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>
Every nonzero complex number $w$ has exactly two distinct square roots (say $z\_1$ and $z\_2$, so that $z\_1^2 = z\_2^2 = w$), and they are opposites (i.e., $z\_2 = -z\_1$). These roots certainly aren't equal; what are equal are their **squares**. Your mistake is calling them both $\sqrt{w}$. It might be more correct to speak of "a square root" instead of "the square root", since there is more than one square root for each nonzero complex number.

With complex numbers, you have to be careful about what you mean by $\sqrt{w}$; it will be one of $z\_1$ or $z\_2$, but you have to decide which is which. If you choose to say that $\sqrt{w} = z\_1$, then you have immediately also determined that $z\_2 = -\sqrt{w}$.

What is really going on is that you are choosing a *branch of the logarithm*, and using the definition that $\sqrt{w} = e^{\frac12\log w}$. This means that you choose a definition of the argument of $w$, so then $\log w = \ln|w| +i\arg w$ (where $\ln$ is the ordinary real-valued function of a positive real variable).

There are several commonly used ranges for $\arg w$:

* $(-\pi,\pi]$, so $\sqrt{w}$ has argument in $(-\pi/2,\pi/2]$
* $[0,2\pi)$, so $\sqrt{w}$ has argument in $[0,\pi)$

In the first case, $\sqrt{w}$ would lie in the right open half-plane or on the positive $y$-axis, so its opposite $-\sqrt{w}$ would lie in the left open half-plane or on the negative $y$-axis.

In the second case, $\sqrt{w}$ would lie in the upper open half-plane or on the positive $x$-axis, so its opposite $-\sqrt{w}$ would lie in the lower open half-plane or on the negative $x$-axis.

Other choices for defining $\arg w$ would lead to other results for which of the two roots is called $\sqrt{w}$ and which is called $-\sqrt{w}$.