MATLAB: How to solve this complex equation

complex numberequations

I have to solve the equation 4Z^4-11Z^2+7-3i=0, where Z is a complex number

My thoughts are to do this:

p=[4,0,-11,0,7-3i]
roots(p)

However, the answer I get is different from the actual answer, which is

2.4849+0.1552i
0.9924-0.2898i
-0.7273+0.1347i

Any help is appreciated!

Best Answer

I believe you are getting the correct answer

% method 1
syms Z
eqn=4*Z^4-11*Z^2+7-3i==0;
Root=vpasolve(eqn)
% method 2
roots(sym2poly(eqn))

Related Solutions

MATLAB: Solving a complicated equation in matlab

I think there is nothing wrong with your code. If I do

f(-0.351409329825233)

the answer is

ans =
   -3.9968e-015

So -0.351409329825233 is a negative root of your equation.

MATLAB: How to solve simultaneous equations

syms x y z
eqn1 = 1500*(20-x)+1.5309E-05*(293^4-(x+273)^4) == z;
eqn2 = 20000*(x-y) == z;
eqn3 = 3600*(y-10)+1.5309E-05*((y+273)^4-100^4) == z;
eqns = [eqn1,eqn2,eqn3];
vars = [x y z];
sol = solve(eqns, vars, 'returnconditions',true)
sol = struct with fields:
             x: [16×1 sym]
             y: [16×1 sym]
             z: [16×1 sym]
    parameters: [1×0 sym]
    conditions: [16×1 sym]

16 solutions.

sol.conditions
ans =

and they are unconditional.

So there are 16 "true" solutions.

Perhaps you only want real-valued ones?

vx = vpa(sol.x); vy = vpa(sol.y); vz = vpa(sol.z);
mask = imag(vx) == 0 & imag(vy) == 0 & imag(vz) == 0;
realx = vx(mask);
realy = vy(mask);
realz = vz(mask);
([realx, realy, realz])
ans =

So there are two "true" solutions that also happen to be real-valued. This was predictable: degree 16 polynomials in real roots always have an even number of real roots.

If you wanted to ignore some of the valid solutions even further, you could further test realx > 0

Best Answer

Related Solutions

MATLAB: Solving a complicated equation in matlab

MATLAB: How to solve simultaneous equations

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