Solving the equation $X^2 + (3 +i)X + 1 +i =0$ I have to find the roots of the equation
$$X^2 + (3 +i)X + 1 +i =0.$$
The first step is to find the discrimant which is $4 + 2i$. Then, I assume that the square root of the discriminant is in the form of $a+bi$, so we have to solve the below system
$$a^2-b^2=4\qquad\text{ and }\qquad ab=1.$$
How can I solve the above system ? Please do not provide another solution (e.g using trigonometric form)
 A: You're work so far is good, and indeed you want to find $a,b\in\Bbb{R}$ such that $(a+bi)^2=4+2i$, i.e.
$$a^2-b^2=4\quad\text{ and }\qquad 2ab=2.$$
The latter shows you that $a$ and $b$ are nonzero, and that $b=a^{-1}$. Plugging this into the former yields
$$a^2-a^{-2}=4,$$
and multiplying everything by $a^2$ and rearranging shows that
$$a^4-4a^2-1=0.$$
This is a quadratic in $a^2$, so by the quadratic formula $a^2=2\pm\sqrt{5}$. Because $a$ is real
$$a=\pm\sqrt{2+\sqrt{5}}.$$
A: $(a^{2}+b^{2})^{2}=(a^{2}-b^{2})^{2}+4a^{2}b^{2}$ so $a^{2}+b^{2}=\sqrt {16+4}$ and and $a^{2}-b^{2}=4$.  Can you continue?
A: Hint:
Multiplying by $a^2$,
$$a^2-b^2=4\land ab=1\implies a^4-a^2b^2=a^4-1=4a^2$$ and you get a biquadratic equation in $a$.

More generally, the square root of $u+iv$ is obtained by solving
$$\begin{cases}a^2-b^2&=u,\\2ab&=v\end{cases}$$ and
$$a^4-ua^2-\frac{v^2}4=0.$$
So,
$$a^2=\frac{\sqrt{u^2+v^2}+u}2$$ (the negative root must be rejected) and similarly
$$b^2=\frac{\sqrt{u^2+v^2}-u}2.$$
After taking the square roots, there are two solutions such that the sign of $ab$ matches that of $v$.
