Question

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

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

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

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

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

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

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

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

Hence smallest positive integer of $n$ is $2$