Question

For positive integers $n_1,n_2$ the value of the expression $(1+i{)}^{n_1}+{\left(1+i^3\right)}^{n_1}+{\left(1+i^5\right)}^{n_2}+{\left(1+i^7\right)}^{n_2}$, where $i=\sqrt{-1}$ 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. $n_1>0,n_2>0$

Answer 1

Answer: (D)

$(1+i{)}^{n_1}+{\left(1+i^3\right)}^{n_1}+{\left(1+i^5\right)}^{n_2}+{\left(1+i^7\right)}^{n_2}$
$=(1+i{)}^{n_1}+(1-i{)}^{n_1}+(1+i{)}^{n_2}+(1-i{)}^{n_2}$
$=2^{n_1/2}(\cos \pi /4+i\sin \pi /4{)}^{n_1}+2^{n_1/2}(\cos \pi /4-i\sin \pi /4{)}^{n_1}$
$+2^{n_2/2}(\cos \pi /+i\sin \pi /4{)}^{n_2}+2^{n_2/2}(\cos \pi /4-i\sin \pi /4{)}^{n_2}$
$[\because 1\pm i=\sqrt{2}(\cos \pi /4\pm i\sin \pi /4)]$
$=2^{n_1/2}\left(\cos \frac{n_1\pi }{4}+i\sin \frac{n_1\pi }{4}\right)+2^{n_1/2}\left(\cos \frac{n_1\pi }{4}-i\sin \frac{n_1\pi }{4}\right)$
$+2^{n_2/2}\left(\cos \frac{n_2\pi }{4}+i\sin \frac{n_2\pi }{4}\right)+2^{n_2/2}\left(\cos \frac{n_2\pi }{4}-i\sin \frac{n_2\pi }{4}\right)$
$=2\cdot 2^{n_1/2}\left(\cos \frac{n_1\pi }{4}\right)+2\cdot 2^{n_2/2}\cos \frac{n_2\pi }{4}$
which is real for all $n_1>0,n_2>0$