Question

The smallest positive integer $n$ for which ${\left(1+i\right)}^{2n}={\left(1-i\right)}^{2n}$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B

$2$

Explanation for the correct answer:

Compute the required number:

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

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

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

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

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

$\Rightarrow$ $i^n={\left(-i\right)}^n$

The smallest positive integer to satisfy this equation is $2$.

Hence the smallest positive integer $n$ is $2$.

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