How to solve a symbolic complex equation with real and imaginary parts?

I have to solve this polynomium: to . So, I know that to solve this equation I have to replace the s to get

Therefore, the real and imaginary parts will be zero and we have and

So, to solve this in matlab, I write this code:

syms s k w
den = s^3 + 9*s^2 + 14*s + 126
assume([ k> 0,w >0])
solw = solve(imag(den)==0, w)
den = subs(den, w, solw)
solk = solve(den==0, k)

Then, this code work well, but if I change the polynomial degree I will have to change the parameters of solve function. So I to know if there is a more generic way to solve this equations

Thanks for reading

Answer 1

[Paul]

I'm not sure what "change the paramters of the solve function" means. If a general apprach is desired, maybe something along the lines of simultaneously solving two equations for two unkonwns:

>> syms s w k
assume(w>=0); assume(k>=0);
D(s) = s^3 + 9*s^2 + 14*s + 2*k;
eqn = [real(D(1j*w))==0, imag(D(1j*w))==0];
sol = solve(eqn,[w k],'ReturnConditions',true);
[sol.w sol.k]
 
ans =
 
[        0,  0]
[ 14^(1/2), 63]

Change D(s) as desired.

[Nélio Dias]

oh thanks. It helped a lot!

[Walter Roberson]

If your polynomial degree is 3 or higher, there is a risk that solve() will decide to return root() objects instead of explicit solutions. You can reduce that problem by using

sol = solve(eqn,[w k],'ReturnConditions',true, 'maxdegree', 4);

which tells it to use explicit formulas up to degree 4.

However, with large enough polynomial degree, solve() will not be able to find an explicit solution, and you will start seeing root() constructs. You can vpa() to get numeric equivalents.

[Nélio Dias]

Thanks for you comment, Walter