Answer
$0$
Work Step by Step
RECALL:
(1) $a^{mn} = (a^m)^n$
(2) $i^2=-1$
(3) $a^{m+n} = a^m \cdot a^n$
Use rule (1) above to obtain:
$=(i^2)^{12}+i^2$
Use rule (2) above to obtain:
$\\=(-1)^{12}+(-1)$
Note that when $-1$ is raised to an even power, the result is $1$.
Thus, the expression above simplifies to:
$=1 + (-1)
\\=0$
