Final answer:
The value of i⁹⁷-i is 0. The value of i⁹⁷ minus i is 0, since the powers of i cycle every 4 powers and i⁹⁷ simplifies down to just i. The correct answer is A.
Step-by-step explanation:
To find the value of i⁹⁷-i, we need to calculate the value of i⁹⁷ and subtract i from it.
The powers of i repeat in a cycle of four: i¹ = i, i² = -1, i³ = -i, and i⁴ = 1.
So, i⁹⁷ will be equal to i⁴ × i⁴ × i⁴ × i⁴... × i⁴ × i⁴ × i. Since there are 24 cycles of four in 97, the value of i⁹⁷ will be 1 × 1 × 1 × 1... × 1 × i which simplifies to i.
Subtracting i from i⁹⁷, we get i - i = 0.
The value of i⁹⁷ minus i is 0, since the powers of i cycle every 4 powers and i⁹⁷ simplifies down to just i.
The value of i⁹⁷ minus i is sought, where i is the imaginary unit such that i² = -1. We can use the fact that powers of i are cyclical: i³ = i² × i = -i and i⁴ = 1. Therefore, i⁹⁷ can be simplified by noting that 97 is 4 more than a multiple of 4 (96). Hence, i⁹⁷ = i¹ = i, and i⁹⁷ - i = i - i = 0.
The value of \(i^{97} - i\) can be determined by understanding the pattern of powers of the imaginary unit \(i\). \(i\) raised to the power of 4 results in 1, and the powers of \(i\) repeat in cycles of four. Therefore, \(i^{97}\) can be expressed as \(i^{96} \times i\). Since \(i^{96}\) is a multiple of 4 and results in 1, the expression simplifies to \(i \times i\), which is \(i^2\). As \(i^2\) equals -1, the overall expression becomes \(-1 \times i\), which is equivalent to \(-i\).
So, the correct option is:
\[ \text{O} \ -i \]
Therefore, the value of \(i^{97} - i\) is \(-i\). This result aligns with the cyclical nature of powers of \(i\) and demonstrates how understanding the patterns of complex numbers can simplify the computation of expressions involving imaginary units.