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Mathematics Question on Complex Numbers and Quadratic Equations

(1+cos⁡(π12)+isin⁡(π12)1+cos⁡(π12)−isin⁡(π12))72\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

Explanation

Solution

Let z=(1+cos⁡π12+isin⁡π121+cos⁡π12−isin⁡π12)72z=\left(\frac{1+\cos \frac{\pi}{12}+i \sin \frac{\pi}{12}}{1+\cos \frac{\pi}{12}-i \sin \frac{\pi}{12}}\right)^{72}
=(2cos⁡2π24+2isin⁡π24cos⁡π242cos⁡2π24−2isin⁡π24cos⁡π24)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}
=(cos⁡π24+isin⁡π24cos⁡π24−isin⁡π24)72=\left(\frac{\cos \frac{\pi}{24}+i \sin \frac{\pi}{24}}{\cos \frac{\pi}{24}-i \sin \frac{\pi}{24}}\right)^{72}
=cos⁡72π24+isin⁡72π24cos⁡72π24−isin⁡72π24=\frac{\cos \frac{72 \pi}{24}+i \sin \frac{72 \pi}{24}}{\cos \frac{72 \pi}{24}-i \sin \frac{72 \pi}{24}}
[∵(cos⁡θ+isin⁡θ)n=cos⁡nθ+isin⁡nθ]\left[\because(\cos \theta+i \sin \theta)^{n}=\cos n \theta+i \sin n \theta\right]
=cos⁡3π+isin⁡3πcos⁡3π−isin⁡3π=\frac{\cos 3 \pi+i \sin 3 \pi}{\cos 3 \pi-i \sin 3 \pi}
=−1+0−1−0=\frac{-1+0}{-1-0}
=1=1