Help me prove $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$ Please help me prove this Leibniz equation: $\sqrt{1+i\sqrt 3}+\sqrt{1-i\sqrt 3}=\sqrt 6$. Thanks!

$\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3} = \sqrt 6$ ?
$1 + i \sqrt 3 = 2 \exp \left( \dfrac {\pi}{3}i + 2 \pi  n i \right)
 \quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} = \sqrt 2 \exp \left( \dfrac {\pi}{6}i + \pi  n i \right)
 \quad \{ n \in \mathbb Z \}$
$\sqrt{1 + i \sqrt 3} =
 \pm \left( \dfrac{\sqrt 6}{2} + \dfrac{\sqrt 2}{2} i \right)$
Similarly
$\sqrt{1 - i \sqrt 3} =
 \pm \left( \dfrac{\sqrt 6}{2} - \dfrac{\sqrt 2}{2} i \right)$
So there are four possible values of $\sqrt{1 + i \sqrt 3} + \sqrt{1 - i\sqrt 3}$
One of then is $\sqrt 6$.
