complex numbers - Is $(-1)^{ab} = (-1)^{ba}$ true? => $(-1)^{ab} = ((-1)^a )^b$ is true?

In general we know $A^{bc} = A^{cb}$ for integer $A$.
I want to extend this to the case $A=-1$.

For integers $a,b$ I guess the above relation holds,
$$\begin{align} (-1)^{2\cdot3} = ((-1)^2)^3 = 1 = ((-1)^3)^2 .\end{align}$$

But if we include the fraction

$$\begin{align} (-1)^{-\frac{1}{2}} &= ((-1)^{-1})^{\frac{1}{2}} = (-1)^{\frac{1}{2}} = i \\ &= ((-1)^{\frac{1}{2}})^{-1}=\frac{1}{i} = -i \end{align}$$
this does not hold any more.

Is something wrong with my computation?

Answer

In complex numbers,
$$(a^b)^c=a^{bc}$$ doesn't hold. You just found a counterexample.

This is due to the $2k\pi$ undeterminacy of the argument. When you take the square root, it becomes a $k\pi$ i.e. a sign indeterminacy.