I am looking for any complex number solutions to the system of equations:

$$\begin{align} |a|^2+|b|^2+|c|^2&=\frac13 \\ \bar{a}b+a\bar{c}+\bar{b}c&=\frac16 (2+\sqrt{3}i). \end{align}$$

Note I put inequality in the tags as I imagine it is an inequality that shows that this has no solutions (as I suspect is the case).

This is connected to my other question... I have found that $(4,1,1)/6$ and $\mu=(2,2+\sqrt{3}i,2-\sqrt{3}i)/6$ are square roots of $(2,1,1)/4$ in $(\mathbb{C}\mathbb{Z}_3,\star)$ but am trying to understand why $\mu$ is not positive in the C*-algebra when, for example, $(14,-6+5i,-6-5i)$ is.

  • $\begingroup$ Does the bar represent conjugate? i.e., $\bar{a} \times b$ ? $\endgroup$ – Mann Jun 16 '15 at 10:08
  • $\begingroup$ @Mann yes... "complex number solutions". $\endgroup$ – JP McCarthy Jun 16 '15 at 10:08

By Cauchy-Schwarz, we must have$$\left(|a|^2+|b|^2+|c|^2 \right) \left(|\overline{c}|^2+|\overline{a}|^2+|\overline{b}|^2 \right) \ge \left| a\overline c + b\overline a + c\overline b\right|^2$$ $$\implies \frac19 \ge \frac1{36}|2+\sqrt3 \; i|^2 = \frac7{4\cdot9}$$ which is obviously not possible...

  • $\begingroup$ $\left(|a|\left|\overline{c}\right|+|b||\overline{a}|+|c| |\overline{b}| \right)^2\neq \left| a\overline c + b\overline a + c\overline b\right|^2$, unless $a,b,c$ are positive reals (WolframAlpha). $\endgroup$ – user26486 Jun 16 '15 at 11:31
  • $\begingroup$ @user26486 and why is that relevant? $\endgroup$ – Macavity Jun 16 '15 at 11:34
  • $\begingroup$ Cauchy-Schwarz says: $\left(|a|^2+|b|^2+|c|^2 \right) \left(|\overline{c}|^2+|\overline{a}|^2+|\overline{b}|^2 \right) \ge \left(|a|\left|\overline{c}\right|+|b||\overline{a}|+|c| |\overline{b}| \right)^2$ $\endgroup$ – user26486 Jun 16 '15 at 11:35
  • $\begingroup$ @user26486 Check proofwiki.org/wiki/Cauchy-Schwarz_Inequality/Complex_Numbers for the statement and proof as applied to Complex numbers. I am sure there are more references. The version you have posted is actually CS inequality for the moduli - which are all real numbers. $\endgroup$ – Macavity Jun 16 '15 at 11:36
  • 2
    $\begingroup$ @JpMcCarthy For equality, $(a, b, c)$ and $(\overline c, \overline a, \overline b)$ must be scalar multiples. Here of course the scalar can be a complex number. $\endgroup$ – Macavity Jun 16 '15 at 12:20

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