Sufficient and necessary conditions for $y_1 = |x_1|^2$, $y_2 = x_1 overline{x_2}$, $y_3 = overline{x_1}...












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.






algebra-precalculus complex-numbers systems-of-equations nonlinear-system







share|cite|improve this question











$endgroup$

















    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.






    algebra-precalculus complex-numbers systems-of-equations nonlinear-system







    share|cite|improve this question











    $endgroup$















      2












      2








      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.






      algebra-precalculus complex-numbers systems-of-equations nonlinear-system







      share|cite|improve this question











      $endgroup$




      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.






      algebra-precalculus complex-numbers systems-of-equations nonlinear-system




      algebra-precalculus complex-numbers systems-of-equations nonlinear-system






      share|cite|improve this question















      share|cite|improve this question













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      edited Jan 7 at 22:20







      B. Groeger

















      asked Jan 7 at 21:18









      B. GroegerB. Groeger

      205




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          $begingroup$

          Your conditions are sufficient with the condition (5) replaced by $y_1,y_4inmathbb{R}_{geq 0}$ (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 $$y_2=bar{y}_3text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=sqrt{y_4}$. All possible solutions $(x_1,x_2)inmathbb{C}timesmathbb{C}$ in this case take the form
          $$(x_1,x_2)=big(0,sqrt{y_4},exp(text{i}theta)big),,$$
          where $thetain[0,2pi)$.



          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
          $$x_1bar{x}_2=sqrt{y_1}left(frac{bar{y}_3}{sqrt{y_1}}right)=bar{y}_3=y_2,.$$
          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 $(x_1,x_2)inmathbb{C}timesmathbb{C}$ are of the form
          $$left(x_1,x_2right)=left(sqrt{y_1},exp(text{i}theta),frac{y_3}{sqrt{y_1}},exp(text{i}theta)right)$$
          for some $thetain[0,2pi)$.






          share|cite|improve this answer











          $endgroup$













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












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          $begingroup$

          Your conditions are sufficient with the condition (5) replaced by $y_1,y_4inmathbb{R}_{geq 0}$ (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 $$y_2=bar{y}_3text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=sqrt{y_4}$. All possible solutions $(x_1,x_2)inmathbb{C}timesmathbb{C}$ in this case take the form
          $$(x_1,x_2)=big(0,sqrt{y_4},exp(text{i}theta)big),,$$
          where $thetain[0,2pi)$.



          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
          $$x_1bar{x}_2=sqrt{y_1}left(frac{bar{y}_3}{sqrt{y_1}}right)=bar{y}_3=y_2,.$$
          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 $(x_1,x_2)inmathbb{C}timesmathbb{C}$ are of the form
          $$left(x_1,x_2right)=left(sqrt{y_1},exp(text{i}theta),frac{y_3}{sqrt{y_1}},exp(text{i}theta)right)$$
          for some $thetain[0,2pi)$.






          share|cite|improve this answer











          $endgroup$













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
















          1












          $begingroup$

          Your conditions are sufficient with the condition (5) replaced by $y_1,y_4inmathbb{R}_{geq 0}$ (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 $$y_2=bar{y}_3text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=sqrt{y_4}$. All possible solutions $(x_1,x_2)inmathbb{C}timesmathbb{C}$ in this case take the form
          $$(x_1,x_2)=big(0,sqrt{y_4},exp(text{i}theta)big),,$$
          where $thetain[0,2pi)$.



          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
          $$x_1bar{x}_2=sqrt{y_1}left(frac{bar{y}_3}{sqrt{y_1}}right)=bar{y}_3=y_2,.$$
          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 $(x_1,x_2)inmathbb{C}timesmathbb{C}$ are of the form
          $$left(x_1,x_2right)=left(sqrt{y_1},exp(text{i}theta),frac{y_3}{sqrt{y_1}},exp(text{i}theta)right)$$
          for some $thetain[0,2pi)$.






          share|cite|improve this answer











          $endgroup$













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














          1












          1








          1





          $begingroup$

          Your conditions are sufficient with the condition (5) replaced by $y_1,y_4inmathbb{R}_{geq 0}$ (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 $$y_2=bar{y}_3text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=sqrt{y_4}$. All possible solutions $(x_1,x_2)inmathbb{C}timesmathbb{C}$ in this case take the form
          $$(x_1,x_2)=big(0,sqrt{y_4},exp(text{i}theta)big),,$$
          where $thetain[0,2pi)$.



          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
          $$x_1bar{x}_2=sqrt{y_1}left(frac{bar{y}_3}{sqrt{y_1}}right)=bar{y}_3=y_2,.$$
          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 $(x_1,x_2)inmathbb{C}timesmathbb{C}$ are of the form
          $$left(x_1,x_2right)=left(sqrt{y_1},exp(text{i}theta),frac{y_3}{sqrt{y_1}},exp(text{i}theta)right)$$
          for some $thetain[0,2pi)$.






          share|cite|improve this answer











          $endgroup$



          Your conditions are sufficient with the condition (5) replaced by $y_1,y_4inmathbb{R}_{geq 0}$ (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 $$y_2=bar{y}_3text{ and }|y_2|^2=|y_3|^2=y_2y_3=y_1y_4=0$$ imply that $y_2=y_3=0$. Hence, we may take $x_1:=0$ and $x_2:=sqrt{y_4}$. All possible solutions $(x_1,x_2)inmathbb{C}timesmathbb{C}$ in this case take the form
          $$(x_1,x_2)=big(0,sqrt{y_4},exp(text{i}theta)big),,$$
          where $thetain[0,2pi)$.



          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
          $$x_1bar{x}_2=sqrt{y_1}left(frac{bar{y}_3}{sqrt{y_1}}right)=bar{y}_3=y_2,.$$
          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 $(x_1,x_2)inmathbb{C}timesmathbb{C}$ are of the form
          $$left(x_1,x_2right)=left(sqrt{y_1},exp(text{i}theta),frac{y_3}{sqrt{y_1}},exp(text{i}theta)right)$$
          for some $thetain[0,2pi)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 7 at 22:11

























          answered Jan 7 at 21:53









          BatominovskiBatominovski

          33.1k33293




          33.1k33293












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


















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
















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




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


















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