For how many natural numbers n such that 1 le | vedclass

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(B) Given the expression $\left(\frac{1+i}{1-i}\right)^n=1$. First,simplify the base $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by

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$505$

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(B) Given the expression $\left(\frac{1+i}{1-i}ight)^n=1$. First,simplify the base $\frac{1+i}{1-i}$ by multiplying the numerator and denominator by the conjugate $(1+i)$: $\frac{1+i}{1-i} 	imes \frac{1+i}{1+i} = \frac{1+i^2+2i}{1-i^2} = \frac{1-1+2i}{1-(-1)} = \frac{2i}{2} = i$. Thus,the equation becomes $i^n = 1$. We know that $i^n = 1$ if and only if $n$ is a multiple of $4$. We need to find the number of multiples of $4$ in the range $1 \leq n \leq 2021$. The multiples are $4, 8, 12, \ldots, 2020$. This is an arithmetic progression where $a = 4$,$d = 4$,and $l = 2020$. Using the formula $l = a + (n-1)d$: $2020 = 4 + (n-1)4$ $\Rightarrow 2016 = (n-1)4$ $\Rightarrow n-1 = 504$ $\Rightarrow n = 505$. Therefore,there are $505$ such natural numbers.

For how many natural numbers $n$ such that $1 \leq n \leq 2021$ is $\left(\frac{1+i}{1-i}\right)^n=1$?

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