Discussion:
z_1 + ... + z_5 = 0
(too old to reply)
quasi
2015-07-11 01:43:06 UTC
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Conjecture:

If z_1,...,z_5 are complex numbers such that

(1) For some positive integer n, (z_k)^n = 1 for all k.

(2) z_1 + ... + z_5 = 0

(3) No two of z_1,...,z_5 sum to zero

then {z_1,...,z_5} is the set of vertices of a regular
pentagon.

quasi
a***@gmail.com
2015-07-11 02:54:21 UTC
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I'll use different coordinates;
the board is in space
Phil Carmody
2015-07-12 11:42:08 UTC
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Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
corresponds to:
(3') No proper subset of the z_i sums to zero

Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)

Phil
--
A well regulated militia, being necessary to the security of a free state,
the right of the people to keep and bear arms, shall be well regulated.
quasi
2015-07-12 20:28:56 UTC
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Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
I think you meant "into 2 triplets".

quasi
Peter Percival
2015-07-13 12:51:45 UTC
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Post by quasi
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
I think you meant "into 2 triplets".
quasi
In the case where 5 = 3, the result is known as the Three Party Theorem,
is it not? So the question arises, for which N is there an N Party
Theorem? I guess "N is the power of a prime" is the answer.
Phil Carmody
2015-07-13 15:00:57 UTC
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Post by quasi
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
I think you meant "into 2 triplets".
Well, when 5's 6, the 3s have to be 2s to compensate!

Phil
--
A well regulated militia, being necessary to the security of a free state,
the right of the people to keep and bear arms, shall be well regulated.
Leon Aigret
2015-07-13 10:56:29 UTC
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Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
If I understand this correctly, one counter-example for both (3) and
(3') would consist of the six numbers exp(r 2 pi i) with r = 1/6, 1/5,
2/5, 3/5, 4/5 amd 5/6.

On the other hand, there are no counter-examples where the n in
condition (1) is a power of a prime number.

Leon
quasi
2015-07-13 11:23:24 UTC
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Post by Leon Aigret
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is
not a counter-example under (3'), as it decomposes into 3
triplets each of which sums to zero.)
If I understand this correctly, one counter-example for both
(3) and (3') would consist of the six numbers exp(r 2 pi i)
with r = 1/6, 1/5, 2/5, 3/5, 4/5 and 5/6.
Nice.
Post by Leon Aigret
On the other hand, there are no counter-examples where the n
in condition (1) is a power of a prime number.
Can you actually prove that, or is it just a conjecture based
on the lack of an obvious counterexample?

At this point, we might as well replace (3) by (3') since
that's what I had in mind anyway.

quasi
Leon Aigret
2015-07-13 12:33:12 UTC
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Post by quasi
Post by Leon Aigret
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is
not a counter-example under (3'), as it decomposes into 3
triplets each of which sums to zero.)
If I understand this correctly, one counter-example for both
(3) and (3') would consist of the six numbers exp(r 2 pi i)
with r = 1/6, 1/5, 2/5, 3/5, 4/5 and 5/6.
Nice.
Post by Leon Aigret
On the other hand, there are no counter-examples where the n
in condition (1) is a power of a prime number.
Can you actually prove that, or is it just a conjecture based
on the lack of an obvious counterexample?
The proof uses the minimal polynomial in Q[X] for exp(2 pi i / p^m) to
show that zero sums only occur for (combinations of) regular p-gons.
Post by quasi
At this point, we might as well replace (3) by (3') since
that's what I had in mind anyway.
quasi
leon
quasi
2015-07-13 12:59:34 UTC
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Post by Leon Aigret
Post by quasi
Post by Leon Aigret
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is
not a counter-example under (3'), as it decomposes into 3
triplets each of which sums to zero.)
If I understand this correctly, one counter-example for both
(3) and (3') would consist of the six numbers exp(r 2 pi i)
with r = 1/6, 1/5, 2/5, 3/5, 4/5 and 5/6.
Nice.
Post by Leon Aigret
On the other hand, there are no counter-examples where the n
in condition (1) is a power of a prime number.
Can you actually prove that, or is it just a conjecture based
on the lack of an obvious counterexample?
The proof uses the minimal polynomial in Q[X] for
exp(2 pi i / p^m) to show that zero sums only occur for
(combinations of) regular p-gons.
The proof is easy for the case where n is prime.

I don't quite see it for the case where n = p^m, m > 1.

Can you give a further hint?

quasi
Leon Aigret
2015-07-13 14:00:42 UTC
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.
.
.
Post by quasi
Post by Leon Aigret
Post by quasi
Post by Leon Aigret
On the other hand, there are no counter-examples where the n
in condition (1) is a power of a prime number.
Can you actually prove that, or is it just a conjecture based
on the lack of an obvious counterexample?
The proof uses the minimal polynomial in Q[X] for
exp(2 pi i / p^m) to show that zero sums only occur for
(combinations of) regular p-gons.
The proof is easy for the case where n is prime.
I don't quite see it for the case where n = p^m, m > 1.
Can you give a further hint?
It's about a polynomial of degree < p^m that is divisible by

X^(p^m - p^(m-1)) + X^(p^m - 2 p^(m-1)) + ... + X^(p^(m-1)) + 1.

That puts an upper limit on the degree of the quotient.

Leon
quasi
2015-07-13 14:21:22 UTC
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Post by Leon Aigret
Post by quasi
Post by Leon Aigret
The proof uses the minimal polynomial in Q[X] for
exp(2 pi i / p^m) to show that zero sums only occur for
(combinations of) regular p-gons.
The proof is easy for the case where n is prime.
I don't quite see it for the case where n = p^m, m > 1.
Can you give a further hint?
It's about a polynomial of degree < p^m that is divisible by
X^(p^m - p^(m-1)) + X^(p^m - 2 p^(m-1)) + ... + X^(p^(m-1)) + 1.
That puts an upper limit on the degree of the quotient.
The quotient must be monic of degree less than p^(m-1).

Thus, when m = 1, then the quotient must be 1 and the result
is immediate.

I still don't see it for the case m > 1.

quasi
Leon Aigret
2015-07-13 15:13:29 UTC
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Post by quasi
Post by Leon Aigret
Post by quasi
Post by Leon Aigret
The proof uses the minimal polynomial in Q[X] for
exp(2 pi i / p^m) to show that zero sums only occur for
(combinations of) regular p-gons.
The proof is easy for the case where n is prime.
I don't quite see it for the case where n = p^m, m > 1.
Can you give a further hint?
It's about a polynomial of degree < p^m that is divisible by
X^(p^m - p^(m-1)) + X^(p^m - 2 p^(m-1)) + ... + X^(p^(m-1)) + 1.
That puts an upper limit on the degree of the quotient.
The quotient must be monic of degree less than p^(m-1).
Thus, when m = 1, then the quotient must be 1 and the result
is immediate.
Apologies for being so vague. It's probably one of those things that
only become obvious after they have been pointed out:

For m > 1 the product of the quotient and the term X^(j p^(m-1)) of

the minimal polynomial contains terms of degree d with

X^(j p^(m-1)) <= d < X^((j+1) p^(m-1)), so the individual products do

not "overlap" and the coefficients of the complete product show a

repetitive pattern with period p^(m-1).

That describes a combination of regular p-gons.

Leon
quasi
2015-07-13 15:34:05 UTC
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Post by Leon Aigret
Post by quasi
Post by Leon Aigret
Post by quasi
Post by Leon Aigret
The proof uses the minimal polynomial in Q[X] for
exp(2 pi i / p^m) to show that zero sums only occur for
(combinations of) regular p-gons.
The proof is easy for the case where n is prime.
I don't quite see it for the case where n = p^m, m > 1.
Can you give a further hint?
It's about a polynomial of degree < p^m that is divisible by
X^(p^m - p^(m-1)) + X^(p^m - 2 p^(m-1)) + ... + X^(p^(m-1)) + 1.
That puts an upper limit on the degree of the quotient.
The quotient must be monic of degree less than p^(m-1).
Thus, when m = 1, then the quotient must be 1 and the result
is immediate.
Apologies for being so vague. It's probably one of those things that
For m > 1 the product of the quotient and the term X^(j p^(m-1)) of
the minimal polynomial contains terms of degree d with
X^(j p^(m-1)) <= d < X^((j+1) p^(m-1)), so the individual products do
not "overlap" and the coefficients of the complete product show a
repetitive pattern with period p^(m-1).
That describes a combination of regular p-gons.
Oh, that's very nice!

I missed the "no overlap" part, which, given the upper bound
for the degree of the quotient, is now quite clear.

Thanks a lot.

quasi
Phil Carmody
2015-07-13 15:04:49 UTC
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Post by Leon Aigret
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
If I understand this correctly, one counter-example for both (3) and
(3') would consist of the six numbers exp(r 2 pi i) with r = 1/6, 1/5,
2/5, 3/5, 4/5 amd 5/6.
Nope, that's 3 pairs that sum to zero, so doesn't satisfy either
(3) or (3').
Post by Leon Aigret
On the other hand, there are no counter-examples where the n in
condition (1) is a power of a prime number.
n=9. r=0/9, 1/9, 3/9, 4/9, 6/9, 7/9 satisfies (3) but not (3').

Phil
--
A well regulated militia, being necessary to the security of a free state,
the right of the people to keep and bear arms, shall be well regulated.
quasi
2015-07-13 15:26:30 UTC
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Post by Phil Carmody
Post by Leon Aigret
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable),
To ease future discussion let's call that variable k.

In other words, z_1,...,z_k, for some choice of k.
Post by Phil Carmody
Post by Leon Aigret
(3') No proper subset of the z_i sums to zero
(3') was what I really had in mind, anyway.

So from this point on, to keep the notation simple, let's
redefine (3) to be (3') (thus no need for the ' symbol).
Post by Phil Carmody
Post by Leon Aigret
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
If I understand this correctly, one counter-example for both
(3) and (3') would consist of the six numbers exp(r 2 pi i)
with r = 1/6, 1/5, 2/5, 3/5, 4/5 and 5/6.
Nope, that's 3 pairs that sum to zero,
Hmmm ...

I see that all 6 numbers sum to zero.

I don't see any pairs that sum to zero.

quasi
Leon Aigret
2015-07-13 16:30:46 UTC
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Post by Phil Carmody
Post by Leon Aigret
Post by Phil Carmody
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
As 5 (which I'm now treating as a variable) <= 5, then (3)
(3') No proper subset of the z_i sums to zero
Which permits the puzzle to be extended to other values of 5.
(There is a counter example when 5 is 6, under (3), which is not
a counter-example under (3'), as it decomposes into 3 triplets
each of which sums to zero.)
If I understand this correctly, one counter-example for both (3) and
(3') would consist of the six numbers exp(r 2 pi i) with r = 1/6, 1/5,
2/5, 3/5, 4/5 amd 5/6.
Nope, that's 3 pairs that sum to zero, so doesn't satisfy either
(3) or (3').
As many old fortan programmers know, strange things can happen when 5
is treated as a variable, but with the conventional interpretation
that counter-example consists of all 5th roots of unity and the
negatives of all 3rd roots of unity, with 1 and -1 removed. I don't
see any pair that sums to zero.
Post by Phil Carmody
Post by Leon Aigret
On the other hand, there are no counter-examples where the n in
condition (1) is a power of a prime number.
n=9. r=0/9, 1/9, 3/9, 4/9, 6/9, 7/9 satisfies (3) but not (3').
I should have stated explicitly that this was only meant to refer to
the (3') case.

Leon
Simon Roberts
2015-07-12 15:12:55 UTC
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Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
|z_k| must be 1
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (1<(360/n)) = 0 (a.)

Can't be.

This is difficult to understand.

does each z_k have a different angle (360/n_k) n = n_k for each k = 1 to 5

It must. Therefore (angles relative to fixed axis)

SUM_1 to 5 (1<(360/n_k)) = 0 (a.)

regular pentagon external angles (not relative to fixed axis) sum to:

SUM_1 to 5 (180-u')) = 360 (c.1)

where u' is the the the internal angle = 108. external angle v' = 72 = 360/5

recall
SUM_1 to 5 (z_k) = 0 ( a given)
or
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)

changing the regular pentagon relative to a
pair of fixed axis with one side parallel to x axis

SUM_k=1 to 5 (1<(k*v')) = 0 (d.1)

that is z_k = 1<((v')*(k)) = 1<((72)*(k))=1<((360/5)*(k)) which doesn't jive with 1<(360/n_k)
u***@gmail.com
2015-07-12 15:27:13 UTC
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Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
|z_k| must be 1
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (1<(360/n)) = 0 (a.)
Can't be.
This is difficult to understand.
does each z_k have a different angle (360/n_k) n = n_k for each k = 1 to 5
It must. Therefore (angles relative to fixed axis)
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
SUM_1 to 5 (180-u')) = 360 (c.1)
where u' is the the the internal angle = 108. external angle v' = 72 = 360/5
recall
SUM_1 to 5 (z_k) = 0 ( a given)
or
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
changing the regular pentagon relative to a
pair of fixed axis with one side parallel to x axis
The base for convenience
Post by Simon Roberts
SUM_k=1 to 5 (1<(k*v')) = 0 (d.1)
that is z_k = 1<((v')*(k)) = 1<((72)*(k))=1<((360/5)*(k)) which doesn't jive with 1<(360/n_k)
u***@gmail.com
2015-07-12 16:40:47 UTC
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Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
|z_k| must be 1
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (1<(360/n)) = 0 (a.)
Can't be.
This is difficult to understand.
does each z_k have a different angle (360/n_k) n = n_k for each k = 1 to 5
It must. Therefore (angles relative to fixed axis)
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
SUM_1 to 5 (180-u')) = 360 (c.1)
where u' is the the the internal angle = 108. external angle v' = 72 = 360/5
recall
SUM_1 to 5 (z_k) = 0 ( a given)
or
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
changing the regular pentagon relative to a
pair of fixed axis with one side parallel to x axis
SUM_k=1 to 5 (1<(k*v')) = 0 (d.1)
I think I know now what you meant:

SUM_k=1 to R (G<(k*360/R)) = 0
where [G<(360/R)]^R = 1 (if G=1)
and 360/R is the external angle for each vertex.
Post by Simon Roberts
that is z_k = 1<((v')*(k)) = 1<((72)*(k))=1<((360/5)*(k)) which doesn't jive with 1<(360/n_k)
u***@gmail.com
2015-07-13 17:01:36 UTC
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Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
|z_k| must be 1
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (1<(360/n)) = 0 (a.)
Can't be.
This is difficult to understand.
does each z_k have a different angle (360/n_k) n = n_k for each k = 1 to 5
It must. Therefore (angles relative to fixed axis)
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
SUM_1 to 5 (180-u')) = 360 (c.1)
where u' is the the the internal angle = 108. external angle v' = 72 = 360/5
recall
SUM_1 to 5 (z_k) = 0 ( a given)
or
SUM_1 to 5 (1<(360/n_k)) = 0 (a.)
changing the regular pentagon relative to a
pair of fixed axis with one side parallel to x axis
SUM_k=1 to 5 (1<(k*v')) = 0 (d.1)
that is z_k = 1<((v')*(k)) = 1<((72)*(k))=1<((360/5)*(k)) which doesn't jive with 1<(360/n_k)
z_k = 1<((v')*(k)) = 1<((72)*(k))=1<((360/5)*(k))

(z_k)^5 = 1. Missed this fact. This no way completes a proof.
Simon Roberts
2017-06-11 15:04:06 UTC
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Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
damn, what do I have here?

I assumed z_k is a complex number.
Post by Simon Roberts
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
This x<y is radius x and angle y. ok.
Post by Simon Roberts
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
wrong. (x<k(pi)/n)^n. x = +/-1.
Post by Simon Roberts
|z_k| must be 1
+/- 1.
Post by Simon Roberts
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (+/-1<(k*360/n)) = 0 (a.)

Five radii of length one a polar graph vectorialy add to zero?

ODD.

no matter what angle.
Peter Percival
2017-06-11 15:19:55 UTC
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Post by Simon Roberts
Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
damn, what do I have here?
I assumed z_k is a complex number.
Post by Simon Roberts
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
This x<y is radius x and angle y. ok.
Post by Simon Roberts
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
wrong. (x<k(pi)/n)^n. x = +/-1.
Post by Simon Roberts
|z_k| must be 1
+/- 1.
How can |anything at all| be negative?
Post by Simon Roberts
Post by Simon Roberts
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (+/-1<(k*360/n)) = 0 (a.)
Five radii of length one a polar graph vectorialy add to zero?
ODD.
no matter what angle.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Simon Roberts
2017-06-11 15:50:06 UTC
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Post by Peter Percival
Post by Simon Roberts
Post by Simon Roberts
Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
damn, what do I have here?
I assumed z_k is a complex number.
Post by Simon Roberts
For some positive integer n, (z_k)^n = 1 for all k implies
z_k = x<y = 1<(360/n) in POLAR NOTAION.
This x<y is radius x and angle y. ok.
Post by Simon Roberts
this is because (z_k)^n = (x<(360/n))^n = x^n<360 =1
wrong. (x<k(pi)/n)^n. x = +/-1.
Post by Simon Roberts
|z_k| must be 1
+/- 1.
How can |anything at all| be negative?
Your right. It is polar notation after all. I see.
Post by Peter Percival
Post by Simon Roberts
Post by Simon Roberts
which in turn implies (since z_1 + ... + z_5 = 0)
SUM_1 to 5 (+/-1<(k*360/n)) = 0 (a.)
Five radii of length one a polar graph vectorialy add to zero?
ODD.
no matter what angle.
--
Do, as a concession to my poor wits, Lord Darlington, just explain
to me what you really mean.
I think I had better not, Duchess. Nowadays to be intelligible is
to be found out. -- Oscar Wilde, Lady Windermere's Fan
Markus Klyver
2017-06-15 01:00:47 UTC
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Is this meant to be a challenge? I don't see any question being asked.
Simon Roberts
2015-07-12 16:56:56 UTC
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Post by quasi
If z_1,...,z_5 are complex numbers such that
(1) For some positive integer n, (z_k)^n = 1 for all k.
(2) z_1 + ... + z_5 = 0
(3) No two of z_1,...,z_5 sum to zero
then {z_1,...,z_5} is the set of vertices of a regular
pentagon.
quasi
I have good days and bad days (mostly bad). I apologize (apologise) for calling you "rude" quasi. -Simon
quasi
2015-07-12 19:52:31 UTC
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Post by Simon Roberts
I have good days and bad days (mostly bad). I apologize
(apologise) for calling you "rude" quasi. -Simon
Sometimes I _am_ rude. Usually it's in a scenario where
I feel a "polite" response would be too weak.

In any case, apology accepted.

quasi
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