Why $i^{-3}$ equals to $i$? What is $i^{-3}$?
Select one:
(a). 0
(b). i
(c). -1
(d). 1
(e). -i
I calculated like this:
$i^{-3}=\frac1{i^3}=\frac1{i^2\times i}=\frac1{-1\times i}=-\frac1i$
And therefore, $i$ is $-\frac1i$.
But the correct answer is: $i$.
How does it calculate?
 A: Because $\frac{-1}{i}=\frac{i^{2}}{i}=i$
A: A geometric approach is to note that exponentation of $i$ is equivalent to rotation of the unit vector in $\mathbb{C}$ (Argand plane):
$$z = |z|e^{i\theta} = |z|(\cos{\theta} + i\sin{\theta})$$
Here, $z=i \implies \theta=\frac{\pi}{2}, |z|=1$
Therefore, $z^{-3} \implies \theta'=\frac{-3\pi}{2} \implies i^{-3} = 0 + i\sin\left(\frac{-3 \pi}{2}\right) = i$
A: $i^{-3} = -\frac 1i$ but you aren't done.  We want to simplify and reduce it.
We need to get rid of the $i$ in the denominator.
$-\frac 1i = \frac 1i \cdot \frac ii = -\frac i{i^2} = -\frac i{-1}= i$.
Another way of recognizing this is $i^2 = -1$
so $i^4 = (-1)^2 = 1$.  $i^{-3}=1\cdot i^{-3}\cdot 1 = i^4 \cdot i^{-3}=i^{4-3} = i$.
One thing to realize is $i^k$ is cyclic.  $i^1 = i; i^2 = -1; i^3 = -i; i^4=1; i^5=i$ etc.   So going in the opposite direction $i^{4} = i^0 = 1;  i^{3} = i^{-1} = -i; i^2 = i^{-2} = -1; i^{1} = i^{-3} = i$.
And to beat this horse to the ground:
We can verify  $\frac 1i = -i$ if and only if $1 = (-i)\cdot i = -(i^2) = 1$.
And $-\frac 1i = i$ if and only if $-1 = i\cdot i$ (which it does)
And $ i^{-3} = i$ if and only if $1 = i\cdot i^3 = i^4 = (i^2)^2 = (-1)^2 =1$ (which it does)
........
And yet another way:
$i^{-3} = x$
$1 = i^3 x$
$1 = (i^2)ix$
$1=-ix$
$-1 = ix$
$-i = i\cdot i x$
$-i = -1\cdot x$
$i = x$.
So $i^{-3} = i$.
