complex numbers

Discussion in 'Math' started by VVS, Oct 15, 2007.

  1. VVS

    Thread Starter Active Member

    Jul 22, 2007
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    Hey I am studying triple E in Imperial College and we are doing complex numbers.
    I dint understand A SINGLE THING :confused: in the latest maths lecture. he started with theroem 1: I am just going to type some lines so that someone of you might recognize this as something and can tell me the name of the theory coz our lecturer just called it "theorem 1" (wth!!!!!!:mad:) I searched the internet for 2 hours and couldnt find anything; anyway here goes:


    if p(z) = a0 + a1 z + a2 z + .........
    where a0,1,2,3,n are real numbers and z may be complex then this polynomial has a solution w, which is complex such that p(w)=0.
    Moreover, if the root w has non zero imaginary part and y is the conjugate of w then p(y)=0.
    If p(w)=0
    then we can write p(z)=(z-w)q(z)

    WHY??????????????? :confused: our lecturer just assumed that!!! anyway then he wrote:

    where q is another polynomial. Moreover the degree of p is n and the degree of q is n-1 Now we can apply Theorem1 to q and find u so that q(u)=0 and so q(z)=(z-u) r(z)
    Hence
    p(z)=(z-w) (z-u) r(z)
    and continuing in the same amnner we can completely factorize
    p(z) in products of linear factors.

    thank you very much for helping in advance. :)
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
    6,960
    144
    Without seeing the whole proof, I would assume from both university experience and the way your question is structured, that q(z) was introduced as one of those 'fudges' to make the subsequent mathematics easier to work with. q(z) can be an arbitrary unknown provided z, w and p(z) are defined. The complex function variable z minus the solution w must have some significance/solution later in the exercise. Does it?

    Dave
     
  3. omnispace

    Member

    Jul 25, 2007
    27
    0
    I think that's just the definition of factoring. It might be easier to see with real numbers. Say p(z) = z^3 - z^2 - 4z + 4. Then we can factor out a (z - 1) and be left with (z^2 - 4). So in this example, w = 1 and q(z) = z^2 - 4.

    Then the factoring process can be continued, factor out (z + 2) so u = -2 and you're left with r(z) = z - 2. Now p(z) has been completely factored. Notice how p is 3rd order, q is 2nd order, and r is 1st order.

    This all applies to imaginary and complex numbers as well. The one thing to keep in mind is that imaginary/complex roots have conjugates, and these are also roots of the equation.
     
  4. Papabravo

    Expert

    Feb 24, 2006
    10,137
    1,786
    Why?
    If a polynomial has real coefficients, all roots with imaginary parts always occur in conjugate pairs. It is the only way that complex factors can produce a real result. If you still don't see it, I suppose there is always time to switch your major to Art History.
     
  5. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    I'm in big trouble. (Or I would be if I were still a student.) I also do not understand why the instructor made this claim. My wife studied Art History and I know darn well I'd never succeed in that subject - too many subtleties and philosophical arguments!

    How do we get p(z)=(z-w)q(z) from p(w)=0?
     
  6. Papabravo

    Expert

    Feb 24, 2006
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    Art History is just a metaphor for something as far away from mathematics and engineering as possible. I could have said French Literature, or Sociology, or Basket Weaving. I thought the "why" was about having a solitary root with an imaginary part.
     
  7. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    I understand now. All of those (except possibly weaving) are subjects I would have trouble with. :rolleyes:

    Hopefully someone knows what the missing steps in this proof might be.
     
  8. Papabravo

    Expert

    Feb 24, 2006
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    I must confess that I am really out of luck when it comes to theorem proving. In a career spanning 38 years I've never had to prove a theorem, but I do have an intuitive sense of things that are right and why. Except for the fact that polynomials with real coefficients require any complex roots to occur in conjugate pairs I'm not going to be much help with the steps of the proof.
     
  9. recca02

    Senior Member

    Apr 2, 2007
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    i concur with Mr omnispace's explanation.
    though i do not know what theorem one is it probably is from factorization.
     
  10. Papabravo

    Expert

    Feb 24, 2006
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    Real roots, when factored out, leave a remainder polynomial with an order that is one less than the original. Complex conjugate roots, when factored out, leave a remainder polynomial whose order has been reduced by two.
     
  11. mogadeet

    Member

    May 1, 2007
    19
    0


    hey vvc

    it sounds like this is what you're supposed to be taking home:

    1) theorem (1): every polynomial has a root in the complex numbers (ie: p(w)=0 for some complex number w). see http://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra

    2) if w = a+b*j let w'= a-b*j be its complex conjugate. then the following can be verified straightforwardly:

    *) (w')+(y') = (w+y)'
    *) (w')*(y') = (w*y)' [in particular (w')^i=(w^i)' for all i]
    *) w' = w for real w [ie: if b=0]

    hence if p(w) = \sum a w^i = 0 then p(w') = \sum a (w')^i = \sum a (w^i)' = \sum (a)' (w^i)' = \sum (a w^i)' = (\sum a w^i)' = 0' = 0

    3) since z^(i+1)-w^(i+1) = z(z^i-w^i)+w^i(z-w), each (z^i-w^i) is divisible by (z-w). hence p(z)-p(w) = \sum a z^i - \sum a w^i = \sum a (z^i-w^i) is divisible by (z-w). therefore if p(w)=0 then (z-w) divides p(z), which is to say that p(z)=(z-w)q(z) for some polynomial q. the degree of q is one smaller than that of p because the highest power of z in q, multiplied by the "z" of (z-w), gives the highest power of z in p.

    hope that helps

    peace
    stm
     
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