Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11
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What is the Smallest Positive Integer N for Which ( 1 + I ) 2 N = ( 1 − I ) 2 N ? - Mathematics

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 

\[\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}\] is 2.
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 13 Complex Numbers
Exercise 13.2 | Q 24 | Page 33
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