I am new to working with complex numbers and am confused about using existing methods of working with indicies.
Consider that:
\begin{equation}\begin{aligned} & x = -1\\ &(\sqrt{x})^{3}=-i\\ \end{aligned}\end{equation}
I am perfectly comfortable with this result, given that using exponential form and raising to a power of 3/2 gives the same result.
My problem is, if you apply the cube before the square root, you get a different result:
\begin{equation}\begin{aligned} &\sqrt{(x)^{3}}=i \end{aligned}\end{equation}
\begin{equation}\begin{aligned} & \text{Since } (-1)^{3} = -1. \end{aligned}\end{equation}
So in this case:
\begin{equation}\begin{aligned} (\sqrt{x})^{3} \ne \sqrt{x^{3}} \end{aligned}\end{equation}
Why is this problem introduced and is there an explanation of how to resolve this other than saying the normal rules don't apply? I want to be sure about when I can apply certain operations when working with complex numbers and to this point any new mathematics I have learned is completely consistent with existing rules (meaning you cannot get a false result).
Short answer: fractional and real exponents applied to complex numbers don't behave the way you expect. Natural-number exponents do. Integer exponents do.
In particular, you've tried to use the rule that $$ x^{\frac a b} = (x^a)^\frac1b, $$ and that one doesn't work in general. That's sort of a pity, but it's also the starting point for a lot of interesting mathematics.
All the other rules you know and love, like $$ x^{a + b}= x^a \cdot x^b $$ work fine, too. But just as, for real numbers, $\log$ is defined only on the positive real-axis, so rules like $$ \log(a^b) = b \log a $$ only make sense for positive values $a$, for the complexes $\log$ is defined on a larger domain...so things have the potential to be more subtle.
If you want to know what rules apply, one possibility is to work with the actual definitions and check.