find the zero of z^3 +729=0 z∈C


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

$f i n d t h e z e r o o f z^{3} + 729 = 0$ $z \in C$
	Answered by Olaf_Thorendsen last updated on 03/Jun/21
	$z^3 = −729 = 9^3 e^(iπ) z = 9e^(i((π/3)+((2kπ)/3))) , k = 0, 1, 2 z ∈ {(9/2)(1±i(√3)), −9}$
	$z^{3} = - 729 = 9^{3} e^{i π}$ $z = 9 e^{i (\frac{π}{3} + \frac{2 k π}{3})}, k = 0, 1, 2$ $z \in {\frac{9}{2} (1 \pm i \sqrt{3}), - 9}$
	Answered by Rasheed.Sindhi last updated on 03/Jun/21

	$z^{3} + 9^{3} = 0$ $(z + 9) (z^{2} - 9 z + 81) = 0$ $z + 9 = 0 ∣ z^{2} - 9 z + 81 = 0$ $z = - 9 ∣ z = \frac{9 \pm \sqrt{81 - 4 (81)}}{2}$ $∣ z = \frac{9 \pm \sqrt{- 3 (81)}}{2}$ $∣ z = \frac{9 \pm 9 i \sqrt{3}}{2} = - 9 (\frac{- 1 \mp i \sqrt{3}}{2})$ $∣ z = - 9 ω, - 9 ω^{2}$ $z = - 9, - 9 ω, - 9 ω^{2}$
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