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.
Best 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