Question

Find the least positive integral value of $n$ for which ${\left(\frac{1+i}{1-i}\right)}^n$ is a real number.

Answer 1

Let us consider the term inside the bracket

$\frac{1+i}{1-i}$

Rationalizing the denominator with its conjugate,

=$\frac{(1+i)(1+i)}{(1-i)(1+i)}$

Simplifying the numerator and denominator, we get

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

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

We know that $i^2$=$-1$

So, the fraction reduces to,

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

=$\frac{2i}{2}$

=$i$

Therefore the term inside bracket is $i$

required term is $i^n$

We know that $i^2$=$-1$ and $-1$ is a real number

Hence least positive value of $n$ is $2$