JEE Main On 16 April 2018 Question 17
Question: The least positive integer n for which $ {{( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )}^{n}}=1, $ is? [JEE Main 16-4-2018]
Options:
A) 2
B) 6
C) 5
D) 3
Show Answer
Answer:
Correct Answer: D
Solution:
First rationalize the number $ ( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )\times ( \frac{1+i\sqrt{3}}{1+i\sqrt{3}} )=( \frac{-2+i2\sqrt{3}}{4} )=( \frac{1-i\sqrt{3}}{-2} ) $ …(1)
$ ( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )\times ( \frac{1-i\sqrt{3}}{1-i\sqrt{3}} )=( \frac{4}{-2-i2\sqrt{3}} )=( \frac{-2}{1+i\sqrt{3}} ) $ …(2)
Using (1)and (2) $ {{( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )}^{3}}=( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )\times ( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )\times ( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} ) $
$ =( \frac{1+i\sqrt{3}}{1-i\sqrt{3}} )\times ( \frac{-2}{1+i\sqrt{3}} )\times ( \frac{1-i\sqrt{3}}{-2} )=1 $
Therefore correct Answer is 3 so correct option is D
