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Question
What is the smallest positive integer n for which \[\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}\] ?
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Solution
\[\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n} \]
\[ \Rightarrow \left[ \left( 1 + i \right)^2 \right]^n = \left[ \left( 1 - i \right)^2 \right]^n \]
\[ \Rightarrow \left( 1^2 + i^2 + 2i \right)^n = \left( 1^2 + i^2 - 2i \right)^n \]
\[ \Rightarrow \left( 1 - 1 + 2i \right)^n = \left( 1 - 1 - 2i \right)^n [ \because i^2 = - 1]\]
\[ \Rightarrow \left( 2i \right)^n = \left( - 2i \right)^n \]
\[ \Rightarrow \left( 2i \right)^n = \left( - 1 )^n (2i \right)^n \]
\[ \Rightarrow ( - 1 )^n = 1\]
\[ \Rightarrow \text { n is a multiple of } 2\]
Thus, the smallest positive integer n for which
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