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Question Number 1559 by 123456 last updated on 19/Aug/15
find complex number α,β such that  α^n =β^m   β^u =α^v   n,m,u,v∈Z  Q1498
findcomplexnumberα,βsuchthatαn=βmβu=αvn,m,u,vZ\boldsymbolQ1498
Answered by Rasheed Soomro last updated on 19/Aug/15
Let gcd(m,u)=k ⇒ lcm(m,u)=((mu)/k)  (α^n )^(u/k) =(β^( m) )^(u/k)  ⇒α^((nu)/k) =β^((mu)/k)             ........(I)   (β^( u) )^(m/k) =(α^v )^(m/k)  ⇒β^( ((mu)/k)) =α^((mv)/k)            .........(II)  [In order to acheive the common exponent(least also)  of β]  From (I) and (II)      α^((nu)/k) =α^((mv)/k) ⇒nu=mv   [This condition is regardless                    of value of α and β]     α^((nu)/k) − α^((mv)/k) =0     α^((mv)/k) (α^(((nu)/k)−((mv)/k)) −1)=0 ⇒   α^((mv)/k) =0 ∨  α^(((nu)/k)−((mv)/k)) −1=0      α=0 ∨ α^(((nu)/k)−((mv)/k)) =1  [You may equally proceed as         α^((nu)/k) (1−α^(((mv)/k)−((nu)/k)) )=0 ⇒α^((nu)/k) =0 ∨ 1−α^(((mv)/k)−((nu)/k)) =0  ⇒ α=0 ∨ α^(((mv)/k)−((nu)/k)) =1⇒α^(((nu)/k)−((mv)/k)) =(1)^(−1) =1 same as above.]    So α=^((nu−mv)/k) (√1)  I−E  α is ( ((nu−mv)/(gcd(m,u))) )th root of unity.  Similar process shows that  β=0 ∨ β  is   ( ((nu−mv)/(gcd(n,v))) )th  root of unity.  Solution for α      {0}∪{1,ω,ω^2 ,......ω^((((nu−mv)/(gcd(m,u)))−1)) }; ω is ( ((nu−mv)/(gcd(m,u))) )th root of unity.  Solution for β      {0}∪{1,μ,μ^2 ,....μ^((((nu−mv)/(gcd(n,v)))−1)) }; μ is ( ((nu−mv)/(gcd(n,v))) )th  root of unity.  Let (α,β)=(ω^( p)  ,μ^q ) such that       α^n =β^( m)  ∧ β^( u) =α^v        (ω^( p) )^n =(μ^q )^m   ∧   (μ^q )^u =(ω^( p) )^v       ω^(np) =μ^(mq)    ∧    μ^(uq) =ω^( vp)   On simplication we get       mv=nu   ,  Which is free of  p  and   q   and  that means  p  and  q may be any integers.  Hence if            A_α ={1,ω,ω^2 ,......ω^((((nu−mv)/(gcd(m,u)))−1)) }    and   A_β ={1,μ,μ^2 ,....μ^((((nu−mv)/(gcd(n,v)))−1)) }  , the solution for pair  (α,β)  is whole  A_α ×A_β
Letgcd(m,u)=klcm(m,u)=muk(αn)uk=(βm)ukαnuk=βmuk..(\boldsymbolI)(βu)mk=(αv)mkβmuk=αmvk(\boldsymbolII)[Inordertoacheivethecommonexponent(leastalso)ofβ]From(\boldsymbolI)and(\boldsymbolII)αnuk=αmvknu=mv[Thisconditionisregardlessofvalueofαandβ]αnukαmvk=0αmvk(αnukmvk1)=0αmvk=0αnukmvk1=0α=0αnukmvk=1[Youmayequallyproceedasαnuk(1αmvknuk)=0αnuk=01αmvknuk=0α=0αmvknuk=1αnukmvk=(1)1=1sameasabove.]Soα=numvk1IEαis(numvgcd(m,u))\boldsymbolthrootofunity.Similarprocessshowsthatβ=0βis(numvgcd(n,v))\boldsymbolthrootofunity.Solutionforα{0}{1,ω,ω2,ω(numvgcd(m,u)1)};ωis(numvgcd(m,u))\boldsymbolthrootofunity.Solutionforβ{0}{1,μ,μ2,.μ(numvgcd(n,v)1)};μis(numvgcd(n,v))\boldsymbolthrootofunity.Let(α,β)=(ωp,μq)suchthatαn=βmβu=αv(ωp)n=(μq)m(μq)u=(ωp)vωnp=μmqμuq=ωvpOnsimplicationwegetmv=nu,Whichisfreeofpandqandthatmeanspandqmaybeanyintegers.HenceifAα={1,ω,ω2,ω(numvgcd(m,u)1)}andAβ={1,μ,μ2,.μ(numvgcd(n,v)1)},thesolutionforpair(α,β)iswholeAα×Aβ
Answered by Rasheed Soomro last updated on 19/Aug/15
α and β may be any complex numbers for                         mv=nu
αandβmaybeanycomplexnumbersformv=nu

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