The complex roots of a biquadratic polynom In my recent post I have a problem with the following function: $x^4-4x^2+16$, and what I need is to find the complex roots.

Here is my answer:
First step, I make the substitution $x^2=y$ which involving $y^2-4y^2+16$, with $x^2=2\pm i\sqrt{12}$. Therfore:

$x^4-4x^2+16=(x^2-2-i\sqrt{12})(x^2-2+i\sqrt{12})
=(x+\sqrt{2+i\sqrt{12}})(x+\sqrt{2-i\sqrt{12}})(x-i\sqrt{2+i\sqrt{12}})(x-\sqrt{2-i\sqrt{12}})$

factorized irreducible over $\mathbb{R}\subset\mathbb{C}$.

Here is author's answer: 
$x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$.
I realize the problem is at factorization: $x^4-4x^2+16=(x^2+bx+c)(x^2+dx+e)$. 
If I pair the complex roots from author's answer I'll obtain: 
$x^4-4x^2+16=(x+\sqrt{3}+i)(x+\sqrt{3}-i)(x-\sqrt{3}-i)(x-\sqrt{3}+i)=(x^2+\sqrt{12}x+4)(x^2-\sqrt{12}x+4)$
factorized irreducible over $\mathbb{Q}\subset\mathbb{R}$, but reducible over $\mathbb{C}$.


My final question: Why $x_1{,_2,_3,_4}=\pm\sqrt{2\pm i\sqrt{12}}$ isn't enough to make the assessment that it is the final answer ? How can I get author's answer $x{_1,_2,_3,_4}=\pm(\sqrt{3}\pm i)$ begining from my complex roots found ? Why we obtain 2 different form of complex roots, who's belong in the same field $\mathbb{C}$ when we change factorization ?


Hint:
with your substitution, the solutions that you have are:
$$
x^2=2(1\pm i\sqrt{3})=4 e^{\pm i \frac{\pi}{3}}
$$
and this gives immediately the autor's answer.
added:
The complex number $a+ib=2+2i\sqrt{3} \rightarrow $ can be write in polar form $\rho e^{i\theta}$ with: 
$$
\rho=\sqrt{a^2+b^2}=\sqrt{4+12}=4
$$
$$
\theta= \arctan\left(\dfrac{b}{a}\right)=\arctan\left(\dfrac{2\sqrt{3}}{2}\right)= \dfrac{\pi}{3} 
$$
And if $x^2=4 e^{\pm i \frac{\pi}{3}}$ we have $x=\pm 2 e^{\pm i \frac{\pi}{6}}$, that, in binomial form, are the solutions.
