Identifying when $i^n$ (for integer $n$) is $1$, $-1$, $i$, $-i$

Question

We have some value $f(n)=i^n$ where n is some integer. It's clear that for even values of $n$, $f$ oscillates between $-1$ and $1$. Likewise, for odd values of $n$, $f$ oscillates between $-i$ and $i$.

I'm interested in narrowing the possibilities precisely. I noticed if $n$ is even it has an even number of product pairs of $i$ (that is to say, $i^2$), then $f(n)=1$. If $n$ has an odd number of product pairs of $i$, then $f(n)=-1$. For example, \begin{align*} \overbrace{i^6}^\text{even n}=&\underbrace{i^2*i^2*i^2}_{\text{3 pairs, odd number of pairs}} = -1 \\ i^8 =& \underbrace{i^2*i^2*i^2*i^2}_{\text{4 pairs, even number of pairs}} = 1 \end{align*}

if $n$ is odd and it has an odd number of pairs, then $f(n)=-i$. If it has an even number of pairs, then $f(n)=i$

My problem is figuring out how to easily find $n$'s factors of $i^2$ to see whether it's even or odd for very large values of $n$, such as $i^{432}$. Let's call the number of pairs here $X$. I tried to argue \begin{align*} \cfrac{i^n}{i^2}=X \end{align*}

At first glance I thought this would work, I mean, I am literally asking "how many times does $i^2$ go into $i^n$?" but, a moment later I realized this would just give me another complex number for most $n$ rather than an integer like I hoped.

So,

How does one find the number of times $i^2$ factors into $i^n$ such that its even or odd factor can be used to argue the result of $i^n$

EDIT: I believe I found my answer, I made a table describing it. If it's wrong then let me know. $$i^n \\ \begin{array}{c|c|c} \hline & n \ \text{is even} & n \ \text{is odd} \\ \hline \lfloor n/2 \rfloor \ \text{is even} & 1 & i \\ \hline \lfloor n/2 \rfloor \ \text{is odd} & -1 & -i \\ \hline \end{array}$$

Answer 1

You should try to figure this out on your own.

But hint. If $i^k = 1$ then $i^{k+m} = i^k*i^m = 1*i^m = i^m$. Do you see what this means?

If you know that $i^k = 1$ and if you have an $n$ so that $n = qk + r$ were $r$ is the remainder of dividing $n$ by $k$ then $i^n = i^{qk}i^r = (i^k)^qi^r = 1^qi^r = i^r$.

Now it's just a matter of figuring out what $i^1, i^2, i^3,.....$ are until you get to $i^k =1$. Do you see how all the rest will follow.

So.... try to work it out yourself.

More hint:

$i^{23} = (i^2)^{11}*i = (-1)^{11}*i= ([-1]^2)^5*(-1)*i = 1^5*(-i) = -i$. Can you see how that all fits in. Hint: $i^4 = (i^2)^2=(-1)^2=1$ and $23 = 4*5 + 3$ so $i^{23}=i^{4*5+3}=(i^4)^5*i^3$....

Answer 2

$i^n=i^{4k+r}\,\,\, Where \,\,\,\,\,\,n=4k+r \,\,\,, \,\,\, r = 0,1,2,3 $ Hence $i^n=i^r\,\,\,\forall r = 0,1,2,3$