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Question Number 17392 by tawa tawa last updated on 05/Jul/17

Find the cube root of z = − 1

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{cube}\:\mathrm{root}\:\mathrm{of}\:\:\mathrm{z}\:=\:−\:\mathrm{1} \\ $$

Answered by mrW1 last updated on 05/Jul/17

−1 −ω=(1/2)+((√3)/2)i −ω^2 =(1/2)−((√3)/2)i

$$−\mathrm{1} \\ $$$$−\omega=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$$$−\omega^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i} \\ $$

Commented by tawa tawa last updated on 05/Jul/17

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Answered by Arnab Maiti last updated on 07/Jul/17

put x^3 =z ∴ x^3 =−1 or, x^3 +1=0 or, (x+1)(x^2 −x+1)=0 ∴ (1) x+1=0 ⇒ x=−1 or (2) x^2 −x+1=0 ∴ x=((1±(√(1−4)))/2) ∴ x=((1±i(√3))/2)= { ((−ω)),((−ω^2 )) :}

$$\mathrm{put}\:\mathrm{x}^{\mathrm{3}} =\mathrm{z} \\ $$$$\therefore\:\mathrm{x}^{\mathrm{3}} =−\mathrm{1} \\ $$$$\mathrm{or},\:\mathrm{x}^{\mathrm{3}} +\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{or},\:\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\therefore\:\left(\mathrm{1}\right)\:\mathrm{x}+\mathrm{1}=\mathrm{0}\:\:\Rightarrow\:\mathrm{x}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{or} \\ $$$$\left(\mathrm{2}\right)\:\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}=\mathrm{0} \\ $$$$\therefore\:\:\mathrm{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{4}}}{\mathrm{2}} \\ $$$$\therefore\:\:\mathrm{x}=\frac{\mathrm{1}\pm{i}\sqrt{\mathrm{3}}}{\mathrm{2}}=\begin{cases}{−\omega}\\{−\omega^{\mathrm{2}} }\end{cases} \\ $$

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