If $z = \frac{\sqrt{3} + i}{2}$, then $z^{101} + z^{103} =$
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Any expression $z^{n-1} + z^{n+1}$ can be factored as $z^n(z + 1/z)$. Since $z + z^{-1} = 2\text{Re}(z) = \sqrt{3}$ and $z^{102} = (z^6)^{17} = (-1)^{17} = -1$, the result is simply $-\sqrt{3}$.