For how many natural numbers ' \(n\) ' such that \(1 \leq n \leq 2021\) is…
For how many natural numbers ' \(n\) ' such that \(1 \leq n \leq 2021\) is \(\left(\frac{1+i}{1-i}\right)^n=1\) ?
504
505
506
503
Solution
It is given that,
\(\begin{aligned}
\left(\frac{1+i}{1-i}\right)^n=1 & \Rightarrow\left(\frac{1+i^2+2 i}{1-i^2}\right)^n=1 \\
\Rightarrow \quad\left(\frac{2 i}{2}\right)^n=1 & \Rightarrow i^n=1
\end{aligned}\)
If \(n\) is a integral multiple of 4, then \(i^n=1\) \(\because 1 \leq n \leq 2021\), so the possible values of \(n\) are \(4,8,12,16, \ldots ., 2020\) and there are 505 values of \(n\). Hence, option (b) is correct.