Quadratic formula for complex numbers Consider the equation 
$$
az^2+bz+c=0, 
$$
where $a,b$ and $c$ are complex numbers and $a\ne 0$. Applying usual operations on $\mathbb{C}$ we have the following:
$$
az^2+bz+c=0\iff z^2+\frac{b}{a}z = -\frac{c}{a}\iff z^2+\frac{b}{a}z + \frac{b^2}{4a^2} = \frac{b^2}{4a^2} -\frac{4ac}{4a^2} \\ \iff \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}.
$$
Now here's where my doubt comes. I know I can take the square root on both sides but wouldn't I have two results on both sides? Moreover, on the denominator of the right side (and on the whole left side) I have $z_0^2$; taking square root of this means I have only one solution ($z_0$)? Because I was told that the solution for the quadratic equation will be the same as for real numbers.
 A: What you've done is correct, right up to the end. Essentially you've rederived the "quadratic formula".
The last step isn't taking the square root of both sides, it's equating the quantity in parentheses on the left to one of the two square roots of the complex number on the right. That will give you the two solutions to the original equation. 
That will be just one solution counted twice if the discriminant is $0$, and will 
recover the usual real roots of real quadratics if the coefficients are real and the discriminant nonzero.
A: Let's call $\sigma$ the square root of $z$ which means that $\sigma$ such as $\sigma^2=z$.
Thus, if $\rho = b^2 - 4ac$ and $\sigma^2 = \rho$ then
\begin{align*}
     & \left(z + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}\\
\iff & \left(z + \frac{b}{2a}\right)^2 = \frac{\rho}{4a^2}\\
\iff & z + \frac{b}{2a} = \frac{\pm\sigma}{2a}\\
\iff & z = \frac{\pm\sigma}{2a} - \frac{b}{2a}\\
\iff & z = \frac{-b\pm\sigma}{2a}
\end{align*}
Let's say we are looking for the square root $x + i y$ of $a + i b$ where $b$ is positive.
It is possible to find the square root of a complex number like this:
\begin{align*}
     & (x + i y)^2 = a + i b \\
\iff & \begin{cases}
        x^2-y^2 &= a\\
        2xy     &= b
        \end{cases}
\iff  \begin{cases}
        x^2-y^2 &= a\\
        y       &= \frac{b}{2x}
        \end{cases}
\iff \begin{cases}
        x^2-\left(\frac{b}{2x}\right)^2 &= a\\
        y^2 &= x^2 - a\\
        2xy     &= b
        \end{cases}\\
\iff & \begin{cases}
        x^2-\frac{b^2}{4x^2} &= a\\
        y^2 &= x^2 - a\\
        2xy     &= b
        \end{cases}
\iff \begin{cases}
        \frac{4x^4-b^2}{4x^2} &= a\\
        y^2 &= x^2 - a\\
        2xy     &= b
        \end{cases}
\iff \begin{cases}
        4x^4-b^2 &= 4ax^2\\
        y^2 &= x^2 - a\\
        2xy     &= b
        \end{cases}\\
\iff & \begin{cases}
        4x^4-4ax^2-b^2 &= 0\\
        y^2 &= x^2 - a\\
        2xy     &= b
        \end{cases}\\
\end{align*}
We're looking for the $x$ such as $4x^4 - 4ax^2 - b^2 = 0$ or if we define $\lambda = x^2$ we have $4 \lambda^2 - 4a\lambda - b^2 = 0$.
$$\rho = (-4a)^2 - 4 (4) (-b^2) = 16 a^2 + 16 b^2 = 16 (a^2 + b^2)$$
$$\lambda = \frac{4a \pm 4\sqrt{a^2 + b^2}}{8} = \frac{a \pm \sqrt{a^2 + b^2}}{2}$$
Since $\lambda = x^2$, we have $\lambda \geq 0$ and thus $\lambda = \frac{a + \sqrt{a^2 + b^2}}{2}$.
It follows that $x = \pm\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}}$.
Knowing that we deduce the value of $y$: $y = \pm\sqrt{x^2-a} = \pm\sqrt{\frac{a + \sqrt{a^2 + b^2}}{2}-a} = \pm\sqrt{\frac{\sqrt{a^2 + b^2} - a}{2}}$.
Also we know that $2xy = b$ which explains why both $x = \pm \ldots$ and $y = \pm \ldots$ only produces 2 solutions instead of 4 (anyway we knew that being complex numbers there were only 2 solutions).
If $b > 0$, then $x$ and $y$ have the same sign and the solutions are $\left( \sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}} , \sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}},-\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$.
If $b < 0$, then $x$ and $y$ have opposite signs and the solutions are $\left(\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}} , -\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}},\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$.
We may conclude that the equation $(x \pm i y)^2 = a \pm i b\mbox{ with } b \mbox{ positive}$ has the following solutions: $\left( \sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}} , \pm\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$ and $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}},\mp\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$.
