Question

what is the value of $i^{97}-i$?
options: $-1$, $0$, $-2i$, (last option unclear)

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cyclic pattern: \(i^1 = i\), \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and then the cycle repeats every 4 powers. So we can find the remainder when the exponent is divided by 4 to simplify \(i^{97}\).

Step2: Find the remainder of 97 divided by 4

Divide 97 by 4: \(97 \div 4 = 24\) with a remainder of 1. So \(i^{97}=i^{4\times24 + 1}=(i^4)^{24}\times i^1\). Since \(i^4 = 1\), this simplifies to \(1^{24}\times i = i\).

Step3: Subtract \(i\) from \(i^{97}\)

Now we calculate \(i^{97}-i\). Substituting \(i^{97}=i\), we get \(i - i = 0\).

Answer:

0