Q61 / 180 · KCET 2009 · Mathematics

The smallest positive integral value of $n$ such that $\left[\frac{1+\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}}{1+\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}}\right]^{\mathrm{n}}$ $n$ is equal to

  • Correct answer
    A
    4
  • B
    3
  • C
    2
  • D
    8

Solution

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

Hence, the correct answer is option: 'A'.

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