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

If $\left(\frac{1+i}{1-i}\right)^x=1$ then
  1. $x=2 n+1$, where $n$ is any positive integer
  2. $x=4 n$, where $n$ is any positive integer
  3. $x=2 n$, where $n$ is any positive integer
  4. $x=4 n+1$, where $n$ is any positive integer

Solution

$\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{x}}=1 \Rightarrow\left[\frac{(1+\mathrm{i})}{1-\mathrm{i}^2}\right]^{\mathrm{x}}=1$ $\left(\frac{1+\mathrm{i}^2+2 \mathrm{i}}{1+1}\right)^{\mathrm{x}}=1 \Rightarrow(\mathrm{i})^{\mathrm{x}}=1 ; \quad \therefore \mathrm{x}=4 \mathrm{n} ; \mathrm{n} \in 1^{+}$

Asked in: JEE Main 2003

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