Question

What is the smallest positive integer n for which $(1+i{)}^{2n}=(1-i{)}^{2n}?$

Answer 1

$(1+i{)}^{2n}=(1-i{)}^{2n}\Rightarrow {\left(\frac{1+i}{1-i}\right)}^{2n}=1\Rightarrow {\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)}^{2n}=1\text{[Rationalizing the denominator]}\Rightarrow {\left(\frac{1+2i-1}{1+1}\right)}^{2n}=1\Rightarrow {\left(\frac{2i}{2}\right)}^{2n}=1\Rightarrow i^{2n}=1\therefore n=2$