Sure, let's solve [tex]\( i^{99} \)[/tex].
To begin, recall that [tex]\( i \)[/tex] is the imaginary unit, defined as [tex]\( i = \sqrt{-1} \)[/tex]. The powers of [tex]\( i \)[/tex] follow a cyclical pattern:
[tex]\[
\begin{align*}
i^1 &= i, \\
i^2 &= -1, \\
i^3 &= -i, \\
i^4 &= 1, \\
i^5 &= i, \\
&\vdots \\
i^6 &= -1, \\
&\vdots
\end{align*}
\][/tex]
This pattern repeats every four powers. Hence, any power of [tex]\( i \)[/tex] can be reduced by identifying its position within this cycle of four terms.
Let's determine the equivalent power of [tex]\( i \)[/tex] within the first four powers for [tex]\( i^{99} \)[/tex]:
[tex]\[
99 \div 4 = 24 \quad \text{remainder} \quad 3
\][/tex]
The remainder when 99 is divided by 4 is 3. This signifies that:
[tex]\[
i^{99} = i^3
\][/tex]
From our earlier cycle:
[tex]\[
i^3 = -i
\][/tex]
Therefore, [tex]\( i^{99} \)[/tex] simplifies to [tex]\(-i\)[/tex].
Thus, the result in this question is [tex]\( -i \)[/tex]. Since [tex]\(-i\)[/tex] is a purely imaginary number, the imaginary part of [tex]\(-i\)[/tex] is:
[tex]\[
-1.0
\][/tex]
So, the final result [tex]\( \boxed{-1.0} \)[/tex] is the correct solution for [tex]\( i^{99} \)[/tex].
