2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $\theta\in[0,2\pi)$.

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

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.