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Question Number 185472 by liuxinnan last updated on 22/Jan/23
if ω^7 =1 (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 ))=?
ifω7=11ω+ω6+1ω2+ω5+1ω3+ω4=?
Commented by Shrinava last updated on 23/Jan/23
Give: ω^7 = 1 ⇒ ω = 1 ▲ ω^(3p+q) =ω^q ⇒ ω^2 =ω^3 =ω^4 =ω^5 =ω^6 =1 ⇒ (1/(ω+ω^6 )) + (1/(ω^2 +ω^5 )) + (1/(ω^3 +ω^4 )) = (1/2) + (1/2) + (1/2) = (3/2) = 1,5 ✓
Give:ω7=1⇒ω=1▴ω3p+q=ωq⇒ω2=ω3=ω4=ω5=ω6=1⇒1ω+ω6+1ω2+ω5+1ω3+ω4=12+12+12=32=1,5✓
Commented by Rasheed.Sindhi last updated on 24/Jan/23
Here ω isn′t cube root of unity. (It′s seventh root of unity.) So ω^(3p+q) ≠ω^q (But ω^(7p+q) =ω^q ) Only for ω=1 your solution is correct. But ω has six other values also.
Hereωisn′tcuberootofunity.(It′sseventhrootofunity.)Soω3p+q≠ωq(Butω7p+q=ωq)Onlyforω=1yoursolutioniscorrect.Butωhassixothervaluesalso.
Answered by mr W last updated on 22/Jan/23
ω^7 =1 ⇒ω=e^((2kπi)/7) (k=0,1,2,...,6) ω+(1/ω)=2 cos ((2kπ)/7) ω^2 +(1/ω^2 )=2 cos ((4kπ)/7) ω^3 +(1/ω^3 )=2 cos ((6kπ)/7) (1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(1/(ω+(1/ω)))+(1/(ω^2 +(1/ω^2 )))+(1/(ω^3 +(1/ω^3 ))) =(1/2)((1/(cos ((2kπ)/7)))+(1/(cos ((4kπ)/7)))+(1/(cos ((6kπ)/7))))= { (((3/2), k=0)),((−2, k=1,2,...,6)) :}
ω7=1⇒ω=e2kπi7(k=0,1,2,…,6)ω+1ω=2cos2kπ7ω2+1ω2=2cos4kπ7ω3+1ω3=2cos6kπ71ω+ω6+1ω2+ω5+1ω3+ω4=1ω+1ω+1ω2+1ω2+1ω3+1ω3=12(1cos2kπ7+1cos4kπ7+1cos6kπ7)={32,k=0−2,k=1,2,…,6
Commented by liuxinnan last updated on 23/Jan/23
I cant understand the last step
Icantunderstandthelaststep
Answered by Rasheed.Sindhi last updated on 22/Jan/23
{ ((ω^7 =1)),((1+ω+ω^2 +...+ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω^2 +ω^5 +ω+ω^6 )/((ω+ω^6 )(ω^2 +ω^5 )))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(ω^3 +ω^6 +ω^8 +ω^(11) ))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(ω^3 +ω^6 +ω+ω^4 ))+(1/(ω^3 +ω^4 )) =((−1−ω^3 −ω^4 )/(−1−ω^2 −ω^5 ))+(1/(ω^3 +ω^4 )) =((1+ω^3 +ω^4 )/(1+ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((1(ω^3 +ω^4 )+(ω^3 +ω^4 )^2 +1+ω^2 +ω^5 )/(ω^3 +ω^4 +ω^5 +ω^6 +ω^8 +ω^9 )) =((ω^3 +ω^4 +ω^6 +2ω^7 +ω^8 +1+ω^2 +ω^5 )/(ω^3 +ω^4 +ω^5 +ω^6 +ω+ω^2 )) =((ω^3 +ω^4 +ω^6 +2+ω+1+ω^2 +ω^5 )/(−1)) =−(2+0)=−2
{ω7=11+ω+ω2+…+ω6=0▸1ω+ω6+1ω2+ω5+1ω3+ω4=ω2+ω5+ω+ω6(ω+ω6)(ω2+ω5)+1ω3+ω4=−1−ω3−ω4ω3+ω6+ω8+ω11+1ω3+ω4=−1−ω3−ω4ω3+ω6+ω+ω4+1ω3+ω4=−1−ω3−ω4−1−ω2−ω5+1ω3+ω4=1+ω3+ω41+ω2+ω5+1ω3+ω4=1(ω3+ω4)+(ω3+ω4)2+1+ω2+ω5ω3+ω4+ω5+ω6+ω8+ω9=ω3+ω4+ω6+2ω7+ω8+1+ω2+ω5ω3+ω4+ω5+ω6+ω+ω2=ω3+ω4+ω6+2+ω+1+ω2+ω5−1=−(2+0)=−2
Answered by Rasheed.Sindhi last updated on 22/Jan/23
{ ((ω^7 =1)),((1+ω+ω^2 +...+ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(ω/(ω^2 +ω^7 ))+(ω^2 /(ω^4 +ω^7 ))+(ω^3 /(ω^6 +ω^7 )) =(ω/(1+ω^2 ))+(ω^2 /(1+ω^4 ))+(ω^3 /(1+ω^6 )) =((1+ω^4 +1+ω^2 )/(1+ω^4 +ω^2 +ω^6 ))−1+1+(ω^3 /(1+ω^6 )) =((1−ω^6 )/(1+ω^4 +ω^2 +ω^6 ))+1+(ω^3 /(1+ω^6 )) =((1−ω^6 )/(−ω−ω^3 −ω^5 ))+(ω^3 /(1+ω^6 ))+1 =((ω^6 −1)/(ω+ω^3 +ω^5 ))+(ω^3 /(1+ω^6 ))+1 =((ω^(12) −1+ω^4 +ω^6 +ω^8 )/(ω+ω^3 +ω^5 +ω^7 +ω^9 +ω^(11) ))+1 =((ω^5 −1+ω^4 +ω^6 +ω)/(ω+ω^3 +ω^5 +1+ω^2 +ω^4 ))+1 =((−1−1−ω^2 −ω^3 )/(−ω^6 ))+1 =((2+ω^2 +ω^3 )/ω^6 )+1 =2ω+ω^3 +ω^4 +1
{ω7=11+ω+ω2+…+ω6=0▸1ω+ω6+1ω2+ω5+1ω3+ω4=ωω2+ω7+ω2ω4+ω7+ω3ω6+ω7=ω1+ω2+ω21+ω4+ω31+ω6=1+ω4+1+ω21+ω4+ω2+ω6−1+1+ω31+ω6=1−ω61+ω4+ω2+ω6+1+ω31+ω6=1−ω6−ω−ω3−ω5+ω31+ω6+1=ω6−1ω+ω3+ω5+ω31+ω6+1=ω12−1+ω4+ω6+ω8ω+ω3+ω5+ω7+ω9+ω11+1=ω5−1+ω4+ω6+ωω+ω3+ω5+1+ω2+ω4+1=−1−1−ω2−ω3−ω6+1=2+ω2+ω3ω6+1=2ω+ω3+ω4+1
Answered by Rasheed.Sindhi last updated on 22/Jan/23
if ω^7 =1 { ((ω^7 =1)),((1+ω+ω^2 +...ω^6 =0)) :} ▶(1/(ω+ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =(1/(ω+ω^6 ))∙((ω−ω^6 )/(ω−ω^6 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω−ω^6 )/(ω^2 −ω^(12) ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω−ω^6 )/(ω^2 −ω^5 ))+(1/(ω^2 +ω^5 ))+(1/(ω^3 +ω^4 )) =((ω^3 +ω^6 −ω^8 −ω^(11) )/(ω^4 −ω^(10) ))+(1/(ω^3 +ω^4 )) =((ω^3 +ω^6 −ω−ω^4 )/(ω^4 −ω^3 ))+(1/(ω^3 +ω^4 )) =((ω^6 +ω^9 −ω^4 −ω^7 +ω^7 +ω^(10) −ω^5 −ω^8 )/(ω^8 −ω^6 )) =((ω^6 +ω^2 −ω^4 −ω^7 +ω^7 +ω^3 −ω^5 −ω)/(ω−ω^6 )) =((ω^6 +ω^3 +ω^2 −ω^4 −ω^5 −ω)/(ω−ω^6 ))+1−1 =((ω^3 +ω^2 −(ω^4 +ω^5 ))/(ω−ω^6 ))−1 =((ω^3 +ω^2 −(−1−ω−ω^2 −ω^3 −ω^6 ))/(ω−ω^6 ))−1 =((ω^3 +ω^2 +1+ω+ω^2 +ω^3 +ω^6 )/(ω−ω^6 ))−1 =((2ω^3 +2ω^2 +1+2ω^6 )/(ω−ω^6 ))
ifω7=1{ω7=11+ω+ω2+…ω6=0▸1ω+ω6+1ω2+ω5+1ω3+ω4=1ω+ω6⋅ω−ω6ω−ω6+1ω2+ω5+1ω3+ω4=ω−ω6ω2−ω12+1ω2+ω5+1ω3+ω4=ω−ω6ω2−ω5+1ω2+ω5+1ω3+ω4=ω3+ω6−ω8−ω11ω4−ω10+1ω3+ω4=ω3+ω6−ω−ω4ω4−ω3+1ω3+ω4=ω6+ω9−ω4−ω7+ω7+ω10−ω5−ω8ω8−ω6=ω6+ω2−ω4−ω7+ω7+ω3−ω5−ωω−ω6=ω6+ω3+ω2−ω4−ω5−ωω−ω6+1−1=ω3+ω2−(ω4+ω5)ω−ω6−1=ω3+ω2−(−1−ω−ω2−ω3−ω6)ω−ω6−1=ω3+ω2+1+ω+ω2+ω3+ω6ω−ω6−1=2ω3+2ω2+1+2ω6ω−ω6
Commented by liuxinnan last updated on 23/Jan/23
thanks
thanks

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