<s>[INST] 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!
 [/INST] Let $x=\sqrt{1+i\sqrt3}+\sqrt{1-i\sqrt3}$.  Then 
$$x^2=(1+i\sqrt3)+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}+(1-i\sqrt3)=2+2\sqrt{1+i\sqrt3}\sqrt{1-i\sqrt3}$$
so
$$(x^2-2)^2=4(1+i\sqrt3)(1-i\sqrt3)=4(1+3)=16$$
so
$$x^4-4x^2-12=(x^2-6)(x^2+2)=0$$
Thus $x\in\{\sqrt6,-\sqrt6,i\sqrt2,-i\sqrt2\}$, so it remains to determine which of these four roots is meant.
The answer depends on which convention you decide to use when computing the square root of a complex number, $\sqrt{a+ib}$ with $b\not=0$.  There are two common conventions:  require $\sqrt{a+ib}$ with $b\not=0$ to have positive real part, or require it to have positve imaginary part.  (In terms of polar coordinates, this amounts to saying $\sqrt{re^{i\theta}}=\sqrt re^{i\theta/2}$, but with $-\pi\lt\theta\le\pi$ for the first convention and $0\le\theta\lt2\pi$ for the second.)
If we assume the first convention, then $x$, being the sum of two square roots, must have positive real part, hence must equal $\sqrt6$.  If we assume the second convention, then $x$ must have positive imaginary part, hence must equal $i\sqrt2$.  Since the desired answer is $\sqrt6$, we see that the problem is tacitly assuming the first convention.
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