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: 
It isn't right to write $\sqrt{8-6i}=3-i$ (the same for the other solution). When you ask, $\sqrt{z}=?$ you mean to say " what **thing** multiplied by itself gives $z$". It just so happens that this question will not yield an answer that is a single number, but the "thing" is the elements present in the set comprised of two numbers. So what is meant as the solution for this problem is
$$z=\sqrt{8-6i} \implies z=\pm(3-i)$$
Some problems even have an infinite number of solutions (even in the real domain), such as trigonometric equations. For example, $\sin{x}=0 $ has an infinite number of solutions on the number line, $x=k\pi$, $k\in \mathbb{Z}$, but it isn't true that they're all the same just because their output through the function "$\sin{x}$" is the same number.  

Moreover you can see that no contradiction arises, because
$(\pm(3-i))^2=8-6i$ (which doesn't mean $3-i=-3+i$.
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.