Prove that he expression √(1+m)+i√(1−m)√(1+m)−i√(1−m)−√(1−m)+i√(1+m)√(1−m)−i√(1+m)(m∈R) simplifies to 2m.
rationalise numerator and denominator in both the terms of expression. on rationalisation we get.
(√(1+m)+i√(1−m))22−(√(1−m)+i√(1+m))22
=(1+m−1+m+2∗i√(1+m)(1−m))−(1−m−1−m+2∗i√(1+m)(1−m))2
=2m