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

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Nélio Dias 2021 年 4 月 6 日
コメント済み: Nélio Dias 2021 年 4 月 7 日
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

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### 採用された回答

Paul 2021 年 4 月 7 日
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.
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Walter Roberson 2021 年 4 月 7 日
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 2021 年 4 月 7 日
Thanks for you comment, Walter

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