Sure, let's discuss the properties of the imaginary unit [tex]\(i\)[/tex] in detail to find which expression is equivalent to [tex]\(i^{16}\)[/tex].
The imaginary unit [tex]\(i\)[/tex] has the property that [tex]\(i^2 = -1\)[/tex]. From this, we can derive the powers of [tex]\(i\)[/tex]:
1. [tex]\(i^1 = i\)[/tex]
2. [tex]\(i^2 = -1\)[/tex]
3. [tex]\(i^3 = i^2 \cdot i = -1 \cdot i = -i\)[/tex]
4. [tex]\(i^4 = i^3 \cdot i = -i \cdot i = -i^2 = -(-1) = 1\)[/tex]
Notice that the powers of [tex]\(i\)[/tex] start repeating every four exponents:
- [tex]\(i^5 = i^4 \cdot i = 1 \cdot i = i\)[/tex]
- [tex]\(i^6 = i^5 \cdot i = i \cdot i = i^2 = -1\)[/tex]
- [tex]\(i^7 = i^6 \cdot i = -1 \cdot i = -i\)[/tex]
- [tex]\(i^8 = i^7 \cdot i = -i \cdot i = -i^2 = -(-1) = 1\)[/tex]
The pattern [tex]\([i, -1, -i, 1]\)[/tex] repeats every four terms. To solve for [tex]\(i^{16}\)[/tex], we observe that [tex]\(16\)[/tex] is a multiple of [tex]\(4\)[/tex] (i.e., [tex]\(16 = 4 \times 4\)[/tex]).
Therefore, we identify that [tex]\(i^{16}\)[/tex] falls at the same place in the pattern as [tex]\(i^4\)[/tex]. Since [tex]\(i^4 = 1\)[/tex], we conclude:
[tex]\[i^{16} = 1\][/tex]
Thus, the correct answer is:
b. 1
