Q75 / 180 · KCET 2021 · Mathematics

If $\left(\frac{1+i}{1-i}\right)^{x}=1$, then

  • A
    $x=4 n+1, n \in N$
  • B
    $x=2 n+1, n \in N$
  • C
    $x=2 n, n \in N$
  • Correct answer
    D
    $x=4 n, n \in N$

Solution

Given, $\left(\frac{1+i}{1-i}\right)^{x}=1$ $\begin{aligned} &\Rightarrow \quad\left[\frac{(1+i)}{(1-i)} \times \frac{(1+i)}{(1+i)}\right]^{x}=1 \\ &\Rightarrow \quad\left[\frac{1+i^{2}+2 i}{1^{2}-i^{2}}\right]^{x}=1 \\ &\Rightarrow \quad\left[\frac{1-1+2 i}{1+1}\right]^{x}=1 \\ &\Rightarrow \quad i^{x}=1 \\ &\therefore \quad 4 n, n \in N . \end{aligned}$

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

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