Question

question 8 of 10
what is ( i^{84} )?

a. -1
b. 1
c. -i
d. i

Explanation:

Step1: Recall the properties of the imaginary unit \( i \)

We know that \( i = \sqrt{-1} \), and the powers of \( i \) follow a cyclic pattern:
\( i^1 = i \)
\( i^2 = -1 \) (because \( i \times i = \sqrt{-1} \times \sqrt{-1} = -1 \))
\( i^3 = i^2 \times i = -1 \times i = -i \)
\( i^4 = i^3 \times i = -i \times i = -i^2 = -(-1) = 1 \)
And then the pattern repeats every 4 powers: \( i^{4n} = 1 \), \( i^{4n + 1} = i \), \( i^{4n + 2} = -1 \), \( i^{4n + 3} = -i \) for any integer \( n \).

Step2: Simplify \( i^{44} \)

We can express 44 as a multiple of 4. Let's find \( n \) such that \( 44 = 4n \). Solving for \( n \), we divide 44 by 4: \( n = \frac{44}{4} = 11 \).
Since \( 44 = 4 \times 11 \), we can use the property \( i^{4n} = 1 \). Substituting \( n = 11 \), we get \( i^{44} = i^{4 \times 11} = (i^4)^{11} \).
We already know that \( i^4 = 1 \), so \( (i^4)^{11} = 1^{11} = 1 \).

Answer:

B. 1