Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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 $\left( 1 + i \right)^{2n} = \left( 1 - i \right)^{2n}$ ?

#### 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.
Is there an error in this question or solution?

#### APPEARS IN

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