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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

$z + \frac{1}{z} = - 1 \Rightarrow z^{2} + 1 = - z \Rightarrow z^{2} + z + 1 = 0$ $Δ = 1 - 4 = - 3 \Rightarrow z_{1} = \frac{- 1 + i \sqrt{3}}{2} = e^{\frac{i 2 π}{3}} a n d z_{2} = \frac{- 1 - i \sqrt{3}}{2} = e^{- \frac{i 2 π}{3}}$ $z = z_{1} \Rightarrow z^{5} + \frac{1}{z^{5}} = e^{\frac{i 10 π}{3}} + e^{- \frac{i 10 π}{3}} = 2 c o s (\frac{10 π}{3}) = 2 c o s (4 π - \frac{2 π}{3})$ $= 2 c o s (\frac{2 π}{3}) = 2 \times (- \frac{1}{2}) = - 1$ $z = z_{2} w e g e t t h e s a m e v a l u e b e c a u s e z_{2} = c o n j (z_{1})$ $f o r z^{11} + \frac{1}{z^{11}}$ $z = z_{1} =\Rightarrow z^{11} + \frac{1}{z^{11}} = e^{\frac{i 22 π}{3}} + e^{- \frac{i 22 π}{3}} = 2 c o s (\frac{22 π}{3})$ $= 2 c o s (8 π - \frac{2 π}{3}) = 2 c o s (\frac{2 π}{3}) = - 1$ $z = z_{2} w e g e t t h e s a m e v a l u e$
	Answered by som(math1967) last updated on 01/Feb/20

	${(z + \frac{1}{z})}^{2} = 1 \Rightarrow z^{2} + \frac{1}{z^{2}} = 1 - 2 = - 1$ $a g a i n {(z + \frac{1}{z})}^{3} = - 1$ $z^{3} + \frac{1}{z^{3}} + 3 (z + \frac{1}{z}) = - 1$ $z^{3} + \frac{1}{z^{3}} = - 1 + 3 = 2$ $∴ (z^{2} + \frac{1}{z^{2}}) (z^{3} + \frac{1}{z^{3}}) = (- 1) (2) = - 2$ $z^{5} + \frac{1}{z^{5}} + z + \frac{1}{z} = - 2$ $∴ z^{5} + \frac{1}{z^{5}} = - 2 + 1 = - 1 (p r o v e d)$ $A g a i n {(z^{3} + \frac{1}{z^{3}})}^{2} = 4$ $z^{6} + \frac{1}{z^{6}} = 4 - 2 = 2$ $∴ (z^{5} + \frac{1}{z^{5}}) (z^{6} + \frac{1}{z^{6}}) = (- 1) (2)$ $z^{11} + \frac{1}{z^{11}} + z + \frac{1}{z} = - 2$ $∴ z^{11} + \frac{1}{z^{11}} = - 2 + 1 = - 1 a n s$
	Answered by mr W last updated on 01/Feb/20
	$z+(1/z)=−1 z^2 +z+1=0 z=((−1±i(√3))/2)=−{cos (±(π/3))+sin (±(π/3)) i}=−e^(±(π/3)i) z^n +(1/z^n )=(−1)^n [e^(±((nπ)/3)i) +e^(∓((nπ)/3)i) ] =2(−1)^n cos (±((nπ)/3)) =2(−1)^n cos (((nπ)/3)) ⇒z^n +(1/z^n )=2(−1)^n cos (((nπ)/3)) n=5: z^5 +(1/z^5 )=2(−1)^5 cos (((5π)/3))=−2×(1/2)=−1 n=11: z^(11) +(1/z^(11) )=2(−1)^(11) cos (((11π)/3))=−2×(1/2)=−1 n=3: z^3 +(1/z^3 )=2(−1)^3 cos (((3π)/3))=−2×(−1)=2 n=12: z^(12) +(1/z^(12) )=2(−1)^(12) cos (((12π)/3))=2×1=2$
	$z + \frac{1}{z} = - 1$ $z^{2} + z + 1 = 0$ $z = \frac{- 1 \pm i \sqrt{3}}{2} = - {\cos (\pm \frac{π}{3}) + \sin (\pm \frac{π}{3}) i} = - e^{\pm \frac{π}{3} i}$ $z^{n} + \frac{1}{z^{n}} = {(- 1)}^{n} [e^{\pm \frac{n π}{3} i} + e^{\mp \frac{n π}{3} i}]$ $= 2 {(- 1)}^{n} \cos (\pm \frac{n π}{3})$ $= 2 {(- 1)}^{n} \cos (\frac{n π}{3})$ $\Rightarrow z^{n} + \frac{1}{z^{n}} = 2 {(- 1)}^{n} \cos (\frac{n π}{3})$ $n = 5 :$ $z^{5} + \frac{1}{z^{5}} = 2 {(- 1)}^{5} \cos (\frac{5 π}{3}) = - 2 \times \frac{1}{2} = - 1$ $n = 11 :$ $z^{11} + \frac{1}{z^{11}} = 2 {(- 1)}^{11} \cos (\frac{11 π}{3}) = - 2 \times \frac{1}{2} = - 1$ $n = 3 :$ $z^{3} + \frac{1}{z^{3}} = 2 {(- 1)}^{3} \cos (\frac{3 π}{3}) = - 2 \times (- 1) = 2$ $n = 12 :$ $z^{12} + \frac{1}{z^{12}} = 2 {(- 1)}^{12} \cos (\frac{12 π}{3}) = 2 \times 1 = 2$
	Commented by jagoll last updated on 01/Feb/20

	$i f x + \frac{1}{x} = - 2$ $t h a t x^{6} + \frac{1}{x^{6}} = 2 {(- 2)}^{6} \times \cos (\frac{6 π}{3})$ $= 128 . t h a t r i g h t s i r$
	Commented by mr W last updated on 01/Feb/20

	${t h a t}^{'} s n o t c o r r e c t s i r!$
	Commented by peter frank last updated on 01/Feb/20

	$t h a n k y o u a l l$
	Commented by mr W last updated on 01/Feb/20

	$i f x + \frac{1}{x} = - 2$ $\Rightarrow x = - 1$ $\Rightarrow x^{3} + \frac{1}{x^{3}} = - 2$ $\Rightarrow x^{6} + \frac{1}{x^{6}} = 2$
	Commented by jagoll last updated on 01/Feb/20

	$s p e c i a l m e a n s o n l y f o r f o r m$ $x + \frac{1}{x} = - 1$
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