Complex Numbers And Quadratic Equations question 406
Question: The smallest positive integer $ n $ for which $ {{(1+i)}^{2n}}={{(1-i)}^{2n}} $ is [Karnataka CET 2004]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: B
Solution:
We have $ {{(1+i)}^{2n}}={{(1-i)}^{2n}} $
$ \Rightarrow {{( \frac{1+i}{1-i} )}^{2n}}=1 $
$ \Rightarrow {{(i)}^{2n}}=1 $
$ \Rightarrow {{(i)}^{2n}}={{(-1)}^{2}} $
$ \Rightarrow {{(i)}^{2n}}={{(i^{2})}^{2}} $
$ \Rightarrow {{(i)}^{2n}}={{(i)}^{4}} $
$ \Rightarrow 2n=4 $
$ \Rightarrow n=2 $ .
BETA
