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Given relations of coefficients and $m$ zeros of a complex polynomial (coefficients are complex), find the polynomial of degree $2n$ and $m \geq n$. For examples, we are finding $P(x)=C_{2n}x^{2n} + C_{2n-1}x^{2n-1}+\cdots+C_1x+C_0$, where $C_k = C_{2n-k}$, or $C_k = C^*_{2n-k}$ (* means conjugate). Totally, there are $2n$ coefficients to be solved and there are $m$ roots and $n$ relations between coefficients. $m+n\geq 2n$

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The $m < n$ from the title seems much more plausible: if $m=1000$ and $n=2$, how are we to get $1000$ roots with only degree $4$? – Erick Wong Jan 19 '13 at 6:27
I need to revise the inequality to $n\leq m \leq 2n$ – user58869 Jan 19 '13 at 11:07
@Chao-LiehChen This still seems very overdetermined. Taking the constraints $C_k = C_{2n-k}$ one has a reciprocal polynomial. Since the $2n$ roots appear in reciprocal pairs, one can only hope to specify at most $n$ of them in generality. – Erick Wong Jan 19 '13 at 18:44
The question should be revised into sub-questions: If this is overdetermined, we may have more than on possibilities for $P(x)$. Then, what are the possible $P(x)$'s when $P(x)$'s are complex reciprocal with m known zeros? What are they when self-reciprocal? What are the additional constraints on $Ci$'s required to have only one possible $P(x)$? – user58729 Jan 21 '13 at 6:45

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