Question 1030811
.
Option C).


<pre>
{{{i^2}}} = -1,

{{{i^3}}} = {{{i^2*i}}} = {{{(-1)*i}}} = {{{-i}}},

{{{i^4}}} = {{{i^2*i^2}}} = {{{(-1)*(-1)}}} = 1,

{{{i^5}}} = {{{i^4^i}}} = {{{i}}},  

and so on, with the cyclic period of 4.

Therefore, {{{i^13}}} = {{{i^12*i}}} = {{{i}}}.
</pre>