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Posers and Puzzles

Posers and Puzzles

  1. 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. Subscriber fiftyonehz
    Member 009
    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. 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. 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. 07 Oct '08 23:11
    Yes, geepamoogle said what I meant when I meant what I said.