Question

question
evaluate:
$i^{89}$
answer
$-1$ $i$
$1$ $-i$

Explanation:

Step1: Recall the cycle of \(i\)

The imaginary unit \(i\) has a cyclic property: \(i^1 = i\), \(i^2 = -1\), \(i^3 = i^2 \cdot i = -i\), \(i^4 = (i^2)^2 = (-1)^2 = 1\), and then the cycle repeats every 4 powers. So we can use the division algorithm to find the remainder when the exponent is divided by 4.

Step2: Divide 89 by 4

We divide 89 by 4: \(89 \div 4 = 22\) with a remainder of 1. Mathematically, \(89 = 4 \times 22 + 1\).

Step3: Simplify \(i^{89}\)

Using the property of exponents \(a^{m + n}=a^m \cdot a^n\) and the cyclicity of \(i\), we can rewrite \(i^{89}\) as \(i^{4\times22 + 1}=(i^4)^{22}\cdot i^1\). Since \(i^4 = 1\), this becomes \(1^{22}\cdot i\). And \(1^{22}=1\), so \(i^{89}=i\).

Answer:

\(i\) (corresponding to the option with \(i\))