Complex Mania

Here,

$\left( \frac { 1+i }{ 1-i } \right) =\left( \frac { 1+i }{ 1-i } \right) \left( \frac { 1+i }{ 1+i } \right) =\frac { { (1+i) }^{ 2 } }{ 1-{ i }^{ 2 } } =i$

It is a well known fact that ${ i }^{ 2 }=-1$ , so $2$ is the least positive integral value of $n$ for which the given expression is real.

$(\dfrac{1+i}{1-i})^{n}=(\dfrac{e^{i\frac{\pi}{4}}}{e^{i\frac{-\pi}{4}}})^{n}$ $=(e^{i\frac{\pi}{2}})^{n} = e^{i\frac{n\pi}{2}}=cis(\frac{n\pi}{2})$ . The least posistive value of $x$ that makes $cis(x)$ a purely real number is $\pi$ . So the least value of $n$ that makes the above expresion a purely real number is $2$