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Roots of f(x)

  1. Oct 15, 2005 #1
    let be the function f(x) so we have that if x is a root also x* is a root, but we have that x is NEVER a pure imaginari number,i mean x is always different from x=ia the my question is if this means that all the roots will be real,the only counterexample i find is:

    [tex]f(x)=(x-x_{0})(x-x_{1})(x-x_{2})........[/tex]

    that is an infinite polynomial that has all its roots in the form a is a root and also a*.
     
  2. jcsd
  3. Oct 15, 2005 #2

    Hurkyl

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    (That's an infinite product, not a polynomial)

    It is easy to construct counterexamples to the claim:

    If f is a function such that
    (1) whenever a is a root of f, then so is a*
    (2) no root of f is purely imaginary
    then the roots of f are real.

    The simplest ideas work. We want f to have a non-imaginary non-real root. Let's pick (1+i).
    Then, f must also have (1-i) as root.
    So we need a function f that is zero at (1+i) and (1-i), and nonzero elsewhere.

    That's very easy to do: let f be the function that is zero at (1+i) and (1-i) and 1 elsewere.
     
    Last edited: Oct 15, 2005
  4. Oct 15, 2005 #3

    shmoe

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    Since I have a guess where you're trying to apply this, let's take the Riemann Zeta function as a counterexample. The roots come in conjugates by the reflection principle yet none of the roots are purely imaginary.

    If you want something with no poles and not a polynomial, take f(x)=(x-1-i)*(x-1+i)*e^x.
     
  5. Oct 15, 2005 #4

    HallsofIvy

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    Am I the only one who gets annoyed at the confusion of "roots" and "zeroes"?

    My understanding is that equations have roots: 2 and 3 are roots of the equation x2- 5x+ 6= 0.
    But that functions have zeroes: 2 and 3 are the zeroes of the function
    x2- 5x+ 6.
     
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