Put into a + bi form.
i^102= ?
i^1802= ?
$\displaystyle i^1 = i$
$\displaystyle i^2 = -1$
$\displaystyle i^3 = -i$
$\displaystyle i^4 = 1$
the pattern repeats as you continue with higher powers of i
$\displaystyle i^{102} = (i^2)^{51} = (-1)^{51} = -1 $
or ...
$\displaystyle i^{102} = i^{100} \cdot i^2 = (i^4)^{25} \cdot (-1) = 1^{25} \cdot (-1) = -1$
so ... $\displaystyle i^{102}$ is a real number (imaginary part is zero).
you try it with $\displaystyle i^{1802}$