Question

Simplify the following :

${\left(\frac{1+i}{1-i}\right)}^{4n+1}$

• A. $1$
• B. $i$
• C. $0$
• D. None of these

Answer 1

The correct option is B $i$

For $n$ is positive integer, $4n+1$ is an odd integer$(5,9,13,....)$

Now,

${\left(\frac{1+i}{1-i}\right)}^{4n+1}$

$⟹{\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)}^{4n+1}$

$⟹{\left(\frac{(1-1+2i}{1-\left(i^2\right)}\right)}^{4n+1}$

$⟹{\left(\frac{2i}{2}\right)}^{4n+1}⟹(i{)}^{4n+1}=i$.................[putting $n=4,9,\mathrm{13.....}$]