So in principle it's exp( ( 1/P(z) )^(Order(RotationalSymmetryGroup)/2) ). And maybe these next: https://www.researchgate.net/figure/227602093_fig5_Figure-5-A-plot-of-the-roots-of-the-symmetric-Hermite-polynomial-H557689z …
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So in principle it's exp( ( 1/P(z) )^(Order(RotationalSymmetryGroup)/2) ). And maybe these next: https://www.researchgate.net/figure/227602093_fig5_Figure-5-A-plot-of-the-roots-of-the-symmetric-Hermite-polynomial-H557689z …