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Question Number 140809 by mathdanisur last updated on 12/May/21
z^5 + (1/z^5 ) = ((205)/(16)) ∙ (z + (1/z))
$$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:=\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right) \\ $$
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
$$\boldsymbol{{z}}^{\mathrm{5}} \:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{5}} }\:−\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right)=\mathrm{0} \\ $$$${a}^{\mathrm{5}} +{b}^{\mathrm{5}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\left({a}+{b}\right)\left({a}^{\mathrm{4}} −{a}^{\mathrm{3}} {b}+{a}^{\mathrm{2}} {b}^{\mathrm{2}} −{ab}^{\mathrm{3}} +{b}^{\mathrm{4}} \right) \\ $$$$\left({z}+\frac{\mathrm{1}}{{z}}\right)\left\{{z}^{\mathrm{4}} −{z}^{\mathrm{3}} \left(\frac{\mathrm{1}}{{z}}\right)+{z}^{\mathrm{2}} \left(\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} −{z}\left(\frac{\mathrm{1}}{{z}}\right)^{\mathrm{3}} +\left(\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\:\frac{\mathrm{205}}{\mathrm{16}}\:\centerdot\:\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right)=\mathrm{0} \\ $$$$\left(\boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}\right)\left\{{z}^{\mathrm{4}} −{z}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{1}}{{z}^{\mathrm{2}} }+\frac{\mathrm{1}}{{z}^{\mathrm{4}} }−\frac{\mathrm{205}}{\mathrm{16}}\right\}=\mathrm{0} \\ $$$$\:^{\bullet} \boldsymbol{{z}}\:+\:\frac{\mathrm{1}}{\boldsymbol{{z}}}=\mathrm{0}\Rightarrow{z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow{z}=\pm{i} \\ $$$$\:^{\bullet} {z}^{\mathrm{4}} +\frac{\mathrm{1}}{{z}^{\mathrm{4}} }−\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\frac{\mathrm{189}}{\mathrm{16}}=\mathrm{0} \\ $$$$\:\:\:\:\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)^{\mathrm{2}} −\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\frac{\mathrm{221}}{\mathrm{16}}=\mathrm{0} \\ $$$$\:\:\:{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\frac{\mathrm{221}}{\mathrm{4}}}}{\mathrm{2}} \\ $$$$\:\:\:{z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }=\frac{\mathrm{1}\pm\frac{\mathrm{15}}{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{2}\pm\mathrm{15}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{17}}{\mathrm{4}},−\frac{\mathrm{13}}{\mathrm{4}} \\ $$$$\:\:\:\:\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} =\frac{\mathrm{17}}{\mathrm{4}}+\mathrm{2},−\frac{\mathrm{13}}{\mathrm{4}}+\mathrm{2}\: \\ $$$$\:\:\:\:\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{4}}\:,−\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:{z}+\frac{\mathrm{1}}{{z}}=\pm\frac{\mathrm{5}}{\mathrm{2}},\frac{\pm{i}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:{z}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{5}}{\mathrm{2}}\:\:\mid\:\:{z}+\frac{\mathrm{1}}{{z}}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{2}{z}^{\mathrm{2}} −\mathrm{5}{z}+\mathrm{2}=\mathrm{0}\:\mid\:\mathrm{2}{z}^{\mathrm{2}} +\mathrm{5}{z}+\mathrm{2}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{4}}\:\:\mid\:{z}=\frac{−\mathrm{5}\pm\sqrt{\mathrm{25}−\mathrm{16}}}{\mathrm{4}} \\ $$$$\:{z}\:=\frac{\mathrm{5}\pm\mathrm{9}}{\mathrm{4}}\:\mid\:{z}=\frac{−\mathrm{5}\pm\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:{z}=\frac{\mathrm{7}}{\mathrm{2}},−\mathrm{1}\:\mid\:{z}=\mathrm{1},−\frac{\mathrm{7}}{\mathrm{2}} \\ $$$${Continue} \\ $$
Commented by mathdanisur last updated on 13/May/21
cool Sit thanks
$${cool}\:{Sit}\:{thanks} \\ $$

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