If $i = \sqrt{-1}$,then $4 + 5\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{334} + 3\left(-\frac{1}{2} + \frac{i\sqrt{3}}{2}\right)^{365}$ is equal to

  • A
    $1 - i\sqrt{3}$
  • B
    $-1 + i\sqrt{3}$
  • C
    $i\sqrt{3}$
  • D
    $-i\sqrt{3}$

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