Question

Find the least value of $n$ for which ${\left(\frac{1+i}{1-i}\right)}^n=1$.

• A. $4$
• B. $3$
• C. $-4$
• D. $1$

Answer 1

Answer: (A), (C)

The correct options are
A $4$
C $-4$

$\because \frac{1+i}{1-i}=\frac{1+i}{1-i}\times \frac{1+i}{1+i}$
$=\frac{(1+i{)}^2}{1-i^2}=\frac{1+2i+i^2}{1-i^2}$
$=\frac{1+2i-1}{1+1}=i$
$\therefore {\left(\frac{1+i}{1-i}\right)}^n=1$
$\Rightarrow i^n=1$
Thus, $i^n$ will be positive integer, if $n=4$.