Can Mathematica reduce this inequality into a domain of z, where {z, r1, r2} are all complex?

Reduce[Abs[z^2 - r1] > Abs[z^3 - r2], z]

The line never returns a result. I also tried just calculating the boundary:

Simplify[Abs[z^2 - r1] == Abs[z^3 - r2]]

But that never returns a result either. I've tried a few other methods that all just crashed or failed. I found a related Question and another related Question but I couldn't figure out how to apply those partial solutions to this problem.

BTW, I realize Solve[] and Reduce[] don't cope with Abs[]. Even this simpler command give a poor result:

Simplify[Abs[z^2] == Abs[z^3]];

(* Abs[z]^2 == Abs[z] *)

And just to be clear, I don't just need the answer to the inequality, I need to know how to make Mathematica do it.

  • $\begingroup$ You should tell Reduce you're working over the reals: Reduce[Abs[z^2 - r1] > Abs[z^3 - r2], z, Reals]. $\endgroup$
    – Chip Hurst
    Nov 13 '15 at 22:36
  • $\begingroup$ @ChipHurst I'm not. I'll add clarification that the variables are complex. $\endgroup$ Nov 13 '15 at 22:40
  • $\begingroup$ How come my tick marks in the title aren't being turned into nice Font? $\endgroup$ Nov 13 '15 at 23:24
  • 1
    $\begingroup$ Titles entry fields don't support all the editing features as the editor pane that accepts the main body of a question. This is a quick of SE that has to be lived with. $\endgroup$
    – m_goldberg
    Nov 14 '15 at 2:07
  • $\begingroup$ What do you think Simplify[Abs[z^2] == Abs[z^3]] should return? Keep in mind z is considered complex unless otherwise stated. $\endgroup$
    – Chip Hurst
    Nov 14 '15 at 2:16

You need to specify all of the variables you want to reduce over:

In[1]:= Reduce[Abs[z^2 - r1] > Abs[z^3 - r2], {z, r1, r2}, Complexes]
Out[1]= (* a huge pile of Im, Re and Sqrt omitted *)
  • $\begingroup$ Thanks for the tip about adding Complexes, but yikes that huge pile result is not a good result. I don't think it should be necessary to explicitly separate Im[z] and Re[z] in the result. $\endgroup$ Nov 13 '15 at 23:08

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