Question

The smallest positive integral value of $'n'$ such that $\left[\frac {1+\sin \frac {\pi}{8}+\,i\,\cos \frac {\pi}{8}}{1+\sin \frac {\pi}{8}-\,i\,\cos \frac {\pi}{8}} \right]^n$ is purely imaginary is, $n$ =

• A. 8
• B. 4
• C. 3
• D. 2

Answer 1

Answer: (B)

$\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{n}$
$=\left[\frac{1+\cos \alpha+i \sin \alpha}{1+\cos \alpha-i \sin \alpha}\right]^{n} \quad\left(\right.$ Put $\left.\alpha=\frac{\pi}{2}-\frac{\pi}{8}\right)$
$=\left[\frac{2 \cos ^{2} \frac{\alpha}{2}+2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^{2} \frac{\alpha}{2}-2 i \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}\right]^{n}$
$=\left[\frac{\cos ^{\alpha}+i \sin \frac{\alpha}{2}}{\cos \frac{\alpha}{2}-i \sin \frac{\alpha}{2}}\right]^{n}$
$=\left(e^{2 i \frac{\alpha}{2}}\right)^{n}=e^{i n \alpha}$
$=e^{i n\left(\frac{3 \pi}{8}\right)}=\cos \frac{3 n \pi}{8}+i \sin \frac{3 n \pi}{8}$
For $n=4$, we get imaginary part.