Question

3.8.3 quiz: imaginary numbers
question 8 of 10
what is (-i)^6?
a. 1
b. -i
c. i
d. -1

Explanation:

Step1: Recall the property of imaginary unit \(i\)

We know that \(i = \sqrt{-1}\), so \(i^2=-1\), \(i^3 = i^2\times i=-i\), \(i^4=(i^2)^2 = (-1)^2 = 1\). The powers of \(i\) repeat every 4.

Step2: Simplify \((-i)^6\)

First, we can rewrite \((-i)^6\) as \((-1)^6\times i^6\). Since \((-1)^6 = 1\), we only need to find \(i^6\).
We can express \(i^6\) as \(i^{4 + 2}=i^4\times i^2\).
We know \(i^4 = 1\) and \(i^2=-1\), so \(i^4\times i^2=1\times(-1)=-1\).
And since \((-1)^6\times i^6 = 1\times(-1)=-1\)? Wait, no, wait: Wait, \((-i)^6=((-1)\times i)^6=(-1)^6\times i^6 = 1\times i^6\). And \(i^6=(i^4)\times(i^2)=1\times(-1)= - 1\)? Wait, but let's check again. Wait, maybe I made a mistake. Wait, \((-i)^6 = [(-1)\times i]^6=(-1)^6\times i^6=1\times i^6\). And \(i^6 = i^{4 + 2}=i^4\times i^2\). Since \(i^4 = 1\) and \(i^2=-1\), so \(i^6=1\times(-1)=-1\)? But wait, let's compute \((-i)^2=(-i)\times(-i)=i^2=-1\), \((-i)^4 = [(-i)^2]^2=(-1)^2 = 1\), \((-i)^6=(-i)^4\times(-i)^2=1\times(-1)=-1\)? Wait, but the options have D as -1. But wait, maybe I messed up. Wait, no, let's do it step by step.

Wait, \((-i)^6\):

First, \((-i)^2=(-i)\times(-i)=i^2=-1\)

\((-i)^4 = [(-i)^2]^2=(-1)^2 = 1\)

\((-i)^6=(-i)^4\times(-i)^2=1\times(-1)=-1\)

Yes, so \((-i)^6=-1\)

Answer:

D. -1