Posers and Puzzles

Posers and Puzzles

  1. Joined
    24 Sep '06
    Moves
    3736
    07 Oct '08 06:07
    Please prove that every complex polynomial function f(x) of degree n > or = 1 has at least one complex zero.

    p.s Kudos to the cat that knows this without internet help.
  2. Subscriberfiftyonehz
    Member 009
    Joined
    24 Jun '05
    Moves
    57223
    07 Oct '08 18:00
    Suppose all zeros are real...

    f(x) = sum_{i=1}^n (x-z_i), where z_i is in R

    then all coefficients are real
  3. Joined
    07 Sep '05
    Moves
    35068
    07 Oct '08 21:08
    Originally posted by fiftyonehz
    Suppose all zeros are real...

    f(x) = sum_{i=1}^n (x-z_i), where z_i is in R

    then all coefficients are real
    I don't think that's what he's getting at - real numbers are complex.
  4. Joined
    15 Feb '07
    Moves
    667
    07 Oct '08 22:13
    Originally posted by mtthw
    I don't think that's what he's getting at - real numbers are complex.
    Complex in this context means a number with a non-zero imaginary component, methinks.

    I don't know if that includes numbers that are purely imaginary or not.

    For the uninitiated, in its broadest sense, a complex number is any number of the form a + b*i, where a and b are any real number, and i is the square root of -1, a number that does not exist within the realm of real numbers.

    a is call the real part and b*i the imaginary part.
  5. Joined
    24 Sep '06
    Moves
    3736
    07 Oct '08 23:11
    Yes, geepamoogle said what I meant when I meant what I said.
Back to Top