Question

simplify each of the following powers of i.
$i^{15}=$
complete
$i^{32}=$
complete
$i^{99}=$
complete
$i^{22}=$

Explanation:

Step1: Recall the powers - cycle of \(i\)

The powers of the imaginary unit \(i\) have a cycle: \(i^1 = i\), \(i^2=-1\), \(i^3 = i^2\times i=-i\), \(i^4=(i^2)^2 = 1\).

Step2: Divide the exponent by 4

To simplify \(i^{22}\), divide 22 by 4. We have \(22\div4 = 5\) with a remainder of 2. That is, \(22 = 4\times5+2\).

Step3: Rewrite \(i^{22}\) using the power - rule

By the power - rule \(a^{m + n}=a^m\times a^n\) and \((a^m)^n=a^{mn}\), \(i^{22}=i^{4\times5 + 2}=(i^4)^5\times i^2\).

Step4: Substitute the known values

Since \(i^4 = 1\) and \(i^2=-1\), then \((i^4)^5\times i^2=1^5\times(-1)=-1\).

Answer:

\(-1\)