Help on solving quadratic equations with complex coefficients Please help me out here, I'm self-studying Complex Numbers and I've gotten to a point where I'm kinda stuck. 
Given a general quadratic equation $az^2 + bz + c = 0$ with $a \not= 0$.
Using the same algebraic manipulation as in the case of real coefficients, we obtain 
$$a\left[\left(z + \frac{b} {2a}\right)^2 - \frac{\Delta} {4a^2}\right] = 0$$
This is equivalent to 
$$\left(z + \frac{b} {2a}\right)^2 = \frac {\Delta} {4a^2}\quad\text{or}\quad \left({2az + b}\right)^2 = \Delta$$
where $\Delta = b^2 - 4ac$ is called the discriminant of the quadratic equation, setting $y= 2az + b$, the expression is reduced to
$$y^2 = \Delta = u + vi$$
where $u$ and $v$ are real numbers.
What I don't understand now is how the author obtained 
$$y_{1,2} = \pm \left(\sqrt { \frac {r + u} {2} } + \operatorname{sgn} (v)\sqrt{ \frac{r - u} {2} }\,i\right)$$
where $r = |\Delta|$ and $\operatorname{sgn}(v)$ is the sign of the real number $v$.
 A: Note that if $y=s+it$ and $y^2=u+iv$ then, by considering separately the real and imaginary parts, we obtain
$$s^2-t^2=u,\quad 2st=v.$$
Now we solve this system with respect to $s$ and $t$. 
From the second $t=v/(2s)$ and plugging it in the first we get
$$s^2-\frac{v^2}{4s^2}=u \Leftrightarrow 4s^4-4us^2-v^2=0$$
and we obtain 
$$s^2=\frac{u+\sqrt{u^2+v^2}}{2}=\frac{|\Delta|+u}{2}$$
Note that we dropped the part with the minus sign because $s^2\geq 0$.
Finally
$$t^2=\frac{v^2}{4s^2}=\frac{v^2}{2(|\Delta|+u)}=\frac{v^2(|\Delta|-u)}{2(|\Delta|^2-u^2)}=\frac{|\Delta|-u}{2}.$$
The we take the square roots keeping in mind that $\mbox{sgn}(s\cdot t)=\mbox{sgn}(v).$
A: Discriminant $\Delta = b^2 - 4ac = \alpha + i\beta$
We need to find the square root of $\Delta$
$\sqrt{\alpha + i\beta} = p + iq$
$\implies \alpha + i\beta = p^2 - q^2 + 2ipq$
$\implies p^2 - q^2 = \alpha$ and $ 2pq = \beta$
$\implies p^2 + q^2 = \sqrt{\alpha^2 + \beta^2}$
$\implies p = \sqrt{\dfrac{\alpha + |\Delta|}{2}}$ and $ q = \sqrt{\dfrac{|\Delta| - \alpha}{2}}$
$\implies \sqrt{\Delta} = \sqrt{\dfrac{\alpha + |\Delta|}{2}} + \dfrac{\beta}{|\beta|} \sqrt{\dfrac{|\Delta| - \alpha}{2}}i$
Now you can use the quadratic formula.
