TS EAMCET · Maths · Complex Number
If \(|Z|=2, Z_1=\frac{Z}{2} e^{i \alpha}\) and \(\theta\) is the \(\operatorname{amp}(Z)\), then \(\frac{Z_1{ }^n-Z_1{ }^{-n}}{Z_1{ }^n+Z_1{ }^{-n}}=\)
- A \(2^n i \tan (n \theta+n \alpha)\)
- B \(i \tan (n \theta-n \alpha)\)
- C \(i \tan (n \theta+n \alpha)\)
- D \(\tan (n \theta+n \alpha)\)
Answer & Solution
Correct Answer
(C) \(i \tan (n \theta+n \alpha)\)
Step-by-step Solution
Detailed explanation
\(Z = |Z|e^{i \theta} = 2e^{i \theta}\) \(Z_1 = \frac{2e^{i \theta}}{2} e^{i \alpha} = e^{i (\theta + \alpha)}\) \(\frac{Z_1{ }^n-Z_1{ }^{-n}}{Z_1{ }^n+Z_1{ }^{-n}} = \frac{e^{in(\theta+\alpha)}-e^{-in(\theta+\alpha)}}{e^{in(\theta+\alpha)}+e^{-in(\theta+\alpha)}}\)…
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