what is the value of $i^{20 + 1}$?\n1\n-1\n-i\ni

Explanation:

Step1: Recall the powers of (i)

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

Step2: Simplify the exponent

For (i^{n}), we can use the formula (n = 4k + r), where (k=\lfloor\frac{n}{4}\rfloor) (the greatest - integer function) and (r=n\bmod{4}) (the remainder when (n) is divided by (4)). For (n = 20 + 1=21), when we divide (21) by (4): (21=4\times5+1). So (i^{21}=i^{4\times5 + 1}). Using the property (i^{4k + r}=(i^{4})^{k}\times i^{r}), since (i^{4}=1), then (i^{4\times5 + 1}=(i^{4})^{5}\times i^{1}).

Step3: Calculate the value

((i^{4})^{5}\times i^{1}=1^{5}\times i) (because (i^{4} = 1)).

Answer:

(i)