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.

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 $\left(-\sqrt{\frac{\sqrt{a^2 + b^2} +  a}{2}},\mp\sqrt{\frac{\sqrt{a^2 + b^2} -  a}{2}}\,\right)$.
