Complex Multiplication

by their own information: {i=-\/-1( rout neg 1)} then wouldn't it be: i i i = i ????

If you use this, then you get $i*i*i=i$ .

Then as $\mathbb{C}$ is a field the cancellative laws hold, so $i*i=1$ which is a contradiction to anything. For example, the fundamental theorem of algebra, $1$ would have three different square roots. Or as $i^2=-1$ , the fact that $1\neq -1$ . And the importance of these facts is such, that you could contradict anything that is based in them. For example if $i*i=1$ and $-1$ at the same time, then $2*i*i=1+(-1)=0$ which contradicts the fact that $\mathbb{R}$ has characteristic zero. Or you could get $i*i=0$ so $1=0$ which contradicts axioms of field for real numbers. Or getting $i*i=0$ would contradict thar the complex field has no zero divisors.

What I mean by this is that there is a mistake in your calculations. This is, that in complex numbers you cannot change the order of exponents and roots in a rational exponent.

This is $(z^{a})^{1/b}\neq (z^{1/b})^a$ . This only holds when $z$ is a nonnegative real number. For example for $z=-1$ , $a=b=2$ , you get in the left side $\sqrt{1}=1$ and in the right side $i^2=-1$ . You could try and see that if you take the negative square root both sides result the same, but then it wouldn't hold for positive numbers. And when doing it in non-real numbers the result is a bit more complicated.

These exponent laws that you have used are defined only for complex numbers. From the complex point of view, use the solution given below.

BIDMAS - you do division before multiplication. [(-1)^(1/2)]^3