Certainly! Let's simplify [tex]\(i^{12}\)[/tex] (where [tex]\(i\)[/tex] is the imaginary unit, defined by [tex]\(i = \sqrt{-1}\)[/tex]) step-by-step.
We need to recognize the pattern in the powers of [tex]\(i\)[/tex]. Let’s observe the calculations for the initial powers:
1. [tex]\( i^1 = i \)[/tex]
2. [tex]\( i^2 = i \times i = -1 \)[/tex]
3. [tex]\( i^3 = i^2 \times i = -1 \times i = -i \)[/tex]
4. [tex]\( i^4 = i^3 \times i = (-i) \times i = (-1) \times (-1) = 1 \)[/tex]
Notice that every fourth power of [tex]\(i\)[/tex] results in 1:
[tex]\[
i^4 = 1
\][/tex]
This implies that:
[tex]\[
i^8 = (i^4)^2 = 1^2 = 1
\][/tex]
[tex]\[
i^{12} = (i^4)^3 = 1^3 = 1
\][/tex]
Hence, [tex]\( i^{12} = 1 \)[/tex].
Therefore, the simplified form of [tex]\( i^{12} \)[/tex] is:
[tex]\[ \boxed{1} \][/tex]
