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!

$\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=\sqrt{6}$
Is an addition of complex conjugates of the form 
$(a+b\sqrt{-1}) + (a-b\sqrt{-1})$
So then
$1+\sqrt{-3}=(a+b\sqrt{-1})^2$
$=a^2+2ab\sqrt{-1}-b^2$
If we equal the real and complex parts we get:
$1=a^2-b^2$  and  $\sqrt{-3}=2ab\sqrt{-1}$
Solving for a in the complex equation gives: 
$a=\frac{\sqrt{3}}{2b}$
Plugging in $a=\frac{\sqrt{3}}{2b}$  into  $1=a^2-b^2$ and multiplying both sides by $b^2$ yields: 
$b^4+b^2-\frac{3}{4}$ which is a quadratic for $b^2$
The quadratic equations then gives $b^2=\frac{-1\pm2}{2}$ 
Using only the positive root means $b=\frac{\sqrt{2}}{2}$
Plug in $b=\frac{\sqrt{2}}{2}$ into $1=a^2-b^2$ to get $a=\frac{\sqrt{6}}{2}$
Therefore
$\sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}=(\frac{\sqrt{6}}{2}+\frac{\sqrt{2}}{2}i)+(\frac{\sqrt{6}}{2}-\frac{\sqrt{2}}{2}i)=\frac{\sqrt{6}}{2}+\frac{\sqrt{6}}{2}=\sqrt{6}$
