TS EAMCET · Maths · Complex Number
If \(z=\cos \alpha+i \sin \alpha ; 0 < \alpha < \frac{\pi}{4}\), then \(\left|\frac{1+z^4}{1-z^3}\right|=\)
- A \(\frac{\cos 2 \alpha}{\sin \frac{3}{2} \alpha}\)
- B \(\frac{\cos \alpha}{\sin \frac{3}{2} \alpha}\)
- C \(\frac{\cos 2 \alpha}{\sin \frac{\alpha}{2}}\)
- D \(\frac{\cos \alpha}{\sin \frac{\alpha}{2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{\cos 2 \alpha}{\sin \frac{3}{2} \alpha}\)
Step-by-step Solution
Detailed explanation
It is given that \(z=\cos \alpha+i \sin \alpha, 0 < \alpha < \frac{\pi}{4}\) So, \(\frac{1+z^4}{1-z^3}=\frac{1+(\cos \alpha+i \sin \alpha)^4}{1-(\cos \alpha+i \sin \alpha)^3}\) \(=\frac{1+\cos 4 \alpha+i \sin 4 \alpha}{1-\cos 3 \alpha-i \sin 3 \alpha}\) (by De-Moivre's theorem)…
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