Mixing
Solve this system of equations with modulus in $mathbb{C}$
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I tried to solve this exercise but failed after some calculations, tough I've extracted some important piece of information from it. The system is:
$$ left{begin{array}{c} x|y| = z^2 \ y|z| = x^2 \ z|x| = y^2 end{array}right. $$
where $x,y,z in mathbb C$.
First I tried substituting $z^2$ with $left(frac{y^2}{|x|}right)^2 = frac{y^4}{xcdot overline{x}}$ but I soon gave this idea up as it caused much harder equations to appear. But the I thought of multiplying them all together and I found the following:
$$ xyz|x||y||z| = x^2y^2z^2 \ xyz|xyz| = (xyz)^2 \ |xyz| = xyz \ Im(xyz) = 0 leftrightarrow xyz in mathbb R $$
After this I am stuck. What can I do next to solve the system?
complex-numbers systems-of-equations
asked yesterday
Stefan Octavian
1345
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up vote
1
down vote
favorite
I tried to solve this exercise but failed after some calculations, tough I've extracted some important piece of information from it. The system is:
$$ left{begin{array}{c} x|y| = z^2 \ y|z| = x^2 \ z|x| = y^2 end{array}right. $$
where $x,y,z in mathbb C$.
First I tried substituting $z^2$ with $left(frac{y^2}{|x|}right)^2 = frac{y^4}{xcdot overline{x}}$ but I soon gave this idea up as it caused much harder equations to appear. But the I thought of multiplying them all together and I found the following:
$$ xyz|x||y||z| = x^2y^2z^2 \ xyz|xyz| = (xyz)^2 \ |xyz| = xyz \ Im(xyz) = 0 leftrightarrow xyz in mathbb R $$
After this I am stuck. What can I do next to solve the system?
complex-numbers systems-of-equations
asked yesterday
Stefan Octavian
1345
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I tried to solve this exercise but failed after some calculations, tough I've extracted some important piece of information from it. The system is:
$$ left{begin{array}{c} x|y| = z^2 \ y|z| = x^2 \ z|x| = y^2 end{array}right. $$
where $x,y,z in mathbb C$.
First I tried substituting $z^2$ with $left(frac{y^2}{|x|}right)^2 = frac{y^4}{xcdot overline{x}}$ but I soon gave this idea up as it caused much harder equations to appear. But the I thought of multiplying them all together and I found the following:
$$ xyz|x||y||z| = x^2y^2z^2 \ xyz|xyz| = (xyz)^2 \ |xyz| = xyz \ Im(xyz) = 0 leftrightarrow xyz in mathbb R $$
After this I am stuck. What can I do next to solve the system?
complex-numbers systems-of-equations
asked yesterday
Stefan Octavian
1345
I tried to solve this exercise but failed after some calculations, tough I've extracted some important piece of information from it. The system is:
$$ left{begin{array}{c} x|y| = z^2 \ y|z| = x^2 \ z|x| = y^2 end{array}right. $$
where $x,y,z in mathbb C$.
First I tried substituting $z^2$ with $left(frac{y^2}{|x|}right)^2 = frac{y^4}{xcdot overline{x}}$ but I soon gave this idea up as it caused much harder equations to appear. But the I thought of multiplying them all together and I found the following:
$$ xyz|x||y||z| = x^2y^2z^2 \ xyz|xyz| = (xyz)^2 \ |xyz| = xyz \ Im(xyz) = 0 leftrightarrow xyz in mathbb R $$
After this I am stuck. What can I do next to solve the system?
complex-numbers systems-of-equations
complex-numbers systems-of-equations
asked yesterday
Stefan Octavian
1345
asked yesterday
Stefan Octavian
1345
asked yesterday
Stefan Octavian
1345
asked yesterday
Stefan Octavian
1345
asked yesterday
Stefan Octavian
1345
1345
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
First note that if $x, y, z$ are solutions to your system, then you must have $|x| = |y| = |z|$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $x = y = z = 0$ -- note that for any solution $x, y, z$, and any positive $alpha$, also $alpha x, alpha y, alpha z$ is a solution. Thus we can take $alpha = 1/|x|$ and reduce to the case where $|x| = |y| = |z|$, which considerably simplifies your equations to
$$
x = z^2,\
y = x^2,\
z = y^2.
$$
Now we can do repeated substitution, and obtain that $x^7 = y^7 = z^7 = 1$. Each of $x, y, z$ is a 7th root of unity. And now you can simply check: if $x$ is any seventh root of unity, and we set $y = x^2$, and $z = y^2$, then the latter two equations are clearly satisfied, and $z^2 = y^4 = x^8 = x$, so therefore so is the first.
Hence the solutions to the equation are $alpha times exp(2pi i tfrac k7)$, for $k in {0,1,2,3,4,5,6}$, and $alpha$ non-negative real.
answered yesterday
Mees de Vries
16.4k12654
add a comment |
up vote
1
down vote
Things to try:
$x|y| = z^2 implies |x| |y| = |z|^2$, which implies $|xyz|=|x|^3=|y|^3=|z|^3$.
$x|y| = z^2 , y|z| = x^2 implies x^3 = y z^2$
answered yesterday
lhf
161k9164383
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First note that if $x, y, z$ are solutions to your system, then you must have $|x| = |y| = |z|$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $x = y = z = 0$ -- note that for any solution $x, y, z$, and any positive $alpha$, also $alpha x, alpha y, alpha z$ is a solution. Thus we can take $alpha = 1/|x|$ and reduce to the case where $|x| = |y| = |z|$, which considerably simplifies your equations to
$$
x = z^2,\
y = x^2,\
z = y^2.
$$
Now we can do repeated substitution, and obtain that $x^7 = y^7 = z^7 = 1$. Each of $x, y, z$ is a 7th root of unity. And now you can simply check: if $x$ is any seventh root of unity, and we set $y = x^2$, and $z = y^2$, then the latter two equations are clearly satisfied, and $z^2 = y^4 = x^8 = x$, so therefore so is the first.
Hence the solutions to the equation are $alpha times exp(2pi i tfrac k7)$, for $k in {0,1,2,3,4,5,6}$, and $alpha$ non-negative real.
answered yesterday
Mees de Vries
16.4k12654
add a comment |
up vote
2
down vote
accepted
First note that if $x, y, z$ are solutions to your system, then you must have $|x| = |y| = |z|$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $x = y = z = 0$ -- note that for any solution $x, y, z$, and any positive $alpha$, also $alpha x, alpha y, alpha z$ is a solution. Thus we can take $alpha = 1/|x|$ and reduce to the case where $|x| = |y| = |z|$, which considerably simplifies your equations to
$$
x = z^2,\
y = x^2,\
z = y^2.
$$
Now we can do repeated substitution, and obtain that $x^7 = y^7 = z^7 = 1$. Each of $x, y, z$ is a 7th root of unity. And now you can simply check: if $x$ is any seventh root of unity, and we set $y = x^2$, and $z = y^2$, then the latter two equations are clearly satisfied, and $z^2 = y^4 = x^8 = x$, so therefore so is the first.
Hence the solutions to the equation are $alpha times exp(2pi i tfrac k7)$, for $k in {0,1,2,3,4,5,6}$, and $alpha$ non-negative real.
answered yesterday
Mees de Vries
16.4k12654
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First note that if $x, y, z$ are solutions to your system, then you must have $|x| = |y| = |z|$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $x = y = z = 0$ -- note that for any solution $x, y, z$, and any positive $alpha$, also $alpha x, alpha y, alpha z$ is a solution. Thus we can take $alpha = 1/|x|$ and reduce to the case where $|x| = |y| = |z|$, which considerably simplifies your equations to
$$
x = z^2,\
y = x^2,\
z = y^2.
$$
Now we can do repeated substitution, and obtain that $x^7 = y^7 = z^7 = 1$. Each of $x, y, z$ is a 7th root of unity. And now you can simply check: if $x$ is any seventh root of unity, and we set $y = x^2$, and $z = y^2$, then the latter two equations are clearly satisfied, and $z^2 = y^4 = x^8 = x$, so therefore so is the first.
Hence the solutions to the equation are $alpha times exp(2pi i tfrac k7)$, for $k in {0,1,2,3,4,5,6}$, and $alpha$ non-negative real.
answered yesterday
Mees de Vries
16.4k12654
First note that if $x, y, z$ are solutions to your system, then you must have $|x| = |y| = |z|$; you can see this by taking the absolute value of each of your equations, and considering the equation where the largest of the three has its square on the right hand side.
In particular -- setting aside the solution $x = y = z = 0$ -- note that for any solution $x, y, z$, and any positive $alpha$, also $alpha x, alpha y, alpha z$ is a solution. Thus we can take $alpha = 1/|x|$ and reduce to the case where $|x| = |y| = |z|$, which considerably simplifies your equations to
$$
x = z^2,\
y = x^2,\
z = y^2.
$$
Now we can do repeated substitution, and obtain that $x^7 = y^7 = z^7 = 1$. Each of $x, y, z$ is a 7th root of unity. And now you can simply check: if $x$ is any seventh root of unity, and we set $y = x^2$, and $z = y^2$, then the latter two equations are clearly satisfied, and $z^2 = y^4 = x^8 = x$, so therefore so is the first.
Hence the solutions to the equation are $alpha times exp(2pi i tfrac k7)$, for $k in {0,1,2,3,4,5,6}$, and $alpha$ non-negative real.
answered yesterday
Mees de Vries
16.4k12654
answered yesterday
Mees de Vries
16.4k12654
answered yesterday
Mees de Vries
16.4k12654
answered yesterday
Mees de Vries
16.4k12654
16.4k12654
add a comment |
add a comment |
up vote
1
down vote
Things to try:
$x|y| = z^2 implies |x| |y| = |z|^2$, which implies $|xyz|=|x|^3=|y|^3=|z|^3$.
$x|y| = z^2 , y|z| = x^2 implies x^3 = y z^2$
answered yesterday
lhf
161k9164383
add a comment |
up vote
1
down vote
Things to try:
$x|y| = z^2 implies |x| |y| = |z|^2$, which implies $|xyz|=|x|^3=|y|^3=|z|^3$.
$x|y| = z^2 , y|z| = x^2 implies x^3 = y z^2$
answered yesterday
lhf
161k9164383
add a comment |
up vote
1
down vote
up vote
1
down vote
Things to try:
$x|y| = z^2 implies |x| |y| = |z|^2$, which implies $|xyz|=|x|^3=|y|^3=|z|^3$.
$x|y| = z^2 , y|z| = x^2 implies x^3 = y z^2$
answered yesterday
lhf
161k9164383
Things to try:
$x|y| = z^2 implies |x| |y| = |z|^2$, which implies $|xyz|=|x|^3=|y|^3=|z|^3$.
$x|y| = z^2 , y|z| = x^2 implies x^3 = y z^2$
answered yesterday
lhf
161k9164383
edited 22 hours ago
answered yesterday
lhf
161k9164383
answered yesterday
lhf
161k9164383
answered yesterday
lhf
161k9164383
161k9164383
add a comment |
add a comment |
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