Reply

PermalinkRaw Message

*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)