Q. For positive integers $n_{1},\ n_{2}$ the value of the expression $(1 + i)^{n_{1}} + \left( 1 + i^{3} \right)^{n_{1}} + \left( 1 + i^{5} \right)^{n_{2}} + \left( 1 + i^{7} \right)^{n_{2}},\ i = \sqrt{- 1}\$ is a real number if and only if
A) $n_{1} = n_{2} + 1$
B) $n_{1} = n_{2} - 1$
C) $n_{1} = n_{2}$
D) $n_{1} > 0,\ n_{2} > 0$
Explanation
(d)
Using $i^{3} = - i^{5}\ and\ i^{7} = - i,$ we can write the given expression as
$(1 + i)^{n_{1}} + (1 - i)^{n_{1}} + (1 + i)^{n_{2}} + (1 - i)^{n_{2}}$
$= 2\ \lbrack^{n_{1}}C_{0} +^{n_{1}}C_{2}i^{2} +^{n_{1}}C_{4^{i^{4}}} + ...\rbrack + 2\ \lbrack^{n_{2}}C_{0} +^{n_{2}}C_{2}i^{2} +^{n_{2}}C_{4}i^{4} + ...\rbrack$
$= 2\ \lbrack^{n_{1}}C_{0} -^{n_{1}}C_{2} +^{n_{1}}C_{4} - ...\rbrack + 2\ \lbrack^{n_{2}}C_{0} -^{n_{2}}C_{2} +^{n_{2}}C_{4} - ...\rbrack$
This is real number, if the values of $n_{1}$and $n_{2}$ is greater than zero
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.