Explaining why $i^n$ can only be equal to $1,i,-1,-i$

Question

How can I explain why $i^n$ can not have any other values than $1,i,-1,-i$ for all $n \in \mathbb{N}$?

Should I write it out as a sequence $i^n= 1,i,-1,-i,1,i,-1,-i,1,...$ and say that it repeats after every 4th term?

Answer 1

Well, $i$ is a primitive fourth root of unity: $i$, $i^2=-1$, $i^3=-i$ and $i^4=1$.

Algebraically, $i$ is a (formal) root of the real polynomial $x^2+1$, i.e., $i^2+1=0.$, i.e., $i^2=-1$. Then $i^3= i^2\cdot i= -i$ and $i^4=i^2\cdot i^2 = (-1)\cdot (-1)=1.$

Answer 2

You might want to think of it this way.

$i$ is just a way to symbolize the vector $(0,1)$ in $\mathbb{R^2}$.

When we consider the product of $(0,1)$ with itself, that is $(0,1)^n=i^n$, what happens in each iteration (that is, each time we multiply) is that the vector's length remains the same while it is rotated by $90^{o}$ anticlockwise*.

There are only four possible positions where we might end up, $(0,1), (0,-1), (1,0), (-1,0)$ which correspond to the values $i, -i, 1, -1$.

*indeed, let $z=x+iy$.

Then $zi=(x,y)*(0,1)=(-y,x)=w$ where $|z|=|w|$ and the points $(x,y), (-y,x)$ are the edges of a right triangle, the other edge being the origin $O$.