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

Explanation:

Step1: Recall the powers of $i$

The imaginary unit $i$ has the following properties: $i = \sqrt{- 1}$, $i^{2}=-1$, $i^{3}=i^{2}\cdot i=-i$, $i^{4}=(i^{2})^{2}=(-1)^{2} = 1$. The powers of $i$ are periodic with a period of 4.

Step2: Divide the exponent by 4

We want to find the value of $i^{201}$. Divide 201 by 4: $201\div4 = 50\cdots\cdots1$. Here, the quotient is 50 and the remainder is 1.

Step3: Simplify $i^{201}$

Since $i^{4k + r}=i^{r}$ where $k$ is an integer and $r$ is the remainder when the exponent is divided by 4, for $k = 50$ and $r = 1$, we have $i^{201}=i^{4\times50+1}=i^{1}=i$.

Answer:

$i$