Question

question 3 of 10
what is the simplified form of ( i^{15} )?

a. -1
b. -i
c. 1
d. i

Explanation:

Step1: Recall the pattern of powers of \(i\)

The imaginary unit \(i\) has a cyclic pattern for its powers:

• \(i^1 = i\)
• \(i^2 = -1\) (by definition, since \(i=\sqrt{-1}\), so \(i^2 = (\sqrt{-1})^2=-1\))
• \(i^3 = i^2\times i=-1\times i = -i\)
• \(i^4=(i^2)^2=(-1)^2 = 1\)

After that, the pattern repeats every 4 powers: \(i^{4n}=1\), \(i^{4n + 1}=i\), \(i^{4n+2}=-1\), \(i^{4n + 3}=-i\) for any integer \(n\).

Step2: Divide the exponent by 4 to find the remainder

We want to simplify \(i^{15}\). Divide 15 by 4:
\(15\div4 = 3\) with a remainder of \(3\) (because \(4\times3 = 12\) and \(15-12 = 3\)). So we can write \(i^{15}=i^{4\times3+3}\).

Step3: Use the pattern of powers of \(i\)

Using the property of exponents \(a^{m + n}=a^m\times a^n\), we have:
\(i^{4\times3+3}=(i^{4})^3\times i^{3}\)
We know that \(i^{4}=1\) and \(i^{3}=-i\) (from the pattern above). Substitute these values:
\((i^{4})^3\times i^{3}=1^3\times(-i)=-i\)

Answer:

B. \(-i\)