Question

Which ONE of the following is the value of \( \frac{1}{i^n} \)?

where \( i = \sqrt{-1} \), and \( n \) is an even positive integer.

• A. +1 or \( -1 \)
• B. \( +i \) or \( -i \)
• C. only +1
• D. only \( -i \)

Hint:

Powers of i repeat every 4 steps: i, -1, -i, 1. For even n, i^n is always +1 or -1.

Answer 1

The Correct Option is A

Step 1: What the problem is asking:
We want $\frac{1}{i^n}$ for even positive n, using $i = \sqrt{-1}$.
A quick way to see the pattern is to just plug in small even values of n and watch what happens.

Step 2: Try small even values of n:
For n = 2: $i^2 = -1$, so $\frac{1}{i^2} = \frac{1}{-1} = -1$.
For n = 4: $i^4 = 1$, so $\frac{1}{i^4} = 1$.
For n = 6: $i^6 = (i^4)(i^2) = 1 \times (-1) = -1$, so $\frac{1}{i^6} = -1$.
For n = 8: $i^8 = (i^4)^2 = 1$, so $\frac{1}{i^8} = 1$.
\[ n = 2, 6, 10, ... \Rightarrow \frac{1}{i^n} = -1, \qquad n = 4, 8, 12, ... \Rightarrow \frac{1}{i^n} = 1 \]

Step 3: Spot the pattern:
The values alternate strictly between +1 and -1 as n runs through the even numbers, they never come out as +i, -i, or anything else.
So no single fixed value works for every even n, the answer has to allow both +1 and -1.
This rules out option (B) (that is for odd n), option (C) (fails at n = 2), and option (D) (never happens for even n).

Final Answer:
Testing even values confirms the result swings between +1 and -1, matching option (A). \[ \boxed{+1 \text{ or } -1} \]