`i^3 = i^2 i = -i` , `i^4 = i^2.i^2 = (-1)(-1) = 1`

`i^5 = i^4i = i` , `i^6 = (i^3)i^3 = (-i)(-i) = -1` etc

`i^(-1) = 1/i xx i/i = i/(-1) = -i` , `i^(-2) = 1/i^2 = 1/(-1) = -1`

`i^(-3) = 1/i^3 = = 1/(-i`)(i/i) = i/1 =i , `i^(-4) = 1/i^4 = 1/1 =1`

In general, for any integer k, `i^(4k) = 1,` ` i^(4k+1) = i,` `i^(4k+2) = -1,` `i^(4k+3) = -i`

`text(Fundas: )`

The Sum of Four Consecutive Powers of i (Iota) is Zero.

If `n in I` and `i = sqrt(-1),` then

`i^n + i^(n+1) + i^(n+2) + i^(n+3) = i^n(1 + i + i^2 + i^3) = i^n( 1 + i - 1 - i ) = 0`

`1. sum_(r = p)^m f(r) = sum_(r=1)^(m-p+1)f(r+p-1)`
`2. sum_(r = -p)^m f(r) = sum_(r=1)^(m+p+1)f(r-p-1)`