Question

Find the least positive integral value of n for which (1+i/1-i)^n is real

Answer 1

$\left(\dfrac{1+\iota}{1-\iota}\right)^n=\left(\dfrac{(1+\iota)(1+\iota)}{(1-\iota)(1+\iota)}\right)^n\\\\\\=\left(\dfrac{1+2\iota-1}{1-\iota^2}\right)^n=\left(\dfrac{2\iota}{2}\right)^n\\\\\\=\iota^n$

Well, it was simply $\iota^n$ !!!

The least positive integral value of $n$ for which $\iota^n$ is real is $\mathbf{n=2}$, because $\iota^2=-1,$ which is real.

Hence 2 is the answer.

[Some may be mistaken by $n = 4$ for $\iota^4=1$ but -1 is also a real number.]

Answer 2

Answer:

Step-by-step explan8ation:

Ok I have done