find-the-zero-of-z-3-729-0-z-C-

Question Number 142623 by Engr_Jidda last updated on 03/Jun/21

$${find}\:{the}\:{zero}\:{of}\:{z}^{\mathrm{3}} +\mathrm{729}=\mathrm{0} \\ $$$${z}\in\mathbb{C} \\ $$

Answered by Olaf_Thorendsen last updated on 03/Jun/21

z^3 = −729 = 9^3 e^(iπ) z = 9e^(i((π/3)+((2kπ)/3))) , k = 0, 1, 2 z ∈ {(9/2)(1±i(√3)), −9}

$${z}^{\mathrm{3}} \:=\:−\mathrm{729}\:=\:\mathrm{9}^{\mathrm{3}} {e}^{{i}\pi} \\ $$$${z}\:=\:\mathrm{9}{e}^{{i}\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{2}{k}\pi}{\mathrm{3}}\right)} ,\:{k}\:=\:\mathrm{0},\:\mathrm{1},\:\mathrm{2} \\ $$$${z}\:\in\:\left\{\frac{\mathrm{9}}{\mathrm{2}}\left(\mathrm{1}\pm{i}\sqrt{\mathrm{3}}\right),\:−\mathrm{9}\right\} \\ $$

Answered by Rasheed.Sindhi last updated on 03/Jun/21

z^3 +9^3 =0 (z+9)(z^2 −9z+81)=0 z+9=0 ∣ z^2 −9z+81=0 z=−9 ∣ z=((9±(√(81−4(81))))/2) ∣ z=((9±(√(−3(81))))/2) ∣ z=((9±9i(√3))/2)=−9(((−1∓i(√3))/2)) ∣ z=−9ω,−9ω^2 z=−9,−9ω,−9ω^2

$${z}^{\mathrm{3}} +\mathrm{9}^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({z}+\mathrm{9}\right)\left({z}^{\mathrm{2}} −\mathrm{9}{z}+\mathrm{81}\right)=\mathrm{0} \\ $$$${z}+\mathrm{9}=\mathrm{0}\:\:\mid\:\:{z}^{\mathrm{2}} −\mathrm{9}{z}+\mathrm{81}=\mathrm{0} \\ $$$${z}=−\mathrm{9}\:\:\mid\:\:{z}=\frac{\mathrm{9}\pm\sqrt{\mathrm{81}−\mathrm{4}\left(\mathrm{81}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{z}=\frac{\mathrm{9}\pm\sqrt{−\mathrm{3}\left(\mathrm{81}\right)}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{z}=\frac{\mathrm{9}\pm\mathrm{9}{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=−\mathrm{9}\left(\frac{−\mathrm{1}\mp{i}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:{z}=−\mathrm{9}\omega,−\mathrm{9}\omega^{\mathrm{2}} \\ $$$${z}=−\mathrm{9},−\mathrm{9}\omega,−\mathrm{9}\omega^{\mathrm{2}} \\ $$

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