Mixing
Sufficient and necessary conditions for $y_1 = |x_1|^2$, $y_2 = x_1 overline{x_2}$, $y_3 = overline{x_1}...
$begingroup$
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
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
$endgroup$
add a comment |
$begingroup$
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
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
$endgroup$
add a comment |
$begingroup$
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
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
$endgroup$
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
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
edited Jan 7 at 22:20
B. Groeger
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
asked Jan 7 at 21:18
B. GroegerB. Groeger
205
205
add a comment |
add a comment |
1 Answer
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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 $thetain[0,2pi)$.
answered Jan 7 at 21:53
BatominovskiBatominovski
1
$endgroup$
$begingroup$
Thanks very much for your efforts! The answer is very insightful.
$endgroup$
– B. Groeger
Jan 7 at 22:10
add a comment |
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1 Answer
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1 Answer
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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 $thetain[0,2pi)$.
answered Jan 7 at 21:53
BatominovskiBatominovski
1
$endgroup$
$begingroup$
Thanks very much for your efforts! The answer is very insightful.
$endgroup$
– B. Groeger
Jan 7 at 22:10
add a comment |
$begingroup$
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 $thetain[0,2pi)$.
answered Jan 7 at 21:53
BatominovskiBatominovski
1
$endgroup$
$begingroup$
Thanks very much for your efforts! The answer is very insightful.
$endgroup$
– B. Groeger
Jan 7 at 22:10
add a comment |
$begingroup$
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 $thetain[0,2pi)$.
answered Jan 7 at 21:53
BatominovskiBatominovski
1
$endgroup$
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 $thetain[0,2pi)$.
answered Jan 7 at 21:53
BatominovskiBatominovski
1
edited Jan 7 at 22:11
answered Jan 7 at 21:53
BatominovskiBatominovski
1
answered Jan 7 at 21:53
BatominovskiBatominovski
1
answered Jan 7 at 21:53
BatominovskiBatominovski
1
1
$begingroup$
Thanks very much for your efforts! The answer is very insightful.
$endgroup$
– B. Groeger
Jan 7 at 22:10
add a comment |
$begingroup$
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
add a comment |
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