Obtain only positive complex solutions

Question

By setting the determinant of a matrix to 0, I obtain 6 complex solutions for w and plot them. However, You can clearly see that the negative solutions mirror the positive solutions as the solution corresponds to the positive and negative roots.

k = 9; l = 12; m = 2; M = 4 ;
mat = {{m*w^2 - 2*k, k, k*Exp[-3 I*ka]}, {k, M*w^2 - (l + k), 
    l}, {-k*Exp[-3 I*ka], l, M*w^2 - (k - l)}};
mydet = ExpToTrig[Det[mat]];
sol = Solve[mydet == 0, w];
funcs = w /. sol;
ReImPlot[funcs, {ka, 0, \[Pi]}, PlotLabel -> "\[Omega](ka) ", 
 PlotLegends -> {{"\[Omega]1", "\[Omega]2", "\[Omega]3", "\[Omega]4", 
    "\[Omega]5", "\[Omega]6"}, "ReIm"}, ReImLabels -> {"Re", "Im"}, 
 Frame -> True]

I would like to extract the positive complex solutions of w to use in the following derivative to find the density of states. I tried plotting for w2, w4, w6 only but it seems that I will be missing some other positive solutions for different ka's.

n = Pi / ka;
dkdw = D[funcs, ka];
dNdw = n/Pi * dkdw;
ReImPlot[dNdw , {ka, 0, \[Pi]}, PlotLabel -> "Density of States ", 
 PlotLegends -> {{"dN/d\[Omega]1", "dN/d\[Omega]2", "dN/d\[Omega]3", 
    "dN/d\[Omega]4", "dN/d\[Omega]5", "dN/d\[Omega]6"}, "ReIm"}, 
 ReImLabels -> {"Re", "Im"}, Frame -> True]

In summary, I want to extract the upper part of the first graph to plot the derivative of those solutions in the second graph.

Answer 1

Just consider the fourth one:

Plot[Re@funcs[[4]], {ka, 0, Pi}, PlotPoints -> 100]

The plot hints the standard Mathematica approach is trajecting across an "implied" branch-cut in a non-analytically-continuous path. Note the function has square and cube roots. I can't imagine you would want this if you plan to take derivatives. If you need analytically-continuous routes over the solutions, you may be able to use this approach: Dealing with branch cuts

Perhaps it would also be useful if I point out, thanks to J.M. in the link above, we could make the substitution $e^{-3 i ka}=z$, then apply GroebnerBasis to the first function for example and obtain: $$ f(z,w)=-432 w^4 + 32 w^6 + 81 (43 - 21 z^2) + 18 w^2 (31 + 18 z^2) $$ Now that's an easy function to route an analytically-continuously path through the solution by letting $z=e^{-3 i ka}$ and following the technique in the link.

Answer 2

Take real numbers

k = 9.; l = 12.; m = 2.; M = 4.;
mat = {{m*w^2 - 2*k, k, k*zeta}, {k, M*w^2 - (l + k), 
l}, {-k*zeta, l, M*w^2 - (k - l)}};
mydet = ExpToTrig[Det[mat]];
sol = Quiet@Solve[mydet == 0, w];
funcs = Chop[w /. sol /. {zeta -> Exp[-3I ka]} // FullSimplify];
Show[Table[Plot[Max[0, (funcs[[k]] + Conjugate[funcs[[k]]])/2], {ka, 0, \[Pi]}, PlotStyle -> Black], {k, 1, Length[funcs]}], PlotRange -> All]
Show[Table[Plot[Max[0, (funcs[[k]] - Conjugate[funcs[[k]]])/(2 I)], {ka, 
 0, \[Pi]}, PlotStyle -> Blue], {k, 1, Length[funcs]}], PlotRange -> All]