solving an equation with a condition

Is there any way to find a value to z that makes the solution of this equation consisting of only negative real numbers or negative real numbers with imaginary numbers?

x^4+z*x^2+z*x+50=0

Answer 1

[Walter Roberson]

syms X Z
eqn = X^4+Z*X^2+Z*X+50==0

eqn = 

syms X_r X_i Z_r Z_i real
assume(X_r < 0 & Z_r < 0)
eqn2 = subs(eqn, [X, Z], [X_r+1i*X_i, Z_r+1i*Z_i])

eqn2 = 

sol = solve(eqn2, [X_r, X_i, Z_r, Z_i], 'returnconditions', true, 'MaxDegree', 4)

sol = struct with fields:
           X_r: [4×1 sym]
           X_i: [4×1 sym]
           Z_r: [4×1 sym]
           Z_i: [4×1 sym]
    parameters: [1×3 sym]
    conditions: [4×1 sym]

Xsol =  sol.X_r + 1i*sol.X_i

Xsol = 

So the x are all different, but follow a typical pattern for roots of a polynomial of degree 4.

Important note for this Xsol and the Zsol: the any x, y, z you see in the outputs are not the x and z of the original question. The x and z of the original question were written into the problem in terms of X and Z (complex valued, capitalized) and X_i X_r Z_i Z_r (capitalized, individual real or imaginary components of their corresponding complex variables.)

Any x, y, z (lower-case) you see in the solutions are auxillary variables generated by MATLAB to help express the solutions.

Zsol = sol.Z_r + 1i*sol.Z_i

Zsol = 

Meanwhile the Z are all the same.

sc = simplify(sol.conditions)

sc = 

So the auxillary variables x and y must be real-valued and the auxillary variable z must be negative (even though it is not specified explicitly that z must be real-valued, the < 0 forces it to be real-valued in context). The ^ means "and". The "i" is sqrt(-1) . The σ are sub-expressions that are broken out to allow the structure of the solution to be expressed more clearly; the individual meaning of the sigma variables is given in the where section.

The sigma variables are messy, and refer to each other. This is common for the solutions to a polynomial of degree 4.

sol.parameters

ans = 

The names of the auxillary variables can change; I list them here

So where does this leave you?

To be honest... not really any further ahead. You are looking for a detailed description of the behaviour of system of two complex variables when the system is a multinomial with maximum degree 4. The solutions to a multinomial of degree 4 are really messy.

[Walter Roberson]

We can proceed graphically to get an idea of how bad the solutions are.

syms X Z
eqn = X^4+Z*X^2+Z*X+50==0

eqn = 

That is a multinomial that is degree 4 in X, so we can solve for X to get 4 roots based upon Z. Then we can look at real() of the roots to see where they are negative. We can do that by breaking Z up into real and imaginary components, and drawing a surface. And because real(z) < 0 by requirement, we can skip plotting for positive real(z)

syms X_r X_i Z_r Z_i real
assume(X_r < 0 & Z_r < 0)
sol = (solve(eqn,X, 'maxdegree', 4));
R = (subs(real(sol),Z,Z_r + 1i*Z_i))

R = 

Messy, but for the moment we only have to plot, not make sense symbolically of the results.

zivec = linspace(-100, 100, 500);
zrvec = linspace(-20,0, 250);
[ZI, ZR] = ndgrid(zivec, zrvec);   %real component < 0

figure(1);
R1fun = matlabFunction(R(1), 'vars', [Z_i, Z_r]);
R1 = R1fun(ZI, ZR);
surface(zrvec, zivec, R1, 'edgecolor', 'none'); xlabel('Z real'); ylabel('Z imag'); zlabel('X real'); title('root 1'); view(3)

Although it is not immediately obvious from the plot, all real(X) < 0 for this first root

figure(2); 
R2fun = matlabFunction(R(2), 'vars', [Z_i, Z_r]);
R2 = R2fun(ZI, ZR);
surface(zrvec, zivec, R2, 'edgecolor', 'none'); xlabel('Z real'); ylabel('Z imag'); zlabel('X real'); title('root 2'); view(3)

Portions of real(X) are > 0 for this second root, but most is negative

figure(3);
R3fun = matlabFunction(R(3), 'vars', [Z_i, Z_r]);
R3 = R3fun(ZI, ZR);
surface(zrvec, zivec, R3, 'edgecolor', 'none'); xlabel('Z real'); ylabel('Z imag'); zlabel('X real'); title('root 3'); view(3)

Portions of real(x) are < 0 for this third root, but most is positive

figure(4);
R4fun = matlabFunction(R(4), 'vars', [Z_i, Z_r]);
R4 = R4fun(ZI, ZR);
surface(zrvec, zivec, R4, 'edgecolor', 'none'); xlabel('Z real'); ylabel('Z imag'); zlabel('X real'); title('root 4'); view(3)

Although it is not completely obvious from this plot, real(X) > 0 for all values for this fourth root.

So:

• when we take the first root of the multinomial of degree 4, and we restrict ourselves to Z that have real(Z) < 0, then all real(X) < 0; so for any given Z with real(Z) < 0, there is a corresponding X that solves the equation and real(x) < 0. The value of X is given by substituting Z into sol(1)
• when we take the fourth root of the multinomial of degree 4, and we restrict ourselves to Z that have real(Z) < 0, then there are no solutions for this case
• when we take the second root of the multinomial of degree 4, and we restrict ourselves to Z that have real(z) < 0, then most real(X) < 0. However, for some real(Z) > -15 or so, there are places where real(X) > 0 which is not desired.
• when we take the third root of the multinomial of degree 4, and we restrict ourselves to Z that have real(Z)<0, then most real(X)>0 . However, for some real(Z) > -15 or so, there are places where real(X) < 0 as desired

A question would then be whether you can come up with an equation that defines the boundaries of the "wings". I would say... Maybe? But it would take a bunch of study of the roots of the multinomial of degree 4 to figure out which subclause is generating the discontinuities in order to get the equations of the boundaries.

Answer 2

[David Hill]

I ran this I could not find any cases where all real root parts are negative.

z=-1e6;
while sum(sign(real(roots([1 0 z z 50]))))~=-4
    z=z+1;
end

[Walter Roberson]

I do not think that is the question. I think the question is to find (x,z) such that real(x)<0 and real(z) < 0 and f(x,z) == 0