Menu Close

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 ∈ Z  If ω^x +ω^y +ω^z =0 prove that 3 ∣ (x+y+z)  Show by an example that the converse is not  true.
Letωiscuberootofunityandx,y,zZIfωx+ωy+ωz=0provethat3(x+y+z)Showbyanexamplethattheconverseisnottrue.
Commented by 112358 last updated on 22/Aug/15
ω^3 −1=0  Cube root of unity⇒ω=e^((2kπi)/3)  ,k=−1,0,1  Let  ψ(x,y,z)=ω^x +ω^y +ω^z =e^((2xπik)/3) +e^((2kπiy)/3) +e^((2kπiz)/3)   ψ(x,y,z)=cos(((2πxk)/3))+cos(((2πyk)/3))+cos(((2πzk)/3))+i[sin(((2kxπ)/3))+sin(((2πyk)/3))+sin(((2πzk)/3))]  If ψ(x,y,z)=0⇒Re(ψ(x,y,z))=0 ,Im(ψ(x,y,z))=0  ∴cos(((2πxk)/3))+cos(((2πyk)/3))+cos(((2πzk)/3))=0  and   sin(((2πxk)/3))+sin(((2πyk)/3))+sin(((2πzk)/3))=0  for k=−1,0,1 and x,y,z∈Z
ω31=0Cuberootofunityω=e2kπi3,k=1,0,1Letψ(x,y,z)=ωx+ωy+ωz=e2xπik3+e2kπiy3+e2kπiz3ψ(x,y,z)=cos(2πxk3)+cos(2πyk3)+cos(2πzk3)+i[sin(2kxπ3)+sin(2πyk3)+sin(2πzk3)]Ifψ(x,y,z)=0Re(ψ(x,y,z))=0,Im(ψ(x,y,z))=0cos(2πxk3)+cos(2πyk3)+cos(2πzk3)=0andsin(2πxk3)+sin(2πyk3)+sin(2πzk3)=0fork=1,0,1andx,y,zZ
Commented by 123456 last updated on 22/Aug/15
x=y=z=0⇒3∣(x+y+z)=0,ω^x +ω^y +ω^z =3≠0
x=y=z=03(x+y+z)=0,ωx+ωy+ωz=30
Answered by 123456 last updated on 23/Aug/15
ω=e^(((2π)/3)ι) =cos ((2π)/3)+ısin ((2π)/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^(φı) =cos φ+ısin φ  z_2 =e^((φ+θ_1 )ı) =cos(φ+θ_1 )+ısin(φ+θ_1 )  z_3 =e^((φ+θ_2 )ı) =cos(φ+θ_2 )+ısin(φ+θ_2 )  z_1 +z_2 +z_3 =0  e^(φı) +e^((φ+θ_1 )ı) +e^((φ+θ_2 )ı) =0  cos φ+ısin φ+cos(φ+θ_1 )+ısin(φ+θ_1 )+cos(φ+θ_2 )+ısin(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   { ((cos φ+cos(φ+θ_1 )+cos(φ+θ_2 )=0)),((sin φ+sin(φ+θ_1 )+sin(φ+θ_2 )=0)) :}  cos(α+β)=cos α cos β−sin α sin β  sin(α+β)=cos α sin β+cos β sin α   { ((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)) :}  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   { ((ax−by=0)),((bx+ay=0)) :}  Δ= determinant ((a,(−b)),(b,a))=a^2 +b^2 =1  Δx= determinant ((0,(−b)),(0,a))=0  Δy= determinant ((a,0),(b,0))=0  x=((Δx)/Δ)=0  y=((Δy)/Δ)=0   { ((1+cos θ_1 +cos θ_2 =0)),((sin θ_1 +sin θ_2 =0)) :}  sin θ_1 =sin −θ_2 ⇔θ_1 =−θ_2 +2πk∨θ_1 =π+θ_2 +2πk,k∈Z  θ_1 =π+θ_2 +2πk⇒1+cos θ_1 +cos θ_2 =1+cos(π+θ_2 +2πk)+cos θ_2 =1−cos θ_2 +cos θ_2 =1≠0  θ_1 =−θ_2 +2πk⇒1+cos θ_1 +cos θ_2 =1+cos(−θ_2 +2πk)+cos θ_2 =1+cos θ_2 +cos θ_2 =1+2cos θ_2 =0  cos θ_2 =−(1/2)  θ_2 =((2π)/3)+2πj∨θ_2 =((4π)/3)+2πj,j∈Z  θ_1 =−((2π)/3)+2π(j+k)∨θ_2 =−((4π)/3)+2π(j+k).  the cube roots of unity are  1,e^(((2π)/3)ı) ,e^(((4π)/3)j) ⇒1,ω,ω^2   ω^n =1,n≡0(mod 3)  ω^n =e^(((2π)/3)ı) ,n≡1(mod 3)  ω^n =e^(((4π)/3)ı) ,n≡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≡0(mod 3)  y≡1(mod 3)  z≡2(mod 3)  them  x+y+z≡0+1+2≡3≡0(mod 3)  wich imply that  3∣(x+y+z)
ω=e2π3ι=cos2π3+ısin2π3somefactthathelptheorem:ifz1,z2,z3are3unitarycomplexnumbersuchthattheirsumis0,thentheanglebetwenthemis120°proof:sincez1,z2,z3areunitary,wecanwritez1∣=∣z2∣=∣z3∣=1z1=eϕı=cosϕ+ısinϕz2=e(ϕ+θ1)ı=cos(ϕ+θ1)+ısin(ϕ+θ1)z3=e(ϕ+θ2)ı=cos(ϕ+θ2)+ısin(ϕ+θ2)z1+z2+z3=0eϕı+e(ϕ+θ1)ı+e(ϕ+θ2)ı=0cosϕ+ısinϕ+cos(ϕ+θ1)+ısin(ϕ+θ1)+cos(ϕ+θ2)+ısin(ϕ+θ2)=0[cosϕ+cos(ϕ+θ1)+cos(ϕ+θ2)]+ı[sinϕ+sin(ϕ+θ1)+sin(ϕ+θ2)]=0{cosϕ+cos(ϕ+θ1)+cos(ϕ+θ2)=0sinϕ+sin(ϕ+θ1)+sin(ϕ+θ2)=0cos(α+β)=cosαcosβsinαsinβsin(α+β)=cosαsinβ+cosβsinα{cosϕ(1+cosθ1+cosθ2)sinϕ(sinθ1+sinθ2)=0sinϕ(1+cosθ1+cosθ2)+cosϕ(sinθ1+sinθ2)=0cosϕ=a,sinϕ=b,1+cosθ1+cosθ2=x,sinθ1+sinθ2=y,a2+b2=1{axby=0bx+ay=0Δ=|abba|=a2+b2=1Δx=|0b0a|=0Δy=|a0b0|=0x=ΔxΔ=0y=ΔyΔ=0{1+cosθ1+cosθ2=0sinθ1+sinθ2=0sinθ1=sinθ2θ1=θ2+2πkθ1=π+θ2+2πk,kZθ1=π+θ2+2πk1+cosθ1+cosθ2=1+cos(π+θ2+2πk)+cosθ2=1cosθ2+cosθ2=10θ1=θ2+2πk1+cosθ1+cosθ2=1+cos(θ2+2πk)+cosθ2=1+cosθ2+cosθ2=1+2cosθ2=0cosθ2=12θ2=2π3+2πjθ2=4π3+2πj,jZθ1=2π3+2π(j+k)θ2=4π3+2π(j+k).thecuberootsofunityare1,e2π3ı,e4π3j1,ω,ω2ωn=1,n0(mod3)ωn=e2π3ı,n1(mod3)ωn=e4π3ı,n2(mod3)thenforωx+ωy+ωz=0asshownabove,themneedtohaveaangleof120°betwenthem,wichimplythatallthethreerootsapear,withoutlossofgeneralytakex0(mod3)y1(mod3)z2(mod3)themx+y+z0+1+230(mod3)wichimplythat3(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 simpler  answer?
THANKS.Fullylogicalandfullofknowledge.Buttootechnical!Longtouroflogic!Couldtherebeasimpleranswer?
Answered by Rasheed Soomro last updated on 23/Aug/15
        x,y,z ∈ Z (given)  All the integers wrt modulo 3 are of three types            3k, 3k+1, 3k+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  ⇒ { ((x, y, z all  are of different types)),(((wrt modulo 3))) :}   [If two or three  are of same type ω^x +ω^y +ω^z ≠0.]           Let one of  x, y, z is equal to 3l, another is equal  to 3m+1 and remaining is equal to 3n+2      ω^x +ω^y +ω^z = ω^(3l) +ω^(3m+1) +ω^(3n+2) =1+ω+ω^2 =0 ⇒  {           x+y+z= 3l+(3m+1)+(3n+2)                           = 3l+3m+3n+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 3k type.                x+y+z=3l+3m+3n=3(l+m+n)  3 ∣ (x+y+z) But ω^x +ω^y +ω^z =ω^(3l) +ω^(3m) +ω^(3n) =1+1+1=3≠0             2) Each of x,y,z is of  3k+1  type.             3) Each of x,y,z is of  3k+2  type.
x,y,zZ(given)Alltheintegerswrtmodulo3areofthreetypes3k,3k+1,3k+2.Sosupposeeachofx,y,zisoneofabovetypes.Itcaneasilybeshownthatωx+ωy+ωz=0{x,y,zallareofdifferenttypes(wrtmodulo3)[Iftwoorthreeareofsametypeωx+ωy+ωz0.]Letoneofx,y,zisequalto3l,anotherisequalto3m+1andremainingisequalto3n+2ωx+ωy+ωz=ω3l+ω3m+1+ω3n+2=1+ω+ω2=0{x+y+z=3l+(3m+1)+(3n+2)=3l+3m+3n+3=3(l+m+n+1)Thisimpliesthat3(x+y+z)}Counterexamplesthatprovethatconverseisnottrue:1)Eachofx,y,zisof3ktype.x+y+z=3l+3m+3n=3(l+m+n)3(x+y+z)Butωx+ωy+ωz=ω3l+ω3m+ω3n=1+1+1=302)Eachofx,y,zisof3k+1type.3)Eachofx,y,zisof3k+2type.

Leave a Reply

Your email address will not be published. Required fields are marked *