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 $k\in\mathbb N$?

  • $\begingroup$ 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

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.

  • $\begingroup$ "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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.