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Question-80209

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Question Number 80209 by peter frank last updated on 01/Feb/20
Commented by mathmax by abdo last updated on 01/Feb/20
z+(1/z) =−1 ⇒z^2 +1=−z ⇒z^2 +z+1=0 Δ=1−4=−3 ⇒z_1 =((−1+i(√3))/2)=e^((i2π)/3) and z_2 =((−1−i(√3))/2)=e^(−((i2π)/3)) z=z_1 ⇒z^5 +(1/z^5 ) =e^((i10π)/3) +e^(−((i10π)/3)) =2cos(((10π)/3))=2cos(4π−((2π)/3)) =2cos(((2π)/3))=2×(−(1/2))=−1 z=z_2 we get the same value because z_2 =conj(z_1 ) for z^(11) +(1/z^(11) ) z=z_1 =⇒z^(11) +(1/z^(11) ) =e^((i22π)/3) +e^(−((i22π)/3)) =2cos(((22π)/3)) =2cos(8π−((2π)/3)) =2cos(((2π)/3))=−1 z=z_2 we get the same value
z+1z=−1⇒z2+1=−z⇒z2+z+1=0Δ=1−4=−3⇒z1=−1+i32=ei2π3andz2=−1−i32=e−i2π3z=z1⇒z5+1z5=ei10π3+e−i10π3=2cos(10π3)=2cos(4π−2π3)=2cos(2π3)=2×(−12)=−1z=z2wegetthesamevaluebecausez2=conj(z1)forz11+1z11z=z1=⇒z11+1z11=ei22π3+e−i22π3=2cos(22π3)=2cos(8π−2π3)=2cos(2π3)=−1z=z2wegetthesamevalue
Answered by som(math1967) last updated on 01/Feb/20
(z+(1/z))^2 =1⇒z^2 +(1/z^2 )=1−2=−1 again (z+(1/z))^3 =−1 z^3 +(1/z^3 ) +3(z+(1/z))=−1 z^3 +(1/z^3 )=−1+3=2 ∴(z^2 +(1/z^2 ))(z^3 +(1/z^3 ))=(−1)(2)=−2 z^5 +(1/z^5 ) +z+(1/z)=−2 ∴z^5 +(1/z^5 )=−2+1=−1 (proved) Again (z^3 +(1/z^3 ))^2 =4 z^6 +(1/z^6 )=4−2=2 ∴(z^5 +(1/z^5 ))(z^6 +(1/z^6 ))=(−1)(2) z^(11) +(1/z^(11) ) +z+(1/z)=−2 ∴z^(11) +(1/z^(11) )=−2+1=−1 ans
(z+1z)2=1⇒z2+1z2=1−2=−1again(z+1z)3=−1z3+1z3+3(z+1z)=−1z3+1z3=−1+3=2∴(z2+1z2)(z3+1z3)=(−1)(2)=−2z5+1z5+z+1z=−2∴z5+1z5=−2+1=−1(proved)Again(z3+1z3)2=4z6+1z6=4−2=2∴(z5+1z5)(z6+1z6)=(−1)(2)z11+1z11+z+1z=−2∴z11+1z11=−2+1=−1ans
Answered by mr W last updated on 01/Feb/20
z+(1/z)=−1 z^2 +z+1=0 z=((−1±i(√3))/2)=−{cos (±(π/3))+sin (±(π/3)) i}=−e^(±(π/3)i) z^n +(1/z^n )=(−1)^n [e^(±((nπ)/3)i) +e^(∓((nπ)/3)i) ] =2(−1)^n cos (±((nπ)/3)) =2(−1)^n cos (((nπ)/3)) ⇒z^n +(1/z^n )=2(−1)^n cos (((nπ)/3)) n=5: z^5 +(1/z^5 )=2(−1)^5 cos (((5π)/3))=−2×(1/2)=−1 n=11: z^(11) +(1/z^(11) )=2(−1)^(11) cos (((11π)/3))=−2×(1/2)=−1 n=3: z^3 +(1/z^3 )=2(−1)^3 cos (((3π)/3))=−2×(−1)=2 n=12: z^(12) +(1/z^(12) )=2(−1)^(12) cos (((12π)/3))=2×1=2
z+1z=−1z2+z+1=0z=−1±i32=−{cos(±π3)+sin(±π3)i}=−e±π3izn+1zn=(−1)n[e±nπ3i+e∓nπ3i]=2(−1)ncos(±nπ3)=2(−1)ncos(nπ3)⇒zn+1zn=2(−1)ncos(nπ3)n=5:z5+1z5=2(−1)5cos(5π3)=−2×12=−1n=11:z11+1z11=2(−1)11cos(11π3)=−2×12=−1n=3:z3+1z3=2(−1)3cos(3π3)=−2×(−1)=2n=12:z12+1z12=2(−1)12cos(12π3)=2×1=2
Commented by jagoll last updated on 01/Feb/20
if x+(1/x)= −2 that x^6 +(1/x^6 ) = 2(−2)^(6 ) ×cos (((6π)/3)) = 128. that right sir
ifx+1x=−2thatx6+1x6=2(−2)6×cos(6π3)=128.thatrightsir
Commented by mr W last updated on 01/Feb/20
that′s not correct sir!
that′snotcorrectsir!
Commented by peter frank last updated on 01/Feb/20
thank you all
thankyouall
Commented by mr W last updated on 01/Feb/20
if x+(1/x)= −2 ⇒x=−1 ⇒x^3 +(1/x^3 )=−2 ⇒x^6 +(1/x^6 )=2
ifx+1x=−2⇒x=−1⇒x3+1x3=−2⇒x6+1x6=2
Commented by jagoll last updated on 01/Feb/20
special means only for form x+(1/x)=−1
specialmeansonlyforformx+1x=−1

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