Cfxtrjtrk

I am very new to Mathematica. I have started learning it only last month. I would like to graph the image of some complex valued polynomials with some provided conditions. For example: $$ p(z_1,z_2,z_3)=z_1z_2^2 +z_2z_3+z_1z_3,$$ given that $|z_1|=1, |z_2|=2=|z_3|$.

On the boundary of the image the Jacobian will be singular:

Clear[r, s, t, u, v, w];

Block[{z1 = Exp[I r], z2 = 2 Exp[I s], z3 = 2 Exp[I t]}, 

 expr = ComplexExpand[ReIm[z1 z2^2 + z2 z3 + z1 z3]]]

(*  {4 Cos[r+2 s]+2 Cos[r+t]+4 Cos[s+t], 4 Sin[r+2 s]+2 Sin[r+t]+4 Sin[s+t]}  *)



sub = {r + t -> u, s + t -> v, r + 2 s -> w};(* see simplified Jacobian *)

jac = D[expr, {{r, s, t}}];                  (* Jacobian is 2 x 3 *)

singRST = Equal @@ Divide @@ jac // Simplify (* Singular if rows are proportional *)

singUVW = singRST /. sub // Simplify

(* Solve cannot solve the system, unless we cut it into bite-size pieces *)

solv = Solve[singUVW[[;; 2]], v] /. C[1] -> 0;

singUW = singUVW[[2 ;;]] /. solv // Simplify;

solu = Solve[#, u] & /@ singUW;

(*

  -((2 Sin[r + 2 s] + Sin[r + t])/(2 Cos[r + 2 s] + Cos[r + t])) ==

   -((2 Sin[r + 2 s] + Sin[s + t])/(2 Cos[r + 2 s] + Cos[s + t])) ==

   -((Sin[r + t] + 2 Sin[s + t])/(Cos[r + t] + 2 Cos[s + t]))



  -((Sin[u] + 2 Sin[w])/(Cos[u] + 2 Cos[w])) ==

   -((Sin[v] + 2 Sin[w])/(Cos[v] + 2 Cos[w])) ==

   -((Sin[u] + 2 Sin[v])/(Cos[u] + 2 Cos[v]))

*)    



(* fix sub so that it works on a general expression *)

invsub = First@Solve[Equal @@@ sub, {u, v, w}];

sub = First@Solve[Equal @@@ invsub, {r, s, t}];

(*some u solutions are complex*)

realu = List /@ Cases[Flatten@solu, _?(FreeQ[#, Complex] &)];



boundaries = PiecewiseExpand /@ 

   Simplify[

    TrigExpand@Simplify[Simplify[expr /. sub] /. solv] /. realu // 

     Flatten[#, 1] &, 0 <= w < 2 Pi];



ParametricPlot[boundaries // Evaluate, {w, 0, 2 Pi}]

Well, it's only a start, since you have to check in the interior boundaries to see whether they might be holes. But @HenrikSchumacher has done that already.

By letting $z_1,z_2,z_3$ trace out circles, we can see some beautiful curves that live within that blob!

 p[z1_, z2_, z3_] := z1 z2^2 + z2 z3 + z1 z3;

 q[t_][a1_, a2_, b1_, b2_, c1_, c2_] := 

 p[Exp[ I (a1 t + a2)], 2 Exp[  I (b1 t + b2)], 2 Exp[ I (c1 t + c2)]];

 Manipulate[

  ParametricPlot[{Re[q[ t][a1, a2, b1, b2, c1, c2]], 

    Im[q[ t][a1, a2, b1, b2, c1, c2]]}, {t, 0, 2 [Pi]}, 

    Axes -> False, Frame -> True, PlotRange -> {{-12, 12},{-12, 12}}], 

    {a1, -5, 5},{a2, 0, 2 [Pi]},{b1, -5, 5},{b2, 0, 2 [Pi]},

    {c1, -5, 5},{c2, 0, 2 [Pi]}]

Here is a look at the analytical form of these curves:

 Manipulate[

  ComplexExpand@ReIm[q[t][a1, a2, b1, b2, c1, c2]], 

  {a1, -5, 5}, {a2, 0, 2 [Pi]}, {b1, -5, 5}, {b2, 0, 2 [Pi]}, 

  {c1, -5, 5}, {c2, 0, 2 [Pi]}]

or

 Manipulate[

  FullSimplify[q[t][a1, a2, b1, b2, c1, c2]], {a1, -5, 5}, {a2, 0,

    2 [Pi]}, {b1, -5, 5}, {b2, 0, 2 [Pi]}, {c1, -5, 5}, {c2, 0, 2 [Pi]}]

Not very elegant, but this might give you a coarse idea.

z1 = Exp[I r];

z2 = 2 Exp[I s];

z3 = 2 Exp[I t];

expr = ComplexExpand[ReIm[z1 z2^2 + z2 z3 + z1 z3]];

f = {r, s, t} [Function] Evaluate[expr];



R = DiscretizeRegion[Cuboid[{-1, -1, -1} Pi, {1, 1, 1} Pi], 

   MaxCellMeasure -> 0.0125];

pts = f @@@ MeshCoordinates[R];

triangles = MeshCells[R, 2, "Multicells" -> True][[1]];

Graphics[{

  Red, Disk[{0, 0}, 10],

  FaceForm[Black], EdgeForm[Thin],

  GraphicsComplex[pts, triangles]

  },

 Axes -> True

 ]

Could be the disk of radius 10...

Here's another numerical approach, similar to @Henrik's, but without the mesh overhead. It can be generalized to more variables easily. It requires some manual intervention to code the constraints on the variables.

poly = z1 z2^2 + z2 z3 + z1 z3;

vars = Variables[poly];

constrVars = Thread[vars -> {1, 2, 2} Array[Exp[I #] &@*Slot, Length@vars]]

(*  {z1 -> E^(I #1), z2 -> 2 E^(I #2), z3 -> 2 E^(I #3)}  *)



polyFN = poly /. constrVars // Evaluate // Function;



Graphics[{

  PointSize[Tiny], 

  polyFN @@ RandomReal[{0, 2 Pi}, {Length@vars, 5 10^4}] // ReIm // Point},

 Frame -> True]

We can see ghosts of some of the boundaries in my other answer.