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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
$${z}+\frac{\mathrm{1}}{{z}}\:=−\mathrm{1}\:\Rightarrow{z}^{\mathrm{2}} \:+\mathrm{1}=−{z}\:\Rightarrow{z}^{\mathrm{2}} \:+{z}+\mathrm{1}=\mathrm{0} \\ $$$$\Delta=\mathrm{1}−\mathrm{4}=−\mathrm{3}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}+{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{1}−{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{−\frac{{i}\mathrm{2}\pi}{\mathrm{3}}} \\ $$$${z}={z}_{\mathrm{1}} \:\Rightarrow{z}^{\mathrm{5}} \:+\frac{\mathrm{1}}{{z}^{\mathrm{5}} }\:={e}^{\frac{{i}\mathrm{10}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\mathrm{10}\pi}{\mathrm{3}}} \:=\mathrm{2}{cos}\left(\frac{\mathrm{10}\pi}{\mathrm{3}}\right)=\mathrm{2}{cos}\left(\mathrm{4}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=\mathrm{2}×\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)=−\mathrm{1} \\ $$$${z}={z}_{\mathrm{2}} \:{we}\:{get}\:{the}\:{same}\:{value}\:{because}\:{z}_{\mathrm{2}} ={conj}\left({z}_{\mathrm{1}} \right) \\ $$$${for}\:{z}^{\mathrm{11}} \:+\frac{\mathrm{1}}{{z}^{\mathrm{11}} } \\ $$$${z}={z}_{\mathrm{1}} =\Rightarrow{z}^{\mathrm{11}} \:+\frac{\mathrm{1}}{{z}^{\mathrm{11}} }\:={e}^{\frac{{i}\mathrm{22}\pi}{\mathrm{3}}} \:+{e}^{−\frac{{i}\mathrm{22}\pi}{\mathrm{3}}} \:=\mathrm{2}{cos}\left(\frac{\mathrm{22}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}{cos}\left(\mathrm{8}\pi−\frac{\mathrm{2}\pi}{\mathrm{3}}\right)\:=\mathrm{2}{cos}\left(\frac{\mathrm{2}\pi}{\mathrm{3}}\right)=−\mathrm{1} \\ $$$${z}={z}_{\mathrm{2}} \:{we}\:{get}\:{the}\:{same}\:{value} \\ $$$$ \\ $$
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
$$\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} =\mathrm{1}\Rightarrow{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\mathrm{1}−\mathrm{2}=−\mathrm{1} \\ $$$${again}\:\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{3}} =−\mathrm{1} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\:+\mathrm{3}\left({z}+\frac{\mathrm{1}}{{z}}\right)=−\mathrm{1} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }=−\mathrm{1}+\mathrm{3}=\mathrm{2} \\ $$$$\therefore\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)=\left(−\mathrm{1}\right)\left(\mathrm{2}\right)=−\mathrm{2} \\ $$$${z}^{\mathrm{5}} +\frac{\mathrm{1}}{{z}^{\mathrm{5}} }\:+{z}+\frac{\mathrm{1}}{{z}}=−\mathrm{2} \\ $$$$\therefore{z}^{\mathrm{5}} +\frac{\mathrm{1}}{{z}^{\mathrm{5}} }=−\mathrm{2}+\mathrm{1}=−\mathrm{1}\:\left({proved}\right) \\ $$$${Again}\:\left({z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }\right)^{\mathrm{2}} =\mathrm{4} \\ $$$${z}^{\mathrm{6}} +\frac{\mathrm{1}}{{z}^{\mathrm{6}} }=\mathrm{4}−\mathrm{2}=\mathrm{2} \\ $$$$\therefore\left({z}^{\mathrm{5}} +\frac{\mathrm{1}}{{z}^{\mathrm{5}} }\right)\left({z}^{\mathrm{6}} +\frac{\mathrm{1}}{{z}^{\mathrm{6}} }\right)=\left(−\mathrm{1}\right)\left(\mathrm{2}\right) \\ $$$$\:{z}^{\mathrm{11}} +\frac{\mathrm{1}}{{z}^{\mathrm{11}} }\:+{z}+\frac{\mathrm{1}}{{z}}=−\mathrm{2} \\ $$$$\therefore{z}^{\mathrm{11}} +\frac{\mathrm{1}}{{z}^{\mathrm{11}} }=−\mathrm{2}+\mathrm{1}=−\mathrm{1}\:{ans} \\ $$
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
$${z}+\frac{\mathrm{1}}{{z}}=−\mathrm{1} \\ $$$${z}^{\mathrm{2}} +{z}+\mathrm{1}=\mathrm{0} \\ $$$${z}=\frac{−\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=−\left\{\mathrm{cos}\:\left(\pm\frac{\pi}{\mathrm{3}}\right)+\mathrm{sin}\:\left(\pm\frac{\pi}{\mathrm{3}}\right)\:{i}\right\}=−{e}^{\pm\frac{\pi}{\mathrm{3}}{i}} \\ $$$${z}^{{n}} +\frac{\mathrm{1}}{{z}^{{n}} }=\left(−\mathrm{1}\right)^{{n}} \left[{e}^{\pm\frac{{n}\pi}{\mathrm{3}}{i}} +{e}^{\mp\frac{{n}\pi}{\mathrm{3}}{i}} \right] \\ $$$$=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \mathrm{cos}\:\left(\pm\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$$=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$$\Rightarrow{z}^{{n}} +\frac{\mathrm{1}}{{z}^{{n}} }=\mathrm{2}\left(−\mathrm{1}\right)^{{n}} \mathrm{cos}\:\left(\frac{{n}\pi}{\mathrm{3}}\right) \\ $$$${n}=\mathrm{5}: \\ $$$${z}^{\mathrm{5}} +\frac{\mathrm{1}}{{z}^{\mathrm{5}} }=\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{5}} \mathrm{cos}\:\left(\frac{\mathrm{5}\pi}{\mathrm{3}}\right)=−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{1} \\ $$$${n}=\mathrm{11}: \\ $$$${z}^{\mathrm{11}} +\frac{\mathrm{1}}{{z}^{\mathrm{11}} }=\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{11}} \mathrm{cos}\:\left(\frac{\mathrm{11}\pi}{\mathrm{3}}\right)=−\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}=−\mathrm{1} \\ $$$${n}=\mathrm{3}: \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }=\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{3}} \mathrm{cos}\:\left(\frac{\mathrm{3}\pi}{\mathrm{3}}\right)=−\mathrm{2}×\left(−\mathrm{1}\right)=\mathrm{2} \\ $$$${n}=\mathrm{12}: \\ $$$${z}^{\mathrm{12}} +\frac{\mathrm{1}}{{z}^{\mathrm{12}} }=\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{12}} \mathrm{cos}\:\left(\frac{\mathrm{12}\pi}{\mathrm{3}}\right)=\mathrm{2}×\mathrm{1}=\mathrm{2} \\ $$
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
$${if}\:{x}+\frac{\mathrm{1}}{{x}}=\:−\mathrm{2}\: \\ $$$${that}\:{x}^{\mathrm{6}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{6}} }\:=\:\mathrm{2}\left(−\mathrm{2}\right)^{\mathrm{6}\:} ×\mathrm{cos}\:\left(\frac{\mathrm{6}\pi}{\mathrm{3}}\right) \\ $$$$\:\:\:\:\:=\:\mathrm{128}.\:{that}\:{right}\:{sir} \\ $$
Commented by mr W last updated on 01/Feb/20
that′s not correct sir!
$${that}'{s}\:{not}\:{correct}\:{sir}! \\ $$
Commented by peter frank last updated on 01/Feb/20
thank you all
$${thank}\:{you}\:{all} \\ $$
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
$${if}\:{x}+\frac{\mathrm{1}}{{x}}=\:−\mathrm{2}\: \\ $$$$\Rightarrow{x}=−\mathrm{1} \\ $$$$\Rightarrow{x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=−\mathrm{2} \\ $$$$\Rightarrow{x}^{\mathrm{6}} +\frac{\mathrm{1}}{{x}^{\mathrm{6}} }=\mathrm{2} \\ $$
Commented by jagoll last updated on 01/Feb/20
special means only for form x+(1/x)=−1
$${special}\:{means}\:{only}\:{for}\:{form}\: \\ $$$${x}+\frac{\mathrm{1}}{{x}}=−\mathrm{1}\: \\ $$

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