Question

The smallest positive integer $n$ for which ${\left(\frac{1+i}{1-i}\right)}^n=1$ is,

• A. $n=8$
• B. $n=16$
• C. $n=12$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct answer:

Compute the required integer:

Given,

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

Multiply and divide numerator by ${\left(1+i\right)}^n$ we get

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

$\Rightarrow$${\left(\frac{1+i^2+2i}{1-i^2}\right)}^n=1$ $\left(\because i^2=-1\right)$

$\Rightarrow$ ${\left(\frac{2i}{2}\right)}^n=1$

$\Rightarrow$ $i^n=1$

Hence, the smallest value of $n$ for this is $4$.

Hence, option $\left(D\right)$ is the correct answer.