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 ?

 A: 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.
A: Note that your answer gives that same result as author's answer, but the more important thing to pay attention with is when you use $\sqrt z$ with $z$ a complex numbers this is not defind so in I would say that your answer is incomplete because you did not find the complex square roots of $2\mp i\sqrt {12}$, and when you did this you will find the same result as author's. 
Another way to see things is that any complex number is of the form $x+iy$ where $x,y\in \Bbb R$ which is the canonical form. and your answer does not give the roots of this form. 
Maybe we can use that:
$$\begin{align}x^4-4x^2+16&=x^2-2\cdot4x^2+16+4x^2\\
&=(x^2-4)^2+4x^2\\
&=(x^2-2ix-4)(x^2+2ix-4)\\
&=((x-i)^2-3)((x+i)^2-3)\\
&=(x-i-\sqrt3)(x-i+\sqrt3)(x+i-\sqrt3)(x+i+\sqrt3)\end{align}$$
this is how the author did it, but you can simplify your expression, for instance:
$$\pm\sqrt{2\pm i\sqrt{12}}=\mp \sqrt{2+\mp 2i\sqrt 3}=\mp \sqrt{3^2+\mp 2i\sqrt 3 +i^2}=\mp \sqrt{(i\mp\sqrt 3)^2}=\pm(\sqrt{3}\pm i)$$
A: To denest the square roots you can employ the simple, memorable algorithm here.
Then $\:2\pm i\sqrt{12}\:$ has norm $= 16.\:$ $\rm\ \color{blue}{subtracting\ out}\ \sqrt{norm}\ = 4\ $ yields $\  {-2}\pm i\sqrt{12}\:$
with $\ {\rm\ \sqrt{trace}}\: =\: \sqrt{-4}\ =\ 2i.\quad\ \ \ \rm \color{brown}{Dividing\ it\ out}\ $ of the above  yields $\quad\ \ \ i\pm\sqrt 3$ 
A: all you have to do is show that $\sqrt{2\pm i\sqrt 12} = \sqrt 3 \pm i$ squaring both side should give you that.
