Question

$\left(\frac{1+\cos\left(\frac{\pi}{12}\right) + i \sin\left(\frac{\pi}{12}\right)}{1+\cos \left(\frac{\pi}{12}\right) - i \sin\left(\frac{\pi}{12}\right)}\right)^{72}$ is equal to

• A. 0
• B. -1
• C. 1
• D. 44563

Answer 1

The Correct Option is C

Let $z=\left(\frac{1+\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}}{1+\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}}\right)^{72}$
$=\left(\frac{2 \cos ^{2} \frac{\pi}{24}+2 i \sin \frac{\pi}{24} \cos \frac{\pi}{24}}{2 \cos ^{2} \frac{\pi}{24}-2 i \sin \frac{\pi}{24} \cos \frac{\pi}{24}}\right)^{72}$
$=\left(\frac{\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}}\right)^{72}$
$=\frac{\cos \frac{72 \pi}{24}+i \sin \frac{72 \pi}{24}}{\cos \frac{72 \pi}{24}-i \sin \frac{72 \pi}{24}}$
$\left[\because(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\right]$
$=\frac{\cos 3 \pi+i \sin 3 \pi}{\cos 3 \pi-i \sin 3 \pi}$
$=\frac{-1+0}{-1-0}$
$=1$