Question

i^{23} = i^{20+3}\
= i^{20} \times i^{3}\
= (i^{4})^{5} \times i^{3}\
= 1^{5} \times i^{3}\
= 1 \times i^{3}\
= -i\
use the example as a model. simplify the expressions.\
\\(i^{37} = \\) \boxed{\downarrow} \
done

Explanation:

Step1: Split exponent

We know that \(37 = 36 + 1\), so \(i^{37}=i^{36 + 1}\).

Step2: Use exponent rule

Using the rule \(a^{m + n}=a^{m}\times a^{n}\), we get \(i^{36+1}=i^{36}\times i^{1}\).

Step3: Rewrite exponent as multiple of 4

Since \(36 = 4\times9\), we can rewrite \(i^{36}\) as \((i^{4})^{9}\).

Step4: Substitute \(i^{4}=1\)

We know that \(i^{4} = 1\), so \((i^{4})^{9}=1^{9}\).

Step5: Simplify and multiply

\(1^{9}=1\), then \(1\times i^{1}=i\).

Answer:

\(i\)