AP EAMCET · Maths · Complex Number
\(\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8\) is equal to
- A \(1\)
- B \(-1\)
- C \(2\)
- D \(\frac{1}{2}\)
Answer & Solution
Correct Answer
(B) \(-1\)
Step-by-step Solution
Detailed explanation
\begin{aligned} & \text { }\left(\frac{1+\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}}{1+\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}}\right)^8 \\ & =\left(\frac{2 \cos ^2 \frac{\pi}{16}-2 i \sin \frac{\pi}{16} \cos \frac{\pi}{16}}{2 \cos ^2 \frac{\pi}{16}+2 i \sin \frac{\pi}{16} \cos…
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