Question

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

Answer 1

$$\begin{gathered} {\left(1+i\right)}^{2n}={\left(1-i\right)}^{2n} \\ \Rightarrow {\left[{\left(1+i\right)}^2\right]}^n={\left[{\left(1-i\right)}^2\right]}^n \\ \Rightarrow {\left(1^2+i^2+2i\right)}^n={\left(1^2+i^2-2i\right)}^n \\ \Rightarrow {\left(1-1+2i\right)}^n={\left(1-1-2i\right)}^n[\because i^2=-1] \\ \Rightarrow {\left(2i\right)}^n={\left(-2i\right)}^n \\ \Rightarrow {\left(2i\right)}^n={\left(-1{)}^n(2i\right)}^n \\ \Rightarrow (-1{)}^n=1 \\ \Rightarrow n\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}2 \end{gathered}$$

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