How do you calculate i^8? I know there is some trick to it. Also i^25.
i^8 = 1
Yes. Definitely there is some trick.
Remember that \(i^2=-1\),
so \(i^4=i^2\cdot i^2=-1\cdot -1=1\)
Therefore \(i^{4n}=(i^4)^n=1^n=1\), for any integer n.
\(i^8=(i^4)^2=1^2=1\)
\(i^{25}=i^{24}\times i=(i^4)^6\times i=1\times i = i\)
