WBJEE · Maths · Complex Number
If \(z=\sin \theta-i \cos \theta,\) then for any integer \(n\)
- A \(z^{n}+\frac{1}{z^{n}}=2 \cos \left(\frac{n \pi}{2}-n \theta\right)\)
- B \(z^{n}+\frac{1}{z^{n}}=2 \sin \left(\frac{n \pi}{2}-n \theta\right)\)
- C \(z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)\)
- D \(z^{n}-\frac{1}{z^{n}}=2 i \cos \left(\frac{n \pi}{2}-n \theta\right)\)
Answer & Solution
Correct Answer
(C) \(z^{n}-\frac{1}{z^{n}}=2 i \sin \left(n \theta-\frac{n \pi}{2}\right)\)
Step-by-step Solution
Detailed explanation
Given, \(z=\sin \theta-i \cos \theta\) \(=\cos \left(\theta-\frac{\pi}{2}\right)+i \sin \left(\theta-\frac{\pi}{2}\right)=e^{i(\theta-\frac{\pi}{2}).}\) Now.…
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