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Question Number 1497 by Rasheed Soomro last updated on 14/Aug/15
Find complex numbers α and β such that  α^2 =β^3   and  β^2 =α^3
Findcomplexnumbersαandβsuchthatα2=β3andβ2=α3
Commented by 123456 last updated on 14/Aug/15
(α,β)=(0,0) is ome of these pairs
(α,β)=(0,0)isomeofthesepairs
Commented by Rasheed Ahmad last updated on 14/Aug/15
(1,1) is an other pair.
(1,1)isanotherpair.
Commented by 123456 last updated on 15/Aug/15
 { ((α^2 =β^3 )),((β^2 =α^3 )) :}  n∈N^∗    { ((α^(4n) =β^(6n) )),((β^(6n) =α^(9n) )) :}  α^(4n) =α^(9n) ⇔α^(4n) (α^(5n) −1)=0   { ((α^(6n) =β^(9n) )),((β^(4n) =α^(6n) )) :}  β^(4n) =β^(9n) ⇔β^(4n) (β^(5n) −1)=0
{α2=β3β2=α3nN{α4n=β6nβ6n=α9nα4n=α9nα4n(α5n1)=0{α6n=β9nβ4n=α6nβ4n=β9nβ4n(β5n1)=0
Commented by Rasheed Soomro last updated on 16/Aug/15
α^2 =β^3   (α^2 )^2 =(β^( 3) )^2   α^4 =(β^( 2) )^3   α^4 =(α^3 )^3       [substituting β^( 2) =α^3 ]  α^4 =α^9   Assuming α≠0,dividing by α^4   α^5 =1⇒α=^5  (√1)  Similarily ,  β^( 5) =1⇒β=^5 (√1)  ∴ α and β both are 5th roots of unity.  We know that   nth root of unity=(cos((2π)/n)+ı sin((2π)/n) )^k   , k=0,1,2,...(n−1)  5th root of unity=(cos((2π)/5)+ı sin((2π)/5) )^k   , k=0,1,2,...4  Let cos((2π)/5)+ı sin((2π)/5)  =ω  5th root of unity=1,ω,ω^2 ,ω^3 ,ω^4       (in order)  Square of ∽         =1,ω^2 ,ω^4 ,ω,ω^3        (in order)  Cube of ∽             =1,w^3 ,ω,ω^4 ,ω^2         (in order)  (𝛚^k_1  )^2  = (𝛚^k_2  )^3      k_1  , k_2  = 0,1,2,3,4  𝛚^(2k_1 (mod 5))  =𝛚^(3k_2 (mod 5))   For 2k_1 (mod 5)=3k_2 (mod 5)  :  (α,β)=(ω^k_1   , ω^k_2  )=(ω^k_2  , ω^k_1  )  (α,β)={(1,1),(ω,ω^4 ),(ω^4 ,ω),(ω^2 ,ω^3 ),(ω^3 ,ω^2 )}
α2=β3(α2)2=(β3)2α4=(β2)3α4=(α3)3[substitutingβ2=α3]α4=α9Assumingα0,dividingbyα4α5=1α=51Similarily,β5=1β=51αandβbothare5throotsofunity.Weknowthatnthrootofunity=(cos2πn+ısin2πn)k,k=0,1,2,(n1)5throotofunity=(cos2π5+ısin2π5)k,k=0,1,2,4Letcos2π5+ısin2π5=ω5throotofunity=1,ω,ω2,ω3,ω4(inorder)Squareof=1,ω2,ω4,ω,ω3(inorder)Cubeof=1,w3,ω,ω4,ω2(inorder)(\boldsymbolωk1)2=(\boldsymbolωk2)3k1,k2=0,1,2,3,4\boldsymbolω2k1(mod5)=\boldsymbolω3k2(mod5)For2k1(mod5)=3k2(mod5):(α,β)=(ωk1,ωk2)=(ωk2,ωk1)(α,β)={(1,1),(ω,ω4),(ω4,ω),(ω2,ω3),(ω3,ω2)}
Commented by 123456 last updated on 15/Aug/15
good approach :)
goodapproach:)
Commented by Rasheed Soomro last updated on 16/Aug/15
Thanks. Please comment, although in negative , on my   answers to Q−1448  and  Q−1466. If  the comments be  positive I will gain confidence and if they are negative  I will learn something. So please.....
\boldsymbolThanks.Pleasecomment,\boldsymbolalthough\boldsymbolin\boldsymbolnegative,onmyanswerstoQ1448andQ1466.IfthecommentsbepositiveIwillgainconfidenceandiftheyarenegativeIwilllearnsomething.Soplease..
Answered by 123456 last updated on 15/Aug/15
 { ((α^2 =β^3 )),((β^2 =α^3 )) :}⇒ { ((α^4 =β^6 )),((β^6 =α^9 )) :}  α^4 =α^9   α^9 −α^4 =0  α^4 (α^5 −1)=0  α=0∨α=(1)^(1/5)   α=0⇒ { ((β^3 =0)),((β^2 =0)) :}⇒β=0  α=1⇒ { ((β^3 =1)),((β^2 =1)) :}⇒ { ((β=1∨β=−(1/2)+((√3)/2)ı∨β=−(1/2)−((√3)/2)ı)),((β=1∨β=−1)) :}⇒β=1  α=e^(72°ı) ⇒ { ((β^3 =e^(144°ı) )),((β^2 =e^(216°ı) )) :}⇒ { ((β=e^(48°ı) ∨β=e^(168°ı) ∨β=e^(288°ı) )),((β=e^(108°ı) ∨β=e^(288°ı) )) :}⇒β=e^(288°ı)   α=e^(144°ı) ⇒ { ((β^3 =e^(288°ı) )),((β^2 =e^(72°ı) )) :}⇒ { ((β=e^(96°ı) ∨β=e^(216°ı) ∨β=e^(336°ı) )),((β=e^(36°ı) ∨β=e^(216°ı) )) :}⇒β=e^(216°ı)   α=e^(216°ı) ⇒ { ((β^3 =e^(72°ı) )),((β^2 =e^(288°ı) )) :}⇒ { ((β=e^(24°ı) ∨β=e^(144°ı) ∨β=e^(264°ı) )),((β=e^(144°ı) ∨β=e^(324°ı) )) :}  α=e^(288°ı) ⇒ { ((β^3 =e^(216°ı) )),((β^2 =e^(144°ı) )) :}⇒ { ((β=e^(72°ı) ∨β=e^(192°ı) ∨β=e^(312°ı) )),((β=e^(72°ı) ∨β=e^(252°ı) )) :}⇒β=e^(72°ı)   S={(0,0),(1,1),(e^(72°ı) ,e^(288°ı) ),(e^(144°ı) ,e^(216°ı) ),(e^(216°ı) ,e^(144°ı) ),(e^(288°ı) ,e^(72°ı) )}
{α2=β3β2=α3{α4=β6β6=α9α4=α9α9α4=0α4(α51)=0α=0α=15α=0{β3=0β2=0β=0α=1{β3=1β2=1{β=1β=12+32ıβ=1232ıβ=1β=1β=1α=e72°ı{β3=e144°ıβ2=e216°ı{β=e48°ıβ=e168°ıβ=e288°ıβ=e108°ıβ=e288°ıβ=e288°ıα=e144°ı{β3=e288°ıβ2=e72°ı{β=e96°ıβ=e216°ıβ=e336°ıβ=e36°ıβ=e216°ıβ=e216°ıα=e216°ı{β3=e72°ıβ2=e288°ı{β=e24°ıβ=e144°ıβ=e264°ıβ=e144°ıβ=e324°ıα=e288°ı{β3=e216°ıβ2=e144°ı{β=e72°ıβ=e192°ıβ=e312°ıβ=e72°ıβ=e252°ıβ=e72°ıS={(0,0),(1,1),(e72°ı,e288°ı),(e144°ı,e216°ı),(e216°ı,e144°ı),(e288°ı,e72°ı)}
Commented by Rasheed Ahmad last updated on 15/Aug/15
(Rasheed Soomro)  Excellent Sir!
(RasheedSoomro)ExcellentSir!
Commented by 123456 last updated on 15/Aug/15
i know  x°=x(π/(180))(°→rad)  e^(xı) =cos x+ısin x(x are mesuared in rad)  e^((x+360°k)ı) =e^(xı) ,k∈Z
iknowx°=xπ180(°rad)exı=cosx+ısinx(xaremesuaredinrad)e(x+360°k)ı=exı,kZ

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