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Here comes a simple scenario

fz = 8 (z + 18 A z + (8 - 48 A) z^3 + 16 z^5);
sol = Solve[fz == 0, z]

As we can see there are five roots. The root (0,0) is always present while the nature of the other four roots strongly depends on the numerical value of A. I found that when $A < -1/18$ we have two real and two purely imaginary roots. For $A > -1/18$ there are three cases:

(i) When $A \in (-1/18, A_1)$ we have 4 purely imaginary roots,

(ii) When $A \in (A_1, A_2)$ we have 4 complex roots,

(iii) When $A > A_2$ we have 4 real roots.

My question: How can I determine the exact values of $A_1$ and $A_2$, whiche delimit the three intervals?

Many thanks in advance!

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  • $\begingroup$ This really is more of a math question than a Mathematica question. In any event, this paper might be of interest. $\endgroup$ – J. M. will be back soon Oct 27 '17 at 2:10
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You can use Discriminant:

Solve[
    Discriminant[8 (z+18 A z+(8-48 A) z^3+16 z^5), z] == 0,
    A
]

{{A -> -(1/18)}, {A -> -(1/18)}, {A -> -(1/18)}, {A -> 0}, {A -> 0}, {A -> 5/ 6}, {A -> 5/6}}

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  • $\begingroup$ Quick and elegant! $\endgroup$ – Vaggelis_Z Oct 26 '17 at 16:28
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fz = 8 (z + 18 A z + (8 - 48 A) z^3 + 16 z^5);

Solve for the real roots and the ConditionalExpression for the root will tell you when it is real

As Root objects

(solR = z /. Solve[fz == 0, z, Reals]) // Column

enter image description here

As radicals

(solR // ToRadicals) // Column

enter image description here

EDIT: For there to be four complex roots then -1/8 < A < 5/6. Further, for all of these four complex roots to be purely imaginary then Re[z] == 0 and there must be 5 roots counting the root at z == 0.

Select[List @@ (Reduce[{fz == 0, Re[z] == 0, -1/8 < A < 5/6}, {A, z}] // 
    ToRadicals), Count[#, z == _, Infinity] == 5 &]

enter image description here

So the interval for four purely imaginary roots is {-1/8, 0}.

For example,

% /. A -> -0.05

(* {z == 0. + 0.0988028 I || z == 0. - 0.0988028 I || z == 0 || 
  z == 0. + 0.800149 I || z == 0. - 0.800149 I} *)
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  • $\begingroup$ Hmmm, With this method we defined that $A_2 = 5/6$. But what about $A_1$? $\endgroup$ – Vaggelis_Z Oct 26 '17 at 16:16

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