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If $z = \frac{\sqrt{3} - i}{2}, then {(i^{101} + z^{101})}^{103}$ equals to

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$z^{2}$

-z

answer is D.

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Detailed Solution

$\begin{array}{l} z = \frac{\sqrt{3} - i}{2} \Rightarrow z = {iω}^{2} \\ {(i^{101} + z^{101})}^{103} = {[i^{101} + {({iω}^{2})}^{101}]}^{103} = (i + iω)^{103} = i^{103} {(- ω^{2})}^{103} \\ = - i (- ω^{2}) = {iω}^{2} = z \end{array}$

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