If [tex]i = \sqrt{-1}[/tex], then [tex]i^2 = -1[/tex], [tex]i^3 = -i[/tex], and [tex]i^4 = 1[/tex]. For larger integer powers of i, the cycle repeats:
[tex]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[/tex]
and so on.
Then
(1)
[tex]i^{8n} = i^{4\cdot2n} = \left(i^4\right)^{2n} = 1^{2n} = 1[/tex]
(2)
[tex]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[/tex]
(3)
[tex]i^{12n+3} = i^{12n}\cdot i^3 = \left(i^4\right)^{3n} \cdot i^3 = 1^{3n}\cdot i^3 = -i[/tex]
(4)
[tex]i^{8n-3} = i^{8n}\cdot i^{-3} = \left(i^4\right)^{2n}\cdot \dfrac1{i^3} = \dfrac{1^{2n}}{i^3} = \dfrac1{i^3} = -\dfrac1i = i[/tex]
