Question

3.8.3 quiz: imaginary numbers
question 10 of 10
what is ( i^{27} )?
a. -1
b. 1
c. ( i )
d. ( -i )

Explanation:

Step1: Recall powers of \(i\)

We know the cycle of powers of the imaginary unit \(i\):
\(i^1 = i\),
\(i^2 = -1\),
\(i^3 = i^2 \cdot i = -1 \cdot i = -i\),
\(i^4 = (i^2)^2 = (-1)^2 = 1\),
and then the cycle repeats every 4 exponents.

Step2: Simplify \(i^{27}\)

To find \(i^{27}\), we divide the exponent 27 by 4 to find the remainder, which tells us the equivalent lower - power exponent in the cycle.
Divide 27 by 4: \(27\div4 = 6\) with a remainder of \(3\) (since \(4\times6 = 24\) and \(27-24 = 3\)).
This means \(i^{27}=i^{4\times6 + 3}\).
Using the property of exponents \(a^{m + n}=a^m\times a^n\), we can rewrite \(i^{4\times6+3}\) as \((i^4)^6\times i^3\).
We know that \(i^4 = 1\), so \((i^4)^6=1^6 = 1\).
And we already found that \(i^3=-i\).
So \((i^4)^6\times i^3=1\times(-i)=-i\).

Answer:

D. \(-i\)