Question

Please anyone help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I would really appreciate it

Answer 1

If
`i = √(-1)`, then
`i^2 = -1`,
`i^3 = -i`, and
`i^4 = 1`. For larger integer powers of i, the cycle repeats:

`i^5 = i^4\cdot i^1 = i \\\\ i^6 = i^4\cdot i^2 = -1 \\\\ i^7 = i^4\cdot i^3 = -i \\\\ i^8 = i^4\cdot i^4 = 1`

and so on.

Then

(1)

`i^(8n) = i^(4\cdot2n) = \left(i^4\right)^(2n) = 1^(2n) = 1`

(2)

`i^(4n+42) = i^(4n+40+2) = i^(4n+40)\cdot i^2 = \left(i^4\right)^(n+10)\cdot i^2 = 1^(n+10)\cdot i^2 = -1`

(3)

`i^(12n+3) = i^(12n)\cdot i^3 = \left(i^4\right)^(3n) \cdot i^3 = 1^(3n)\cdot i^3 = -i`

(4)

`i^(8n-3) = i^(8n)\cdot i^(-3) = \left(i^4\right)^(2n)\cdot \frac1{i^3} = (1^(2n))/(i^3) = \frac1{i^3} = -\frac1i = i`