how to find complex polynomial solution

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halil hasan
halil hasan 2019-7-30
评论： Walter Roberson 2019-7-30
I want to find roots of CDP, T's are constant real values
CDP1 =T_6*(1i*w).^6 +T_5*(1i*w).^5 +T_4*(1i*w).^4 +T_3*(1i*w).^3 +T_2*(1i*w).^2 +T_1*(1i*w) +T_0;
CDP2 =Tp_4*(1i*w)^.4 +Tp_3*(1i*w).^3 +Tp_2*(1i*w).^2 +Tp_1*(1i*w) +Tp_0;
CDP= CDP1*(1i*w).^1.4 +CDP2;

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Alex Mcaulley
Alex Mcaulley 2019-7-30
Then, after defining all the constant values:
syms w
CDP1 = T_6*(1i*w).^6 + T_5*(1i*w).^5 + T_4*(1i*w).^4 + T_3*(1i*w).^3 + T_2*(1i*w).^2 + T_1*(1i*w) + T_0;
CDP2 = Tp_4*(1i*w)^.4 + Tp_3*(1i*w).^3 + Tp_2*(1i*w).^2 + Tp_1*(1i*w) + Tp_0;
CDP = CDP1*(1i*w).^1.4 + CDP2;
sol = double(solve(CDP))
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halil hasan
halil hasan 2019-7-30
I think
if I can replace equvalent of i powers, to a powerless i ; it may solve my problem.
Walter Roberson
Walter Roberson 2019-7-30
w^10 is okay. You then multiply by i and raise the result to 0.9 or 4.9. By the power law, (A*B)^C is A^C*B^C so (i*w^10)^0.9 is i^0.9 * (w^10)^0.9 and that second part is not polynomial
In the case where w is nonnegative real if you are willing to treat 0.9 as 9/10 exactly (which it is not) then you could multiply out to get w^9. But if that is what you want then you need to code it: with the 1i in there, matlab would never compute it that way. You would be getting a different branch of 0.9 power.

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