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i^{23}=i^{20 + 3}=i^{20}\times i^{3}=(i^{4})^{5}\times i^{3}=1^{5}\time…

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.

Explanation:

Step1: Rewrite the exponent

We know that \(i^{82}=i^{80 + 2}\) since \(80\) is a multiple of \(4\) and \(82=4\times20+2\). Then \(i^{82}=i^{80}\times i^{2}\) according to the rule \(a^{m + n}=a^{m}\times a^{n}\).

Step2: Simplify \(i^{80}\)

Since \(i^{4}=1\), and \(80 = 4\times20\), then \(i^{80}=(i^{4})^{20}=1^{20}=1\).

Step3: Simplify the expression

We have \(i^{82}=i^{80}\times i^{2}\), substituting \(i^{80} = 1\) and \(i^{2}=- 1\), we get \(i^{82}=1\times(-1)=-1\).

Answer:

\(-1\)