relationship between complex numbers
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Consider the following:
- Two equilateral triangles inscribed in a circle.
- The vertices of the large triangle are the geometric images of the three cubic roots of $z$ (a complex number).
- The small triangle vertices are the midpoints of the sides of the larger triangle.
- The vertices of the small triangle are the geometric images of the three cubic roots of $w$ (another complex number).
What is the relationship between $w$ and $z$?
geometry complex-numbers
edited Jun 19 '14 at 15:41
sds
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
add a comment |Â
up vote
4
down vote
favorite
Consider the following:
- Two equilateral triangles inscribed in a circle.
- The vertices of the large triangle are the geometric images of the three cubic roots of $z$ (a complex number).
- The small triangle vertices are the midpoints of the sides of the larger triangle.
- The vertices of the small triangle are the geometric images of the three cubic roots of $w$ (another complex number).
What is the relationship between $w$ and $z$?
geometry complex-numbers
edited Jun 19 '14 at 15:41
sds
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Consider the following:
- Two equilateral triangles inscribed in a circle.
- The vertices of the large triangle are the geometric images of the three cubic roots of $z$ (a complex number).
- The small triangle vertices are the midpoints of the sides of the larger triangle.
- The vertices of the small triangle are the geometric images of the three cubic roots of $w$ (another complex number).
What is the relationship between $w$ and $z$?
geometry complex-numbers
edited Jun 19 '14 at 15:41
sds
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
Consider the following:
- Two equilateral triangles inscribed in a circle.
- The vertices of the large triangle are the geometric images of the three cubic roots of $z$ (a complex number).
- The small triangle vertices are the midpoints of the sides of the larger triangle.
- The vertices of the small triangle are the geometric images of the three cubic roots of $w$ (another complex number).
What is the relationship between $w$ and $z$?
geometry complex-numbers
geometry complex-numbers
edited Jun 19 '14 at 15:41
sds
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
edited Jun 19 '14 at 15:41
sds
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
edited Jun 19 '14 at 15:41
sds
3,3981127
edited Jun 19 '14 at 15:41
sds
3,3981127
edited Jun 19 '14 at 15:41
sds
3,3981127
3,3981127
asked Jun 19 '14 at 15:28
Jorge
1685
asked Jun 19 '14 at 15:28
Jorge
1685
asked Jun 19 '14 at 15:28
Jorge
1685
1685
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Let $a^3=z$, $b^3=w$, and $r^3=1$, where $r = expfrac2ipi3$ is the primitive cubic root of unity.
Then the vertices of the outer triangle are $a, ar, ar^2$, those of the inner triangle are $b, br, br^2$, and, wlog, $b=fraca+ar2$.
Now, $$w=b^3=a^3(frac1+r2)^3=zfrac1+3r+3r^2+r^38=-fracz8$$
because $r^3=1$ and $1+r+r^2=frac1-r^31-r=0$.
answered Jun 19 '14 at 15:39
sds
3,3981127
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let $a^3=z$, $b^3=w$, and $r^3=1$, where $r = expfrac2ipi3$ is the primitive cubic root of unity.
Then the vertices of the outer triangle are $a, ar, ar^2$, those of the inner triangle are $b, br, br^2$, and, wlog, $b=fraca+ar2$.
Now, $$w=b^3=a^3(frac1+r2)^3=zfrac1+3r+3r^2+r^38=-fracz8$$
because $r^3=1$ and $1+r+r^2=frac1-r^31-r=0$.
answered Jun 19 '14 at 15:39
sds
3,3981127
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
add a comment |Â
up vote
3
down vote
accepted
Let $a^3=z$, $b^3=w$, and $r^3=1$, where $r = expfrac2ipi3$ is the primitive cubic root of unity.
Then the vertices of the outer triangle are $a, ar, ar^2$, those of the inner triangle are $b, br, br^2$, and, wlog, $b=fraca+ar2$.
Now, $$w=b^3=a^3(frac1+r2)^3=zfrac1+3r+3r^2+r^38=-fracz8$$
because $r^3=1$ and $1+r+r^2=frac1-r^31-r=0$.
answered Jun 19 '14 at 15:39
sds
3,3981127
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let $a^3=z$, $b^3=w$, and $r^3=1$, where $r = expfrac2ipi3$ is the primitive cubic root of unity.
Then the vertices of the outer triangle are $a, ar, ar^2$, those of the inner triangle are $b, br, br^2$, and, wlog, $b=fraca+ar2$.
Now, $$w=b^3=a^3(frac1+r2)^3=zfrac1+3r+3r^2+r^38=-fracz8$$
because $r^3=1$ and $1+r+r^2=frac1-r^31-r=0$.
answered Jun 19 '14 at 15:39
sds
3,3981127
Let $a^3=z$, $b^3=w$, and $r^3=1$, where $r = expfrac2ipi3$ is the primitive cubic root of unity.
Then the vertices of the outer triangle are $a, ar, ar^2$, those of the inner triangle are $b, br, br^2$, and, wlog, $b=fraca+ar2$.
Now, $$w=b^3=a^3(frac1+r2)^3=zfrac1+3r+3r^2+r^38=-fracz8$$
because $r^3=1$ and $1+r+r^2=frac1-r^31-r=0$.
answered Jun 19 '14 at 15:39
sds
3,3981127
edited Jun 19 '14 at 16:10
answered Jun 19 '14 at 15:39
sds
3,3981127
answered Jun 19 '14 at 15:39
sds
3,3981127
answered Jun 19 '14 at 15:39
sds
3,3981127
3,3981127
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
add a comment |Â
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
Depending on how $a$ and $b$ are located, you may also get $b=rfraca+ar2$ or $b= r^2fraca+ar2$, but the term not influence the result anyway, since you take the third power afterwards.
– mlk
Jun 19 '14 at 15:46
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
I'm lost. Why $r^3=1$ and $1+r+r^2=0$?
– Jorge
Jun 19 '14 at 15:55
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
$r$ is the primitive cubic root of unity. See edit.
– sds
Jun 19 '14 at 15:56
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
Thanks. This statement is true? for a square the primitive root of unity should be r=exp $Pi i over 4$
– Jorge
Jun 19 '14 at 16:03
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
The $n$-th primitive root of unity if $expfrac2ipin$.
– sds
Jun 19 '14 at 16:09
add a comment |Â
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