If the complex cube roots of $(- i)$ are $α, β, γ,$ then $α^{2} + β^{2} + γ^{2} =$

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– i

answer is D.

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Detailed Solution

$\begin{array}{rcl} - i & = & \cos \frac{π}{2} - i \sin \frac{π}{2} \end{array}$

$\begin{array}{rcl} {(- i)}^{1 / 3} & = & {(\cos \frac{π}{2} - i \sin \frac{π}{2})}^{1 / 3} \end{array}$

$\begin{array}{rcl} = & \cos \frac{\frac{π}{2} + 2 kπ}{3} - i \sin \frac{2 kπ + \frac{π}{2}}{3}; k = 0, 1, 2 \end{array}$

$\begin{array}{rcl} α & = & \cos \frac{π}{6} - i \sin \frac{π}{6}, β = \cos \frac{5 π}{6} - i \sin \frac{5 π}{6}, \end{array}$

$\begin{array}{rcl} γ & = & \cos \frac{9 π}{6} - i \sin \frac{9 π}{6} \end{array}$

$\begin{array}{rcl} α^{2} + β^{2} + γ^{2} \end{array}$

$\begin{array}{rcl} = & {(\cos \frac{π}{6} - i \sin \frac{π}{6})}^{2} + {(\cos \frac{5 π}{6} - i \sin \frac{5 π}{6})}^{2} + {(\cos \frac{3 π}{2} - i \sin \frac{3 π}{2})}^{2} \end{array}$

$\begin{array}{rcl} = & \cos \frac{π}{3} - i \sin \frac{π}{3} + \cos \frac{5 π}{3} - i \sin \frac{5 π}{3} + {(i)}^{2} \end{array}$

$\begin{array}{rcl} = \end{array} \begin{array}{rcl} \frac{1}{2} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} \frac{i \sqrt{3}}{2} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{1}{2} \end{array} \begin{array}{rcl} + \end{array} \begin{array}{rcl} \frac{i \sqrt{3}}{2} \end{array} \begin{array}{rcl} - \end{array} \begin{array}{rcl} 1 \end{array} = 0$