Important Key points

Example 2.1
Simplify the following
(i)$i^7$
(ii)$i^{1729}$
(iii)$i^{-1924} = i^{2018}$
(iv)$\sum_{n=1}^{102}i^n$
(v)$i\ i^2\ i^3\ ...\ i^{40}$

Solution :
(i). $(i)^7 = (i)^{4+3} = (i)^4\ (i)^3 = 1.(i^3) = (i)^3 = -i$
(ii). $(i)^{1729} = (i)^{1728+1} = (i)^{1728}(i)^1 = 1.(i) = i$
(iii). $(i)^{-1924} + (i)^{2018} = (i)^{-1924+0} + (i)^{2016+2}$
$\hspace{0.5in}= (i)^{-1924}(i)^0 + (i)^{2016}(i)^2 = 1.(1)+1. (-1) = 1-1 = 0$ \href{https://shorturl.at/A08CN} $\url{https://shorturl.at/A08CN}$