Question

compute $i^{82}$.

Explanation:

Step1: Recall the powers of $i$

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

Step2: Divide the exponent by 4

Divide 82 by 4: $82\div4 = 20$ with a remainder. $82 = 4\times20+2$.

Step3: Rewrite $i^{82}$

$i^{82}=i^{4\times20 + 2}=(i^4)^{20}\times i^2$.
Since $i^4 = 1$, then $(i^4)^{20}=1^{20}=1$. And $i^2=-1$. So $(i^4)^{20}\times i^2=1\times(-1)=-1$.

Answer:

$-1$