“Converse” to the theorem “sum of roots of unity equal 0”
6
$begingroup$
It is well known that sum of roots of unity equal 0. However, if $sum_j exp(i phi_j)=0$, can we say something about the relation between the $exp(i phi_j)$'s? For example, we can rotate one of them to the position of 1, can we say that the other units, at least a subgroup of them, fall on the roots of unity?
Suppose we take the angles $0, pi/3, pi, 4 pi/3$. Then $exp(0) + exp(i pi/3) + exp(i pi) + exp(i 4pi/3) = 0$, but we can separate these units into the sets ${0, pi}$ and ${pi/3, 4pi/3}$, the former of which correspond to roots of unity and the latter correspond to roots of unity after rotation.
My feeling is that there is an obvious counter-example, but so far I haven't found it. Any thoughts are appreciated.
roots-of-unity
edited Dec 21 '18 at 12:57
Did
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
$endgroup$
add a comment |
6
$begingroup$
It is well known that sum of roots of unity equal 0. However, if $sum_j exp(i phi_j)=0$, can we say something about the relation between the $exp(i phi_j)$'s? For example, we can rotate one of them to the position of 1, can we say that the other units, at least a subgroup of them, fall on the roots of unity?
Suppose we take the angles $0, pi/3, pi, 4 pi/3$. Then $exp(0) + exp(i pi/3) + exp(i pi) + exp(i 4pi/3) = 0$, but we can separate these units into the sets ${0, pi}$ and ${pi/3, 4pi/3}$, the former of which correspond to roots of unity and the latter correspond to roots of unity after rotation.
My feeling is that there is an obvious counter-example, but so far I haven't found it. Any thoughts are appreciated.
roots-of-unity
edited Dec 21 '18 at 12:57
Did
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
$endgroup$
$begingroup$
I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
$endgroup$
– YiFan
Dec 21 '18 at 13:56
$begingroup$
By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
$endgroup$
– enochk.
Dec 21 '18 at 15:02
$begingroup$
I see. Sorry, I misunderstood on my first reading.
$endgroup$
– YiFan
Dec 21 '18 at 15:03
add a comment |
6
6
6
2
$begingroup$
It is well known that sum of roots of unity equal 0. However, if $sum_j exp(i phi_j)=0$, can we say something about the relation between the $exp(i phi_j)$'s? For example, we can rotate one of them to the position of 1, can we say that the other units, at least a subgroup of them, fall on the roots of unity?
Suppose we take the angles $0, pi/3, pi, 4 pi/3$. Then $exp(0) + exp(i pi/3) + exp(i pi) + exp(i 4pi/3) = 0$, but we can separate these units into the sets ${0, pi}$ and ${pi/3, 4pi/3}$, the former of which correspond to roots of unity and the latter correspond to roots of unity after rotation.
My feeling is that there is an obvious counter-example, but so far I haven't found it. Any thoughts are appreciated.
roots-of-unity
edited Dec 21 '18 at 12:57
Did
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
$endgroup$
It is well known that sum of roots of unity equal 0. However, if $sum_j exp(i phi_j)=0$, can we say something about the relation between the $exp(i phi_j)$'s? For example, we can rotate one of them to the position of 1, can we say that the other units, at least a subgroup of them, fall on the roots of unity?
Suppose we take the angles $0, pi/3, pi, 4 pi/3$. Then $exp(0) + exp(i pi/3) + exp(i pi) + exp(i 4pi/3) = 0$, but we can separate these units into the sets ${0, pi}$ and ${pi/3, 4pi/3}$, the former of which correspond to roots of unity and the latter correspond to roots of unity after rotation.
My feeling is that there is an obvious counter-example, but so far I haven't found it. Any thoughts are appreciated.
roots-of-unity
roots-of-unity
edited Dec 21 '18 at 12:57
Did
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
edited Dec 21 '18 at 12:57
Did
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
edited Dec 21 '18 at 12:57
Did
249k23228466
edited Dec 21 '18 at 12:57
Did
249k23228466
edited Dec 21 '18 at 12:57
Did
249k23228466
249k23228466
asked Dec 21 '18 at 11:27
enochk.enochk.
461
asked Dec 21 '18 at 11:27
enochk.enochk.
461
asked Dec 21 '18 at 11:27
enochk.enochk.
461
461
$begingroup$
I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
$endgroup$
– YiFan
Dec 21 '18 at 13:56
$begingroup$
By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
$endgroup$
– enochk.
Dec 21 '18 at 15:02
$begingroup$
I see. Sorry, I misunderstood on my first reading.
$endgroup$
– YiFan
Dec 21 '18 at 15:03
add a comment |
$begingroup$
I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
$endgroup$
– YiFan
Dec 21 '18 at 13:56
$begingroup$
By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
$endgroup$
– enochk.
Dec 21 '18 at 15:02
$begingroup$
I see. Sorry, I misunderstood on my first reading.
$endgroup$
– YiFan
Dec 21 '18 at 15:03
$begingroup$
I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
$endgroup$
– YiFan
Dec 21 '18 at 13:56
$begingroup$
I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
$endgroup$
– YiFan
Dec 21 '18 at 13:56
$begingroup$
By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
$endgroup$
– enochk.
Dec 21 '18 at 15:02
$begingroup$
By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
$endgroup$
– enochk.
Dec 21 '18 at 15:02
$begingroup$
I see. Sorry, I misunderstood on my first reading.
$endgroup$
– YiFan
Dec 21 '18 at 15:03
$begingroup$
I see. Sorry, I misunderstood on my first reading.
$endgroup$
– YiFan
Dec 21 '18 at 15:03
add a comment |
1 Answer
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active
oldest
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3
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Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, exp(i phi_1), exp(i phi_1) + exp(i phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
$endgroup$
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
$begingroup$
@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
$endgroup$
– Qiaochu Yuan
Dec 21 '18 at 23:22
$begingroup$
Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
$endgroup$
– enochk.
Dec 22 '18 at 0:06
1
$begingroup$
@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
$endgroup$
– Andreas Blass
Dec 22 '18 at 0:24
$begingroup$
Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
$endgroup$
– enochk.
Dec 22 '18 at 1:14
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, exp(i phi_1), exp(i phi_1) + exp(i phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
$endgroup$
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
$begingroup$
@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
$endgroup$
– Qiaochu Yuan
Dec 21 '18 at 23:22
$begingroup$
Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
$endgroup$
– enochk.
Dec 22 '18 at 0:06
1
$begingroup$
@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
$endgroup$
– Andreas Blass
Dec 22 '18 at 0:24
$begingroup$
Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
$endgroup$
– enochk.
Dec 22 '18 at 1:14
add a comment |
3
$begingroup$
Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, exp(i phi_1), exp(i phi_1) + exp(i phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
$endgroup$
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
$begingroup$
@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
$endgroup$
– Qiaochu Yuan
Dec 21 '18 at 23:22
$begingroup$
Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
$endgroup$
– enochk.
Dec 22 '18 at 0:06
1
$begingroup$
@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
$endgroup$
– Andreas Blass
Dec 22 '18 at 0:24
$begingroup$
Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
$endgroup$
– enochk.
Dec 22 '18 at 1:14
add a comment |
3
3
3
$begingroup$
Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, exp(i phi_1), exp(i phi_1) + exp(i phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
$endgroup$
Geometrically such a sum corresponds to a polygon (not necessarily convex, possibly self-intersecting) all of whose sides have length $1$; the vertices of the polygon are $0, exp(i phi_1), exp(i phi_1) + exp(i phi_2)$, etc. The angles involved need not be roots of unity; for example it could be a rhombus. The case of all of the roots of unity summing to zero corresponds to a regular $n$-gon.
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
edited Dec 21 '18 at 23:01
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
answered Dec 21 '18 at 22:55
Qiaochu YuanQiaochu Yuan
282k32597944
282k32597944
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
$begingroup$
@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
$endgroup$
– Qiaochu Yuan
Dec 21 '18 at 23:22
$begingroup$
Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
$endgroup$
– enochk.
Dec 22 '18 at 0:06
1
$begingroup$
@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
$endgroup$
– Andreas Blass
Dec 22 '18 at 0:24
$begingroup$
Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
$endgroup$
– enochk.
Dec 22 '18 at 1:14
add a comment |
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
$begingroup$
@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
$endgroup$
– Qiaochu Yuan
Dec 21 '18 at 23:22
$begingroup$
Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
$endgroup$
– enochk.
Dec 22 '18 at 0:06
1
$begingroup$
@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
$endgroup$
– Andreas Blass
Dec 22 '18 at 0:24
$begingroup$
Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
$endgroup$
– enochk.
Dec 22 '18 at 1:14
$begingroup$
Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
$endgroup$
– enochk.
Dec 21 '18 at 23:17
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Thanks for the comment. First, if self-intersecting, rearrange until not self intersecting. Second, for $exp(iphi_1)+exp(iphi_2)$ to be a vertex, $phi_2$ is uniquely defined by $phi_1$ if vertices are ordered counterclockwise. Third, the vertices do not have to correspond to one complete set of roots of unity ( that is $exp(2pi i n /N)$ for some $N$) but they might be a complete set after a rotation, or the may be made up of several complete sets of roots of unity for different $N$'s. Fourth, the only cyclic rhombus is a square, which correspond to the roots of unity of N=4.
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– enochk.
Dec 21 '18 at 23:17
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@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
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– Qiaochu Yuan
Dec 21 '18 at 23:22
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@enoch: I don't follow your last comment. There's no need for the polygon to be cyclic. Very explicitly, consider $1 + z + (-1) + (-z)$ for any $z = exp(i phi)$.
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– Qiaochu Yuan
Dec 21 '18 at 23:22
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Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
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– enochk.
Dec 22 '18 at 0:06
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Your example is a cyclic polygon and not technically a rhombus, but I get your point. The polygon must be cyclic by definition because its vertices are on the circle. Also what you proposed is made of two sets ${1,-1}$ and ${z, -z}$. They are both roots of unity for $N=2$ only that the second set requires a rotation. You can see that I used the same example with $z=exp(i pi/3)$ in the main post.
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– enochk.
Dec 22 '18 at 0:06
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@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
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– Andreas Blass
Dec 22 '18 at 0:24
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@enochk. You seem to have misunderstood what polygon Qiaochu is referring to. When a sum $a_0+a_1+dots+a_n$ is $0$, the intended polygon's vertices are not $a_1,a_2,dots,a_n$ but rather $0, a_1,a_1+a_2,a_1+a_2+a_3,dots,a_1+a_2+dots+a_{n-1}$. So the example in his comment is indeed a rhombus and, for most values of $z$, not cyclic.
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– Andreas Blass
Dec 22 '18 at 0:24
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Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
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– enochk.
Dec 22 '18 at 1:14
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Thanks for the clarification. However, the example given is still not a counter-example to my main query as the example is a sum of two complete sets of roots of unity for $N=2$; $z$ and $-z$ are simply a rotated set of square roots of unity. Expanding on Qiaochu's visualisation for a sum of 3 terms, if $exp(i phi_1)+exp(i phi_2) + exp(i phi_3)=0$, the partial sums must correspond to the vertices of an equilateral triangle, hence taking WLOG $phi_1=0$, the other points must be $zeta_3$ and $zeta_3^2$.
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– enochk.
Dec 22 '18 at 1:14
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I'm confused. $exp(ipi/3)$ and $exp(4ipi/3)$ are both cube roots of unity. Are you trying to restrict to square roots?
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– YiFan
Dec 21 '18 at 13:56
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By sum of roots of unity, of course it refers to the set of all roots of unity corresponding to some number $n$, so in this case I mean that $pi/3$ and $4pi/3$ belong to a complete set of roots of unity corresponding to 2, after a rotation. Of course it needn't only be square roots.
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– enochk.
Dec 21 '18 at 15:02
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I see. Sorry, I misunderstood on my first reading.
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– YiFan
Dec 21 '18 at 15:03