\(\begin{array}{l}{i^3} = {i^2}.i = - 1.i = - i\\{i^4} = {i^3}.i = - i.i = - {i^2} = 1\\{i^5} = {i^4}.i = 1.i = i\end{array}\).
Ta có:
\(\begin{array}{l}{i^1} = i\\{i^2} = - 1\\{i^3} = - i\\{i^4} = 1\\{i^5} = i\\{i^6} = - 1\end{array}\)
Vậy tổng quát lên ta có: Nếu \(n = 4q + r, 0 ≤ r < 4\) thì
\(\begin{array}{l}{i^{4q}} = {i^0} = 1\\{i^{4q + 1}} = {i^1} = i\\{i^{4q + 2}} = {i^2} = - 1\\{i^{4q + 3}} = {i^3} = - i\end{array}\)