Show that if a set of complex numbers $z_1,z_2,\ldots,z_n$ satisfy $$z_1^l+z_2^l+\cdots+z_n^l=0$$ for every odd $l$, then for any $z_i$ we can always find some $z_j$ such that $z_i+z_j=0$.

The question has been answered here for real numbers but not for complex numbers

  • $\begingroup$ what has been tried ? $\endgroup$ – user451844 Aug 17 '17 at 23:09
  • $\begingroup$ This and this. $\endgroup$ – Astor Aug 17 '17 at 23:25
  • $\begingroup$ can you show it for the integers ? $\endgroup$ – user451844 Aug 17 '17 at 23:28
  • $\begingroup$ It is shown here for the reals. $\endgroup$ – Astor Aug 17 '17 at 23:30
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    $\begingroup$ @ross-millikan It is not a duplicate! This is for complex numbers! $\endgroup$ – Astor Aug 17 '17 at 23:36

Let $\displaystyle P(z) = \sum_{k=0}^n (-1)^ke_k z^{n-k}$ be the $n^{th}$ degree polynomial with roots $z_k \mid k=1,\cdots,n$, where by Vieta's formulas $e_k$ are the elementary symmetric polynomials.

Let $\displaystyle p_i=\sum_{k=1}^n z_k^i\,$, where it is given that $p_l=0$ for all odd $l$.

From Newton's identities $\displaystyle k e_k = \sum_{i=1}^k (-1)^{i-1}e_{k-i}p_i$ it follows (by induction, for example) that $e_l=0$ for all odd $l$. Therefore, the polynomial $P(z)$ has every other coefficient $0$, so it contains either only even powers of $z$, or only odd powers of $z$, depending on the parity of $n$. In the first case $P(z)$ is an even function, in the second case an odd one. In both cases $P(z)=0 \iff P(-z)=0$ so the roots of $P(z)$ can be grouped in pairs of mutually opposites.

  • $\begingroup$ neat proof! last line is a bit problematic, better use P even(odd) to its fullest. $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 3:28
  • $\begingroup$ @orangeskid Thanks. Note that the question states for any z_i we can always find some z_j such that z_i + z_j =0. There is no condition $i \ne j\,$, and in fact the statement would no longer hold true with such an additional condition. That covers the case where a $0$ root is its own opposite. $\endgroup$ – dxiv Aug 18 '17 at 3:39
  • $\begingroup$ what i had in mind is polynomials like $[z-1)^2(z+1)$ $\endgroup$ – Orest Bucicovschi Aug 18 '17 at 3:43
  • $\begingroup$ @orangeskid That would be $\,z_1=z_2=1, z_3=-1\,$, but $\,z_1+z_2+z_3 \ne 0\,$ so it doesn't satisfy the premises of OP's question. $\endgroup$ – dxiv Aug 18 '17 at 3:57

We prove instead the following Lemma. Using this Lemma, your claim follows by an immediate (strong) induction.

Lemma Let $a_1,.., a_k$ be complex numbers, not all of them zero, and $z_1,..,z_k$ non-zero, pairwise distinct complex numbers. If $$a_1z_1^l+...+a_kz_k^l=0$$ for all odd integers $l$, then, there exists some $i\neq j$ such that $$z_i+z_j=0$$


Consider the determinant $$\Delta=\begin{vmatrix} z_1 & z_2 & z_3 &...&z_k \\ z_1^3& z_2^3 & z_3^3 &...&z_k^3 \\ z_1^5 & z_2^5 & z_3^5 &...&z_k^5 \\ ...&...&...&....&... \\ z_1^{2k-1} & z_2^{2k-1} & z_3^{2k-1} &...&z_k^{2k-1} \\ \end{vmatrix}$$

First, since $a_1 \mbox{col 1}+...+a_k \mbox{col k}=0$ we get $\Delta=0$. Next, using the Vandermonde formula, we get $$0=\Delta=z_1z_2...z_k\begin{vmatrix} 1 & 1 & 1 &...&1 \\ z_1^2& z_2^2 & z_3^2 &...&z_k^2 \\ z_1^4 & z_2^4 & z_3^4 &...&z_k^4 \\ ...&...&...&....&... \\ z_1^{2k-2} & z_2^{2k-2} & z_3^{2k-2} &...&z_k^{2k-2} \\ \end{vmatrix}\\=z_1z_2..z_k \prod_{1 \leq i <j \leq k} (z_j^2-z_i^2)$$

It follows from the hypothesis that there exists some $i <j$ such that $z_j^2-z_i^2=0$ and hence $z_i+z_j=0$.

  • $\begingroup$ Very neat! But in your final determinant, shouldn't the cubes be fourth powers? Looks like it may be a typo. $\endgroup$ – awkward Aug 21 '17 at 11:18
  • $\begingroup$ @awkward ty fixed. $\endgroup$ – N. S. Aug 21 '17 at 13:18

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