Could somebody explain to me why i^-5= -i and not 1/i
Thanks
$\displaystyle i^{-5} = i^{-4} \cdot i^{-1} = i^{-1}$
However, if you multiply by i/i (which happens to be the conjugate) you get $\displaystyle i^{-1} \cdot \dfrac{i}{i} = \dfrac{i}{i^2} = -i$
$\displaystyle \dfrac{1}{i}$ is a correct intermediate step but we tend to want a real denominator so we multiply through by the complex conjugate which is i/i in this case
another way of looking at it is this:
suppose we had some number (which we don't know what it is), let's call it x, and -x = 1/x.
then (-x)(-x) = (1/x)(-x) (nothing funny going on here).
but (-x)(-x) = (-1)(x)(-1)(x) = (-1)(-1)(x)(x) = x^2, while (1/x)(-x) = -x/x = (-1)(x/x) = (-1)(1) = -1.
therefore, for such an x, x^2 = -1.
so x is a square root of -1. this tells us not only that -i = 1/i, but also that -(-i) = 1/(-i), or: i = 1/(-i).
in geometric terms: if you go 1/4 the way around a circle, the way back home is the same way you get 1/4 the way around the circle in the opposite direction.