To solve [tex]\(i^{32}\)[/tex], we need to understand the properties of the imaginary unit [tex]\(i\)[/tex]. By definition:
- [tex]\(i = \sqrt{-1}\)[/tex]
- [tex]\(i^2 = -1\)[/tex]
- [tex]\(i^3 = -i\)[/tex]
- [tex]\(i^4 = 1\)[/tex]
Notice that every power of [tex]\(i\)[/tex] follows a cyclical pattern with a period of 4:
1. [tex]\(i^1 = i\)[/tex]
2. [tex]\(i^2 = -1\)[/tex]
3. [tex]\(i^3 = -i\)[/tex]
4. [tex]\(i^4 = 1\)[/tex]
5. [tex]\(i^5 = i\)[/tex]
6. [tex]\(i^6 = -1\)[/tex]
7. [tex]\(i^7 = -i\)[/tex]
8. [tex]\(i^8 = 1\)[/tex]
And this cycle repeats every 4 exponents. To simplify [tex]\(i^{32}\)[/tex], we should express 32 in terms of this cycle:
[tex]\[ 32 \mod 4 = 0 \][/tex]
This means that [tex]\(i^{32}\)[/tex] corresponds to the same value as [tex]\(i^0\)[/tex] in the cycle since 32 is a multiple of 4.
Therefore,
[tex]\[ i^{32} = i^{4 \times 8} = (i^4)^8 = 1^8 = 1 \][/tex]
Thus,
[tex]\[ i^{32} = 1 \][/tex]
