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AP EAMCET · Maths · Complex Number

\(\frac{\left(\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}\right)^8}{\left(\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}\right)^8}\) is equal to

  1. A \(i\)
  2. B \(-i\)
  3. C 1
  4. D 2
Verified Solution

Answer & Solution

Correct Answer

(C) 1

Step-by-step Solution

Detailed explanation

\text { } \begin{aligned} & \frac{\left(\sin \frac{\pi}{8}+i \cos \frac{\pi}{8}\right)^8}{\left(\sin \frac{\pi}{8}-i \cos \frac{\pi}{8}\right)^8} \\ & =\frac{i^8\left(\cos \frac{\pi}{8}-i \sin \frac{\pi}{8}\right)^8}{(-i)^8\left(\cos \frac{\pi}{8}+i \sin \frac{\pi}{8}\right)^8}…