To solve for [tex]\( i^{32} \)[/tex], we need to use the properties of the imaginary unit [tex]\( i \)[/tex]. The imaginary unit [tex]\( i \)[/tex] satisfies the equation [tex]\( i^2 = -1 \)[/tex].
The powers of [tex]\( i \)[/tex] follow a cyclical pattern, repeating every four values:
- [tex]\( i^1 = i \)[/tex]
- [tex]\( i^2 = -1 \)[/tex]
- [tex]\( i^3 = -i \)[/tex]
- [tex]\( i^4 = 1 \)[/tex]
This cycle repeats for higher powers of [tex]\( i \)[/tex]:
- [tex]\( i^5 = i \)[/tex]
- [tex]\( i^6 = -1 \)[/tex]
- [tex]\( i^7 = -i \)[/tex]
- [tex]\( i^8 = 1 \)[/tex]
- and so on...
To determine [tex]\( i^{32} \)[/tex], we find the remainder when 32 is divided by the cycle length of 4.
[tex]\[
32 \div 4 = 8 \quad \text{with a remainder of} \quad 0
\][/tex]
Since the remainder is 0, [tex]\( i^{32} \)[/tex] corresponds to [tex]\( i^0 \)[/tex] in the cycle.
[tex]\[
i^0 = 1
\][/tex]
Thus, the final answer is:
B. 1
