Powers of Imaginary Unit i

We now use the defining property $i^2=-1$ to work out every other integer power of $i$:

and the negative powers

So $i^n$ only ever takes four possible values, and $i^4=1$ means the pattern repeats every $4$ steps.

General rule. For any integer $n$, write $n=4q+k$ with $q,k$ integers and $0\le k<4$ (the division algorithm — $q$ and $k$ are the quotient and remainder of $n$ divided by $4$). Then

So $i^n$ always equals one of $1,i,-1,-i$, according to the remainder $k$ that $n$ leaves on division by $4$.

Result. Any four consecutive integer powers of $i$ sum to zero:

since these four exponents leave the four different remainders $0,1,2,3$ (in some order) on division by $4$, so the four terms are $1,i,-1,-i$ in some order, and $1+i-1-i=0$. This is the standard trick for collapsing a long sum or product of powers of $i$: group the terms into blocks of four consecutive exponents (each block sums to $0$) and simplify whatever is left over. The same idea handles a product $i\cdot i^2\cdot i^3\cdots i^N=i^{1+2+\cdots+N}=i^{N(N+1)/2}$ — add the exponents first, then reduce that single exponent modulo $4$. …