Question

The smallest positive integral value of ${}^{\prime }{n}^{\prime }$ 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)

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