Question
If $${n_1},{n_2}$$ are positive integers then $${\left( {1 + i} \right)^{{n_1}}} + {\left( {1 + {i^3}} \right)^{{n_1}}} + {\left( {1 + {i^5}} \right)^{{n_2}}} + {\left( {1 + {i^7}} \right)^{{n_2}}}$$ is a real number if and only if
A.
$${n_1} = {n_2} + 1$$
B.
$${n_1} + 1 = {n_2}$$
C.
$${n_1} = {n_2}$$
D.
$${n_1},{n_2}$$ are any two positive integers
Answer :
$${n_1},{n_2}$$ are any two positive integers
Solution :
Expression $$ = {\left( {1 + i} \right)^{{n_1}}} + {\left( {1 - i} \right)^{{n_1}}} + {\left( {1 + i} \right)^{{n_2}}} + {\left( {1 - i} \right)^{{n_2}}}$$
$$\eqalign{
& = {2^{\frac{{{n_1}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }}} \right)^{{n_1}}} + {2^{\frac{{{n_1}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} - i\frac{1}{{\sqrt 2 }}} \right)^{{n_1}}} + {2^{\frac{{{n_2}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} + i\frac{1}{{\sqrt 2 }}} \right)^{{n_2}}} + {2^{\frac{{{n_2}}}{2}}}{\left( {\frac{1}{{\sqrt 2 }} - i\frac{1}{{\sqrt 2 }}} \right)^{{n_2}}} \cr
& = {2^{\frac{{{n_1}}}{2}}}\left\{ {{{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}^{{n_1}}} + {{\left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)}^{{n_1}}}} \right\} + {2^{\frac{{{n_2}}}{2}}}\left\{ {{{\left( {\cos \frac{\pi }{4} + i\sin \frac{\pi }{4}} \right)}^{{n_2}}} + {{\left( {\cos \frac{\pi }{4} - i\sin \frac{\pi }{4}} \right)}^{{n_2}}}} \right\} \cr
& = {2^{\frac{{{n_1}}}{2}}}\left\{ {\cos \frac{{{n_1}\pi }}{4} + i\sin \frac{{{n_1}\pi }}{4} + \cos \frac{{{n_1}\pi }}{4} - i\sin \frac{{{n_1}\pi }}{4}} \right\} + {2^{\frac{{{n_2}}}{2}}}\left\{ {\cos \frac{{{n_2}\pi }}{4} + i\sin \frac{{{n_2}\pi }}{4} + \cos \frac{{{n_2}\pi }}{4} - i\sin \frac{{{n_2}\pi }}{4}} \right\},\,\,{\text{if }}{n_1},{n_2}\,{\text{are integers}} \cr
& = {2^{\frac{{{n_1}}}{2}}} \cdot 2\cos \frac{{{n_1}\pi }}{4} + {2^{\frac{{{n_2}}}{2}}} \cdot 2\cos \frac{{{n_2}\pi }}{4} = {\text{real}}{\text{.}} \cr} $$