Q80 / 200 · Manipal (MET) 2010 · Mathematics
If $n_1, n_2$ are positive integers, then $(1+i)^{n_1}+\left(1+i^3\right)^{n_1}+\left(1+i^5\right)^{n_2}+\left(1+i^7\right)^{n_2}$ is a real number if and only if
- A$n_1=n_2+1$
- B$n_1=n_2$
- C$n_1, n_2$ are any two negative integers
- Correct answerD$n_1, n_2$ are both any positive integers
Solution
Expression can be written as $=(1+i)^{n_1}+(1-i)^{n_1}+(1+i)^{n_2}+(1-i)^{n_2}$
$=2^{n_1 / 2}\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)^{n_1}+2^{n_1 / 2}\left(\frac{1}{\sqrt{2}}-i \frac{1}{\sqrt{2}}\right)^{n_1}+2^{n_2 / 2}\left(\frac{1}{\sqrt{2}}+i \frac{1}{\sqrt{2}}\right)^{n_2}+2^{n_2 / 2}\left(\frac{1}{\sqrt{2}}-i \frac{1}{\sqrt{2}}\right)^{n_2}$
$=2^{n_1 / 2}\left\{\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{n_1}+\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^{n_1}\right\}+2^{n_2 / 2}\left\{\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)^{n_2}+\left(\cos \frac{\pi}{4}-i \sin \frac{\pi}{4}\right)^{n_2}\right\}$
$+2^{n_2 / 2}\left\{\cos \frac{n_2 \pi}{4}+i \sin \frac{n_2 \pi}{4}\right.\left.+\cos \frac{n_2 \pi}{4}-\sin \frac{n_2 \pi}{4}\right\}$
$=2^{n_1 / 2} \cdot 2 \cos \frac{n_1 \pi}{4}+2^{n_2 / 2} \cdot 2 \cos \frac{n_2 \pi}{4}=\mathrm{real}$
ie, $n_1, n_2$ are any two positive integers.
Hence, the correct answer is option: 'D'.