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I have the following :

 w[n_] := E^((2*I*\[Pi])/n)
 Phi[n_, k_, x_] := (1/n)*Sum[(w[n]^(-j*k)*E^(x*w[n]^j)), {j, 0, n - 1}]
 NEuler[n_, x_] := (1/Phi[n, 0, x])
 NEulerPoly[n_, x_, z_] := (E^(x*(z - 1/n)))/Phi[n, 0, x/n]
 NEPoly[n_, z_, r_, M_] := Expand[FullSimplify[Coefficient[r!Normal[Series[NEulerPoly[n, x, z], {x, 0, M}]], x,  r]]]

The polynomials in z are indexed by r and form a sequence $\{z\}_{r\in\mathbb{N}}$. I would like to look at the roots of these complex polynomials by extracting the real and imaginary parts and plotting them as an ordered pair, or by some other method. HOwever, I have no idea how to do this. I don't know mathematica very good and am trying to learn the commands and functions that can handle this type of procedure.

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  • $\begingroup$ Plot[ReIm@(your function),(your domain)]? $\endgroup$
    – Feyre
    Jan 16, 2017 at 16:33
  • $\begingroup$ That seems to plot the actual polynomial, not the complex roots of the polynomial... $\endgroup$ Jan 16, 2017 at 16:36
  • $\begingroup$ You can find appropriate ideas here: Finding real roots of negative numbers $\endgroup$
    – Artes
    Jan 16, 2017 at 16:39
  • $\begingroup$ When I try to evaluate NEPoly, I get all zeroes: NEPoly[n, z, r, #] & /@ Range[20] $\endgroup$
    – bill s
    Jan 16, 2017 at 16:44
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    $\begingroup$ Here's a start: ListPlot[{Re[z], Im[z]} /. Table[Solve[0 == NEPoly[2, z, r, 20], z], {r, 1, 10}] // N]. $\endgroup$
    – march
    Jan 16, 2017 at 16:51

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