Let $\mathbb C$ and $\mathbb R$ denote the fields of complex and real numbers, respectively. Suppose $x_1, x_2 \in \mathbb C$, and \begin{align} y_1 & = |x_1|^2 \tag 1 \\ y_2 & = x_1 \overline{x_2} \tag 2 \\ y_3 & = \overline{x_1} x_2 \tag 3 \\ y_4 & = |x_2|^2 \tag 4 \end{align} where $\overline{x}$ denotes the complex conjugate of $x$. Are there any sufficient and necessary conditions, in terms of $y_1$, $y_2$, $y_3$ and $y_4$, for all the above conditions to hold?

The following are among the necessary conditions. \begin{align} y_1, y_4 & \in \mathbb R_+ \tag 5 \\ y_2 & = \overline{y_3} \tag 6 \\ y_1 y_4 & = y_2 y_3 \tag 7 \end{align} where $\mathbb R_+$ denotes the nonnegative subset of $\mathbb R$. I think something is still missing, although I could not find a counterexample.


Your conditions are sufficient with the condition (5) replaced by $y_1,y_4\in\mathbb{R}_{\geq 0}$ (I just noticed that you had fixed this condition). To show that your conditions are also sufficient, we first deal with the trivial cases $y_1=0$ or $y_4=0$. Without loss of generality, let $y_1=0$. Then $$y_2=\bar{y}_3\text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=\sqrt{y_4}$. All possible solutions $(x_1,x_2)\in\mathbb{C}\times\mathbb{C}$ in this case take the form $$(x_1,x_2)=\big(0,\sqrt{y_4}\,\exp(\text{i}\theta)\big)\,,$$ where $\theta\in[0,2\pi)$.

We now assume that $y_1>0$ and $y_4>0$. Take $x_1:=\sqrt{y_1}$ and $x_2:=\dfrac{y_3}{\sqrt{y_1}}$. Ergo, (1) and (3) follow immediately. To show (2) we have $$x_1\bar{x}_2=\sqrt{y_1}\left(\frac{\bar{y}_3}{\sqrt{y_1}}\right)=\bar{y}_3=y_2\,.$$ Additionally, $$y_1y_4=y_2y_3=|y_3|^2$$ implies that $$y_4=\frac{|y_3|^2}{y_1}=\left(\frac{\bar{y}_3}{\sqrt{y_1}}\right)\left(\frac{y_3}{\sqrt{y_1}}\right)=\bar{x}_2x_2=|x_2|^2\,.$$ In fact, all solutions $(x_1,x_2)\in\mathbb{C}\times\mathbb{C}$ are of the form $$\left(x_1,x_2\right)=\left(\sqrt{y_1}\,\exp(\text{i}\theta),\frac{y_3}{\sqrt{y_1}}\,\exp(\text{i}\theta)\right)$$ for some $\theta\in[0,2\pi)$.

  • $\begingroup$ Thanks very much for your efforts! The answer is very insightful. $\endgroup$ – B. Groeger Jan 7 at 22:10

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