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\begin{tabular}{|r|r|r|}
\hline
[tex]$i$[/tex] & \multicolumn{1}{|c|}{[tex]$\sqrt{-1}$[/tex]} & [tex]$i$[/tex] \\
\hline
[tex]$i^2$[/tex] & [tex]$i \times i = \sqrt{-1} \sqrt{-1}$[/tex] & -1 \\
\hline
[tex]$i^3$[/tex] & [tex]$i^2 \times i = -1 \times i$[/tex] & [tex]$-i$[/tex] \\
\hline
[tex]$i^4$[/tex] & [tex]$i^2 \times i^2 = (-1)(-1)$[/tex] & 1 \\
\hline
[tex]$i^5$[/tex] & [tex]$i^4 \times i = 1 \times i$[/tex] & [tex]$i$[/tex] \\
\hline
[tex]$i^6$[/tex] & [tex]$i^4 \times i^2 = 1 \times (-1)$[/tex] & -1 \\
\hline
[tex]$i^7$[/tex] & [tex]$i^4 \times i^3 = 1 \times (-i)$[/tex] & [tex]$-i$[/tex] \\
\hline
[tex]$i^8$[/tex] & [tex]$i^4 \times i^4 = 1 \times 1$[/tex] & 1 \\
\hline
\end{tabular}

Simplify [tex]$i^{12}$[/tex]:
A. [tex]$-1$[/tex]
B. [tex]$-i$[/tex]
C. 1



Answer :

GinnyAnswer
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]

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