z^5 + (1/z^5 ) = ((205)/(16)) ∙ (z + (1/z))


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Question Number 140809 by mathdanisur last updated on 12/May/21

$z^{5} + \frac{1}{z^{5}} = \frac{205}{16} \cdot (z + \frac{1}{z})$
	Answered by Rasheed.Sindhi last updated on 19/May/21
	$z^5 + (1/z^5 ) − ((205)/(16)) ∙ (z + (1/z))=0 a^5 +b^5 =(a+b)(a^4 −a^3 b+a^2 b^2 −ab^3 +b^4 ) (z+(1/z)){z^4 −z^3 ((1/z))+z^2 ((1/z))^2 −z((1/z))^3 +((1/z))^4 } − ((205)/(16)) ∙ (z + (1/z))=0 (z + (1/z)){z^4 −z^2 +1−(1/z^2 )+(1/z^4 )−((205)/(16))}=0 ^• z + (1/z)=0⇒z^2 +1=0⇒z=±i ^• z^4 +(1/z^4 )−(z^2 +(1/z^2 ))−((189)/(16))=0 (z^2 +(1/z^2 ))^2 −(z^2 +(1/z^2 ))−((221)/(16))=0 z^2 +(1/z^2 )=((1±(√(1+((221)/4))))/2) z^2 +(1/z^2 )=((1±((15)/2))/2)=((2±15)/4) =((17)/4),−((13)/4) (z+(1/z))^2 =((17)/4)+2,−((13)/4)+2 (z+(1/z))^2 =((25)/4) ,−(5/4) z+(1/z)=±(5/2),((±i(√5))/2) z+(1/z)=(5/2) ∣ z+(1/z)=−(5/2) 2z^2 −5z+2=0 ∣ 2z^2 +5z+2=0 z=((5±(√(25−16)))/4) ∣ z=((−5±(√(25−16)))/4) z =((5±9)/4) ∣ z=((−5±9)/4) z=(7/2),−1 ∣ z=1,−(7/2) Continue$
	$z^{5} + \frac{1}{z^{5}} - \frac{205}{16} \cdot (z + \frac{1}{z}) = 0$ $a^{5} + b^{5}$ $= (a + b) (a^{4} - a^{3} b + a^{2} b^{2} - {a b}^{3} + b^{4})$ $(z + \frac{1}{z}) {z^{4} - z^{3} (\frac{1}{z}) + z^{2} {(\frac{1}{z})}^{2} - z {(\frac{1}{z})}^{3} + {(\frac{1}{z})}^{4}}$ $- \frac{205}{16} \cdot (z + \frac{1}{z}) = 0$ $(z + \frac{1}{z}) {z^{4} - z^{2} + 1 - \frac{1}{z^{2}} + \frac{1}{z^{4}} - \frac{205}{16}} = 0$ $^{∙} z + \frac{1}{z} = 0 \Rightarrow z^{2} + 1 = 0 \Rightarrow z = \pm i$ $^{∙} z^{4} + \frac{1}{z^{4}} - (z^{2} + \frac{1}{z^{2}}) - \frac{189}{16} = 0$ ${(z^{2} + \frac{1}{z^{2}})}^{2} - (z^{2} + \frac{1}{z^{2}}) - \frac{221}{16} = 0$ $z^{2} + \frac{1}{z^{2}} = \frac{1 \pm \sqrt{1 + \frac{221}{4}}}{2}$ $z^{2} + \frac{1}{z^{2}} = \frac{1 \pm \frac{15}{2}}{2} = \frac{2 \pm 15}{4}$ $= \frac{17}{4}, - \frac{13}{4}$ ${(z + \frac{1}{z})}^{2} = \frac{17}{4} + 2, - \frac{13}{4} + 2$ ${(z + \frac{1}{z})}^{2} = \frac{25}{4}, - \frac{5}{4}$ $z + \frac{1}{z} = \pm \frac{5}{2}, \frac{\pm i \sqrt{5}}{2}$ $z + \frac{1}{z} = \frac{5}{2} ∣ z + \frac{1}{z} = - \frac{5}{2}$ $2 z^{2} - 5 z + 2 = 0 ∣ 2 z^{2} + 5 z + 2 = 0$ $z = \frac{5 \pm \sqrt{25 - 16}}{4} ∣ z = \frac{- 5 \pm \sqrt{25 - 16}}{4}$ $z = \frac{5 \pm 9}{4} ∣ z = \frac{- 5 \pm 9}{4}$ $z = \frac{7}{2}, - 1 ∣ z = 1, - \frac{7}{2}$ $C o n t i n u e$
	Commented by mathdanisur last updated on 13/May/21

	$c o o l S i t t h a n k s$
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