Question

The smallest positive integral value of $n$ for which $(1+i{)}^{2n}=(1-i{)}^{2n}$ is

• A. $4$
• B. $8$
• C. $2$
• D. $12$

Answer 1

The correct option is C $2$

We have,
$(1+i{)}^{2n}=(1-i{)}^{2n}$
$\Rightarrow {\left[\right(1+i{)}^2]}^n={\left[\right(1-i{)}^2]}^n\Rightarrow {(1+i^2+2i)}^n={(1+i^2-2i)}^n\Rightarrow (1-1+2i{)}^n=(1+(-1)-2i{)}^n\Rightarrow (2i{)}^n=(-2i{)}^n\Rightarrow {\left(\frac{2i}{-2i}\right)}^n=1\Rightarrow (-1{)}^n=1$

Hence, the smallest positive integral value of $n$ is $2.$