Let-is-cube-root-of-unity-and-x-y-z-Z-If-x-y-z-0-prove-that-3-x-y-z-Show-by-an-example-that-the-converse-is-not-true-

Question Number 1585 by Rasheed Soomro last updated on 22/Aug/15

Let ω is cube root of unity and x, y, z \in Z

If ω^{x} + ω^{y} + ω^{z} = 0 prove that 3 ∣ (x + y + z)

Show b y a n e x a m p l e that the converse is n o t true .

Commented by 112358 last updated on 22/Aug/15

ω^{3} - 1 = 0

C u b e r o o t o f u n i t y \Rightarrow ω = e^{\frac{2 k π i}{3}}, k = - 1, 0, 1

L e t

ψ (x, y, z) = ω^{x} + ω^{y} + ω^{z} = e^{\frac{2 x π i k}{3}} + e^{\frac{2 k π i y}{3}} + e^{\frac{2 k π i z}{3}}

ψ (x, y, z) = c o s (\frac{2 π x k}{3}) + c o s (\frac{2 π y k}{3}) + c o s (\frac{2 π z k}{3}) + i [s i n (\frac{2 k x π}{3}) + s i n (\frac{2 π y k}{3}) + s i n (\frac{2 π z k}{3})]

I f ψ (x, y, z) = 0 \Rightarrow R e (ψ (x, y, z)) = 0, I m (ψ (x, y, z)) = 0

∴ c o s (\frac{2 π x k}{3}) + c o s (\frac{2 π y k}{3}) + c o s (\frac{2 π z k}{3}) = 0

a n d

s i n (\frac{2 π x k}{3}) + s i n (\frac{2 π y k}{3}) + s i n (\frac{2 π z k}{3}) = 0

f o r k = - 1, 0, 1 a n d x, y, z \in Z

Commented by 123456 last updated on 22/Aug/15

x = y = z = 0 \Rightarrow 3 ∣ (x + y + z) = 0, ω^{x} + ω^{y} + ω^{z} = 3 \neq 0

Answered by 123456 last updated on 23/Aug/15

ω = e^{\frac{2 π}{3} ι} = \cos \frac{2 π}{3} + ı \sin \frac{2 π}{3}

some fact that help

theorem :

if z_{1}, z_{2}, z_{3} are 3 unitary complex number

such that their sum is 0, then the angle

betwen them is 120 °

proof :

since z_{1}, z_{2}, z_{3} are unitary, we can write

∣ z_{1} ∣=∣ z_{2} ∣=∣ z_{3} ∣= 1

z_{1} = e^{ϕ ı} = \cos ϕ + ı \sin ϕ

z_{2} = e^{(ϕ + θ_{1}) ı} = \cos (ϕ + θ_{1}) + ı \sin (ϕ + θ_{1})

z_{3} = e^{(ϕ + θ_{2}) ı} = \cos (ϕ + θ_{2}) + ı \sin (ϕ + θ_{2})

z_{1} + z_{2} + z_{3} = 0

e^{ϕ ı} + e^{(ϕ + θ_{1}) ı} + e^{(ϕ + θ_{2}) ı} = 0

\cos ϕ + ı \sin ϕ + \cos (ϕ + θ_{1}) + ı \sin (ϕ + θ_{1}) + \cos (ϕ + θ_{2}) + ı \sin (ϕ + θ_{2}) = 0

[\cos ϕ + \cos (ϕ + θ_{1}) + \cos (ϕ + θ_{2})] + ı [\sin ϕ + \sin (ϕ + θ_{1}) + \sin (ϕ + θ_{2})] = 0

{\begin{cases} \cos ϕ + \cos (ϕ + θ_{1}) + \cos (ϕ + θ_{2}) = 0 \\ \sin ϕ + \sin (ϕ + θ_{1}) + \sin (ϕ + θ_{2}) = 0 \end{cases}

\cos (α + β) = \cos α \cos β - \sin α \sin β

\sin (α + β) = \cos α \sin β + \cos β \sin α

{\begin{cases} \cos ϕ (1 + \cos θ_{1} + \cos θ_{2}) - \sin ϕ (\sin θ_{1} + \sin θ_{2}) = 0 \\ \sin ϕ (1 + \cos θ_{1} + \cos θ_{2}) + \cos ϕ (\sin θ_{1} + \sin θ_{2}) = 0 \end{cases}

\cos ϕ = a, \sin ϕ = b, 1 + \cos θ_{1} + \cos θ_{2} = x, \sin θ_{1} + \sin θ_{2} = y, a^{2} + b^{2} = 1

{\begin{cases} a x - b y = 0 \\ b x + a y = 0 \end{cases}

Δ = | \begin{matrix} a & - b \\ b & a \end{matrix} | = a^{2} + b^{2} = 1

Δ x = | \begin{matrix} 0 & - b \\ 0 & a \end{matrix} | = 0

Δ y = | \begin{matrix} a & 0 \\ b & 0 \end{matrix} | = 0

x = \frac{Δ x}{Δ} = 0

y = \frac{Δ y}{Δ} = 0

{\begin{cases} 1 + \cos θ_{1} + \cos θ_{2} = 0 \\ \sin θ_{1} + \sin θ_{2} = 0 \end{cases}

\sin θ_{1} = \sin - θ_{2} \Leftrightarrow θ_{1} = - θ_{2} + 2 π k \lor θ_{1} = π + θ_{2} + 2 π k, k \in Z

θ_{1} = π + θ_{2} + 2 π k \Rightarrow 1 + \cos θ_{1} + \cos θ_{2} = 1 + \cos (π + θ_{2} + 2 π k) + \cos θ_{2} = 1 - \cos θ_{2} + \cos θ_{2} = 1 \neq 0

θ_{1} = - θ_{2} + 2 π k \Rightarrow 1 + \cos θ_{1} + \cos θ_{2} = 1 + \cos (- θ_{2} + 2 π k) + \cos θ_{2} = 1 + \cos θ_{2} + \cos θ_{2} = 1 + 2 \cos θ_{2} = 0

\cos θ_{2} = - \frac{1}{2}

θ_{2} = \frac{2 π}{3} + 2 π j \lor θ_{2} = \frac{4 π}{3} + 2 π j, j \in Z

θ_{1} = - \frac{2 π}{3} + 2 π (j + k) \lor θ_{2} = - \frac{4 π}{3} + 2 π (j + k) .

the cube roots of unity are

1, e^{\frac{2 π}{3} ı}, e^{\frac{4 π}{3} j} \Rightarrow 1, ω, ω^{2}

ω^{n} = 1, n \equiv 0 (\mod 3)

ω^{n} = e^{\frac{2 π}{3} ı}, n \equiv 1 (\mod 3)

ω^{n} = e^{\frac{4 π}{3} ı}, n \equiv 2 (\mod 3)

then for

ω^{x} + ω^{y} + ω^{z} = 0

as shown above, them need to have a angle

of 120 ° betwen them, wich imply that

all the three roots apear, without loss of generaly

take

x \equiv 0 (\mod 3)

y \equiv 1 (\mod 3)

z \equiv 2 (\mod 3)

them

x + y + z \equiv 0 + 1 + 2 \equiv 3 \equiv 0 (\mod 3)

wich imply that

3 ∣ (x + y + z)

Commented by Rasheed Soomro last updated on 23/Aug/15

THANKS . Fully logical and full of knowledge .

But too technical! Long tour of logic!

Could there be a simple r answer ?

Answered by Rasheed Soomro last updated on 23/Aug/15

x, y, z \in Z (given)

All the integers wrt modulo 3 are of three types

3 k, 3 k + 1, 3 k + 2 .

So suppose each of x, y, z is one of above types .

It can easily be shown that

ω^{x} + ω^{y} + ω^{z} = 0 \Rightarrow {\begin{cases} x, y, z all are of different types \\ (wrt modulo 3) \end{cases}

[If two or three are of same type ω^{x} + ω^{y} + ω^{z} \neq 0 .]

Let one of x, y, z is equal to 3 l, another is equal

to 3 m + 1 and remaining is equal to 3 n + 2

ω^{x} + ω^{y} + ω^{z} = ω^{3 l} + ω^{3 m + 1} + ω^{3 n + 2} = 1 + ω + ω^{2} = 0 \Rightarrow

{

x + y + z = 3 l + (3 m + 1) + (3 n + 2)

= 3 l + 3 m + 3 n + 3 = 3 (l + m + n + 1)

This implies that 3 ∣ (x + y + z)

}

Counter examples that prove that converse is not true :

1) Each of x, y, z is of 3 k type .

x + y + z = 3 l + 3 m + 3 n = 3 (l + m + n)

3 ∣ (x + y + z) But ω^{x} + ω^{y} + ω^{z} = ω^{3 l} + ω^{3 m} + ω^{3 n} = 1 + 1 + 1 = 3 \neq 0

2) Each of x, y, z is of 3 k + 1 type .

3) Each of x, y, z is of 3 k + 2 type .