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?

• There won't be a nice form. If you Plot3D it, you'll see that the region you seek is really rather a horrible shape. Do you have any more information about $p, q$? – Patrick Stevens Aug 28 '15 at 7:47
• I can consider p and q each having modulus less than 1. How to get the 3Dplot ? Could you please provide me the code? – Sk Sarif Hassan Aug 28 '15 at 8:18
• If you look in the documentation for Plot3D, you'll find out. Plot3D[Abs[(4 p)/(-1 + Sqrt[1 + 4 p + 4 q])^2] + Abs[(4 q)/(-1 + Sqrt[1 + 4 p + 4 q])^2], {p, -5, 5}, {q, -5, 5}] This is of course for real $p, q$ only. – Patrick Stevens Aug 28 '15 at 8:20
• But even the reduction is taking a lot of time. Is there any way to get it faster? – Sk Sarif Hassan Aug 28 '15 at 8:30
• @belisarius $p, q$ are possibly complex, I think. – Patrick Stevens Aug 28 '15 at 9:11

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]


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

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.