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Question Number 124320 by sogol last updated on 02/Dec/20

z+(1/z)=1 z^3 =??

$${z}+\frac{\mathrm{1}}{{z}}=\mathrm{1} \\ $$$${z}^{\mathrm{3}} =?? \\ $$

Answered by Dwaipayan Shikari last updated on 02/Dec/20

z^2 −z+1=0⇒z=((1±i(√3))/2)=e^(±((iπ)/3)) z^3 =e^(±iπ) =−1

$${z}^{\mathrm{2}} −{z}+\mathrm{1}=\mathrm{0}\Rightarrow{z}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}={e}^{\pm\frac{{i}\pi}{\mathrm{3}}} \\ $$$${z}^{\mathrm{3}} ={e}^{\pm{i}\pi} =−\mathrm{1} \\ $$

Answered by $@y@m last updated on 02/Dec/20

(z+(1/z))^3 =1^3 z^3 +(1/z^3 )+3(z+(1/z))=1 z^3 +(1/z^3 )+3=1 z^3 +(1/z^3 )+2=0 (z^3 )^2 +2z^3 +1=0 (z^3 +1)^2 =0 z^3 =−1

$$\left({z}+\frac{\mathrm{1}}{{z}}\right)^{\mathrm{3}} =\mathrm{1}^{\mathrm{3}} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }+\mathrm{3}\left({z}+\frac{\mathrm{1}}{{z}}\right)=\mathrm{1} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }+\mathrm{3}=\mathrm{1} \\ $$$${z}^{\mathrm{3}} +\frac{\mathrm{1}}{{z}^{\mathrm{3}} }+\mathrm{2}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{3}} \right)^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\left({z}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${z}^{\mathrm{3}} =−\mathrm{1} \\ $$

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