Question

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

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

Answer 1

The correct option is C $x=4n$

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

$\Rightarrow {[\frac{1+i}{1-i}\times \frac{1+i}{1+i}]}^x=1$

$\Rightarrow {\left[\frac{1+i^2+2i}{1^2-i^2}\right]}^x=1$

$\Rightarrow {\left[\frac{1-1+2i}{1+1}\right]}^x=1$ $[\because i^2=-1]$

$\Rightarrow (i{)}^x=1$

$\therefore x=4n,n\in N$