Q171 / 270 · NDA 2017 (Phase 2) · Mathematics

The smallest positive integer $n$ for which $\left(\frac{1+i}{1-i}\right)^n=1$, is

  • A
    $1$
  • Correct answer
    B
    $4$
  • C
    $8$
  • D
    $16$

Solution

$\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}=1$ We know, $\mathrm{i}^2=-1$ and $\mathrm{i}^4=1$. Now, rationalise the denominator. $\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^n=\left(\frac{1+2 i+i^2}{1-i^2}\right)^n=\left(\frac{1+2 i-i}{1-(-1)}\right)^n=\left(\frac{2 i}{2}\right)^n=-i^n$ The smallest positive integer for which $\mathrm{i}^{\mathrm{n}}=1$ is 4

Hence, the correct answer is option: 'B'.

Relevant Questions