Alright, let's simplify [tex]\(i^{38}\)[/tex].
First, recall that the powers of the imaginary unit [tex]\(i\)[/tex] are cyclical with a period of 4. Specifically:
[tex]\[
\begin{align*}
i^1 &= i, \\
i^2 &= -1, \\
i^3 &= -i, \\
i^4 &= 1, \\
\end{align*}
\][/tex]
and this pattern repeats every four exponents. That means [tex]\(i^5\)[/tex] would be [tex]\(i\)[/tex], [tex]\(i^6\)[/tex] would be [tex]\(-1\)[/tex], and so on.
To find [tex]\(i^{38}\)[/tex], we can reduce the exponent 38 modulo 4, because [tex]\(i^{n}\)[/tex] has the same value as [tex]\(i^{(n \mod 4)}\)[/tex]:
[tex]\[
38 \mod 4 = 2
\][/tex]
This tells us that:
[tex]\[
i^{38} = i^2
\][/tex]
From our cyclic pattern above, we know that:
[tex]\[
i^2 = -1
\][/tex]
Hence, the simplified form of [tex]\(i^{38}\)[/tex] is [tex]\(-1\)[/tex].
