equation solving - Reduction of an Inequality in $mathbb{C}$

Reduce[
  Abs[-((4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] +
  Abs[-((4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] < 1 , Abs[p]]

It is taking lot of time. It is running. Can any one help to reduce the inequality?

Answer

Because the question seeks an expression for the modulus of p, it makes sense to express p and q in terms of the moduli and phases.

sim = Simplify[(Abs[-((4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2)] + 
                Abs[-((4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2)]) /. 
          {p -> pm Exp[I pp], q -> qm Exp[I qp]}, pm >= 0 && qm >= 0 && (pp | qp) ∈ Reals]
(* (4 (pm + qm))/Abs[-1 + Sqrt[1 + 4 E^(I pp) pm + 4 E^(I qp) qm]]^2 *)

In what follows, we explore sim <= 1 instead of the question's sim < 1 in order to obtain solutions at the boundary, sim == 1, which is where most solutions seem to lie. Although

Reduce[sim <= 1 && pm >= 0 && qm >= 0, pm]

still produced no answer, even after 19 hours, the special case of setting qp to π did. Some hand-holding was required, however.

Reduce[(sim /. {qp -> Pi}) <= 1 && 2 π > pp >= 0 && pm >= 0 && qm >= 0, pm]

returned unevaluated with the message

Reduce::nsmet: This system cannot be solved with the methods available to Reduce. >>

However,

Reduce[FullSimplify[Reduce[(sim /. {qp -> Pi}) <= 1 && pm >= 0 && qm >= 0, pm], 
    2 Pi > pp >= 0 && pm >= 0 && qm >= 0] && 2 Pi > pp >= 0 && pm >= 0 && qm >= 0, pm]

did produce a meaningful answer.

(* (0 <= pp < 2 π && qm >= 1/4 && pm == 0) || (pp == π && qm >= 1/4 && pm >= 0) || 
   (pp == π && ((qm > 1/4 && pm >= 0) || (0 <= qm <= 1/4 && pm >= 1/4 (1 - 4 qm)))) *)

Note that, except for the solution pm == 0, all these solutions require pp == π.

In summary, solutions are available for qp -> π and perhaps other cases. Whether a solution can be obtained in general within several hours of computation is unknown.

Comments

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