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!

Changing into polar form we have $ 1+ i \sqrt{3} = 2 e^{i\pi/3}$ and $1 - i \sqrt{3} = 2e^{-i\pi/3}$ so the left hand side is $$ \sqrt{2} \left( e^{i\pi/6} + e^{-i\pi/6} \right)= 2 \sqrt{2} \cos(\pi/6)= 2\sqrt{2} \cdot \frac{\sqrt{3}}{2}= \sqrt{6}. $$

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.