Question: If we look at the equation
\begin{align}
 z = \sqrt{ 8 - 6 i },
 \end{align}
we will find the solutions
\begin{align}
 z\_1 = -3+i
 \end{align}
\begin{align}
 z\_2 = 3 - i
 \end{align}

How can they be both correct if we can substitute
\begin{align}
 \sqrt{ 8 - 6 i } = -3+i
 \end{align}
\begin{align}
 \sqrt{ 8 - 6 i } = 3-i
 \end{align}
and if
\begin{align}
 \sqrt{ 8 - 6 i } = \sqrt{ 8 - 6 i }
 \end{align}
then
\begin{align}
 3-i = -3+i
 \end{align}
and that is of course false. I can't work out where I'm going wrong.

Answer: 
This is because taking square roots in the complex plane is not an injective function. In the case of positive real numbers, the square root function is uniquely defined as the positive square root of the number. Here is a quick explanation of branches in the complex plane: <https://plus.maths.org/content/maths-minute-choosing-square-roots>
This is precisely why the use of the radical function is restricted to *positive real numbers*. When looking for the square **rootS** (notice the plural) of a complex number $a$, you actually look for the solutionS of the following equation :
$$ \boxed{z^2 = a}$$

For $a \neq 0$, there are exactly $2$ distinct solutions to the equation.
Notice that if you already found a solution $z\_1$, then $z\_2 := - z\_1$ is a solution as well since it satisfies ${z\_2}^2 = (-z\_1)^2 = {z\_1}^2 = a$.

This is why the equation above does **not** substitute into $z = \sqrt{a}$.

> 
> Writing "$z^2 = a \Rightarrow z = \sqrt{a}$" is hence **not a valid argument**.
> 
> 
> 

For example :

* $a > 0$ a positive real, eg $a = 2$  

The equation $z^2 = 2$ has two solutions : $\sqrt{2}$ **and** $- \sqrt{2}$.
* $a < 0$ a negative real, eg $a = - 2$  

The equation $z^2 = - 2$ has two solutions : $\sqrt{2} \cdot i $ **and** $- \sqrt{2} \cdot i$.
* $a \notin \mathbb{R}$ a strictly complex number, eg $a = 8 - 6 \cdot i$  

The equation $z^2 = 8 - 6 \cdot i$ has two solutions :
$$\boxed{-3 + i \textbf{ and } 3 - i}$$

All in all, remember that $\sqrt{a}$ only "works" if $a$ is a positive real number, and even so it provides you anly one of the two solutions of $z^2 = a$, namely the positive one.