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Suppose $z$ is a complex number with $\bar{z}$ denoting its conjugate. Does there exist real numbers $\{a_1,\ldots, a_n\}$ such that

$$z^k+\bar z^k= a_1^k+a_2^k+\cdots+a_n^k,$$

for all $k\in\mathbb N$?

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  • $\begingroup$ By considering $k=0$ we see that $n$ must be $2$ ... $\endgroup$ – Henning Makholm Feb 16 '17 at 22:56
  • $\begingroup$ @HenningMakholm Thanks. I guess this implies that the $a$'s do not exist. Since $k=1$ and $k=2$ would determine $a_1$ an $a_2$, and then the other ones can't be satisfied. $\endgroup$ – mzp Feb 16 '17 at 23:03
  • $\begingroup$ Why does $k=2\Rightarrow n\leq 2$ $\endgroup$ – Stella Biderman Feb 16 '17 at 23:13
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Suppose $z$ is complex. Let us ask whether any real numbers $a_i$ exist for $1 \leq i \leq n$ such that
$$z^k+\bar z^k = a_1^k+a_2^k+\cdots+a_n^k \tag1$$ for every positive integer $k.$ This is almost the same as the original question, except that it avoids the simple argument in which setting $k=0$ shows that $n=2.$ In fact, the new conditions are slightly weaker.

If $z$ is real then of course for any $n\geq2$ we can set $a_1 = a_2 = z$ and $a_3 = \cdots = a_n = 0.$

But in the case where $z$ is not real, I will prove by contradiction that there is no set of real numbers $a_i$ satisfying Equation $1.$

Assume $z$ is not real and Equation $1$ is true. Write $z = re^{i(\theta + m\pi)}$ where $0 < \lvert\theta\rvert < \frac\pi2$ and $m$ is an integer, and consider the following two cases:

Case $0 < \lvert\theta\rvert \leq \frac\pi4.$ Then $0 < \lvert 2\theta \rvert \leq \frac\pi2$ and there exists some positive integer $p$ such that $\frac\pi2 \leq p\lvert2\theta\rvert \leq \pi.$ Then $z^{2p} = r^{2p}e^{i(2p\theta + 2pm\pi)} = r^{2p}e^{i(2p\theta)}$ and $\Re(z^{2p}) \leq 0.$

Case $\frac\pi4 < \lvert\theta\rvert \leq \frac\pi2.$ Then $z^2 = r^2e^{i(2\theta + 2m\pi)} = r^2e^{i(2\theta)}$ where $\frac\pi2 < \lvert2\theta\rvert \leq \pi,$ so $\Re(z^2) < 0.$

Combining these two cases, if $z$ is not real there is some positive integer $p$ such that $\Re(z^{2p}) \leq 0$ and therefore $z^{2p} + \bar z^{2p} \leq 0$. On the other hand, $a_1^{2p}+a_2^{2p}+\cdots+a_n^{2p} \geq 0$ for any positive integer $p,$ with equality only if $a_1 = \cdots = a_n = 0.$ Therefore either $z^{2p} + \bar z^{2p} < a_1^{2p}+a_2^{2p}+\cdots+a_n^{2p},$ contradicting the assumption, or $z^k+\bar z^k = 0$ for all positive integers $k.$ But $z + \bar z=0$ implies $z = ir$ where $r$ is real, which implies $z^2 + \bar z^2 = -2r^2,$ which implies $r=0,$ which contradicts the assumption that $z$ is not real. By contradiction, no such set of real numbers $a_i$ exists when $z$ is not real.

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  • $\begingroup$ "in the case where $z$ is not real, then either $\Re(z^2) \leq 0$ or $\Re(z^4) \leq 0$" Why? $\endgroup$ – dxiv Feb 16 '17 at 23:30
  • $\begingroup$ @dxiv Oops, $4$ is not always enough. We may need to use a much higher even power of $z.$ I think this is fixed now. $\endgroup$ – David K Feb 17 '17 at 4:00
  • $\begingroup$ Looks good to me now, +1. $\endgroup$ – dxiv Feb 17 '17 at 4:07
  • $\begingroup$ @dvix I forgot to mention, thanks for pointing out the gaping holes in the earlier "proof." And by "gaping" I mean a total of $\frac\pi2$ radians. $\endgroup$ – David K Feb 17 '17 at 4:13

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