Question

i^{467}=
i^{17}=
\sqrt{-25}=
9i =

Explanation:

For \(i^{467}\):

Step1: Recall \(i\) cycle (\(i^1 = i, i^2 = -1, i^3 = -i, i^4 = 1\), cycle of 4)

Divide exponent 467 by 4: \(467 \div 4 = 116\) remainder \(3\) (since \(4\times116 = 464\), \(467 - 464 = 3\))

Step2: Use remainder to find \(i^{467}\)

Since remainder is 3, \(i^{467}=i^3=-i\)

For \(i^{17}\):

Step1: Divide 17 by 4 (cycle of \(i\))

\(17 \div 4 = 4\) remainder \(1\) (since \(4\times4 = 16\), \(17 - 16 = 1\))

Step2: Use remainder to find \(i^{17}\)

Since remainder is 1, \(i^{17}=i^1 = i\)

For \(\sqrt{-25}\):

Step1: Recall \(\sqrt{-a}=\sqrt{a}\times i\) (for \(a>0\))

Here \(a = 25\), so \(\sqrt{-25}=\sqrt{25}\times i\)

Step2: Simplify \(\sqrt{25}\)

\(\sqrt{25}=5\), so \(\sqrt{-25}=5i\)

For \(9i\):

Answer:

\(i^{467}=\boldsymbol{-i}\)
\(i^{17}=\boldsymbol{i}\)
\(\sqrt{-25}=\boldsymbol{5i}\)
\(9i=\boldsymbol{9i}\)