### Home > A2C > Chapter Ch9 > Lesson 9.2.2 > Problem 9-92

9-92.

*i*^{0} = 1 Can you see the pattern?

*i*^{1} = *i*

*i*^{2} = −1

*i*^{3} = (*i*)(*i*^{2}) = −*i*

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

*i*^{5} = (*i*^{2})(*i*^{2})(*i*) = *i*

The pattern repeats 1, *i*, −1, and −*i*

16 is a multiple of 4.

25 is on more than a multiple of 4.

39 is one less than a multiple of 4.

100 is a multiple of 4.

*i*^{16} = 1*i*^{25} = *i**i*^{39} = −*i**i*^{100} = 1

4*n* is a multiple of 4.

Refer to (a), (b), and (c).

*i*^{396} = *i*^{4·99} = 1

*i*^{397} = *i*^{4·99+1} = ?

*i*^{398} = *i*^{4·99+2} = ?

*i*^{399} = ?