of [itex]x^2 + 1 = 0[/itex], it's not that we haven't figured out a sufficiently clever way to solve it. It doesn't have a real-number solution. There is nothing "apparently" about it.
     
  10. Dec 29, 2013 #9

    bobie

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    I do not see your point, steven, not all equations have solutions.

    I have showed that in the case of the cubic equations we do not have imaginary numbers and you can get the root with a simple formula that does not even need regular conjugate numbers.
    and, if you want to use conjugates, i is useless anyway: (b=10, c=18, a=3, h (hypothenuse)=√10 ,√10-9 = 1) :

    (3+1) + (3-1)= (3+3) + (1-1 =0) = 6 = x
    (3+1) * (3-1) = (3*3 - 1*-1= ) 10 + (3*1+3*-1 = 0) = b
    (3+1)2= (3+1)*(3+1) = (3*3 -1*1=) 8 + (3*1+3*1=) 6; h = 8 2+62= 100 = b2
    (3+1)3= (3+1)*(3+1)*(3+1) = 18, 26; h = 182+262=1000= b3; √1000-324 = 26

    x =√40*.9487 = 6 independently of all this, anyway

    To solve the problem that produced imaginary numbers do we need i? is it a grounded claim?
     
    Last edited: Dec 29, 2013
  11. Dec 29, 2013 #10

    stevendaryl

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    But once you introduce [itex]i[/itex], the equation [itex]x^2 + 1 = 0[/itex] does.

    More generally, once imaginary numbers have been introduced, every polynomial equation has a solution. More useful for some purposes is this: Every polynomial equation can be factored. Given an equation of the form [itex]A_n x^n + A_{n-1} x^{n-1} + ... + A_0[/itex], we can factor it as:

    [itex]A (x - x_1) (x - x_2) ... (x - x_n)[/itex]

    for some complex constants [itex]A, x_1, x_2, ..., x_n[/itex]

    To solve the equation [itex]x^2 + 1 = 0[/itex], you certainly have to introduce imaginary numbers.
     
  12. Dec 29, 2013 #11

    Nugatory

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    I didn't say it couldn't be done, I said that if you try it you'll see why it isn't done.
     
  13. Dec 29, 2013 #12

    stevendaryl

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    Why would you want to?
     
  14. Jan 5, 2014 #13
    If you are referring to 500 BC, the existence of negative numbers was not understood at that time. A subtraction problem giving a negative number would have been considered absurd. There would have been no concern about roots of negative numbers because there were no negative numbers.

    I don't understand what you mean by "the negative square root is taken as an absolute value".
     
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