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Roshe theorem questiom.

  1. Jul 14, 2010 #1
    find the number of roots on the right side of the field
    for
    [tex]f(z)=z^6+7z^4-3z^2+z+1[/tex]

    ??

    usually i have stuff like 1<|z|<4

    but here instead they say "on the right side of the field"
    meaning on the right frim the imaginary axes

    i cant interpret it into a circles like i am use to
     
  2. jcsd
  3. Jul 14, 2010 #2
    First, just plot the roots using Mathematica:

    Code (Text):
    myeqn = z^6 + 7 z^4 - 3 z^2 + z - 1;
    myzeros = N[z /. Solve[myeqn == 0, z]]
    mypoints = Point @@ {{Re[#], Im[#]} & /@ myzeros}
    Show[{Graphics[{{Red, PointSize[0.01], mypoints},
        Circle[{0, 0}, 1]}]}, PlotRange -> {{-5, 5}, {-5, 5}},
     Axes -> True]
     
    Not doing that to act smart. I think it's crucial to learn Mathematica and use it in your math education. So we know where they are. Now prove it using the Argument Principle:

    First, what would happen if I just plug in the value [itex]ai[/itex] for real a into that expression. Will it ever be zero? Next consider the change in argument of

    [tex]f(z)=z^6+7z^4-3z^2+z-1=z^6\left(1+7/z^2-3/z^4+1/z^5-1/z^6\right)[/tex]

    along the closed half-circle contour in the right half-plane and allow R to go to infinity. Where do you start with such a thing? Well, [itex]\arg(zw)=\arg(z)+\arg(w)[/itex] so that as R goes to infinity, how will the argument change for [itex]z^6[/itex] along the curved part of the half-circle from [itex]-\pi/2[/itex], around to [itex]\pi/2[/itex]? You can do that. And as R goes to infinity, how will the argument change for that other expression as you go around the half-circle? Finally, how will the argument change from the point say [itex]f(a i)[/itex] down the imaginary axis, to the point [itex]f(-ai)[/itex] as a goes to infinity? Now, does your work with the Argument Principle agree with the plot above?
     
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