Question

If \( |Z|=2 \), \( Z_1 = \frac{Z}{2}e^{i\alpha} \) and \( \theta \) is the 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. \( \tan(n\theta + n\alpha) \)

Hint:

Recall that \( \frac{e^{ix} - e^{-ix}}{e^{ix} + e^{-ix}} = \frac{2i\sin x}{2\cos x} = i\tan x \). Recognizing this hyperbolic-like structure with complex exponents allows for instant simplification.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:

We express the complex number \( Z_1 \) in Euler's form \( re^{i\phi} \) and then use De Moivre's Theorem to simplify the expression \( Z_1^n \).
Step 2: Detailed Explanation:

Given: 1. \( |Z| = 2 \) and \( \text{amp}(Z) = \theta \). So, \( Z = 2e^{i\theta} \). 2. \( Z_1 = \frac{Z}{2}e^{i\alpha} \). Substitute \( Z \) into the equation for \( Z_1 \): \[ Z_1 = \frac{2e^{i\theta}}{2} e^{i\alpha} = e^{i(\theta + \alpha)} \] Let \( \phi = \theta + \alpha \). Then \( Z_1 = e^{i\phi} \). Now, find \( Z_1^n \) and \( Z_1^{-n} \): \[ Z_1^n = (e^{i\phi})^n = e^{in\phi} = \cos(n\phi) + i\sin(n\phi) \] \[ Z_1^{-n} = e^{-in\phi} = \cos(n\phi) - i\sin(n\phi) \] Now substitute these into the required expression: \[ E = \frac{Z_1^n - Z_1^{-n}}{Z_1^n + Z_1^{-n}} \] Numerator: \[ (\cos(n\phi) + i\sin(n\phi)) - (\cos(n\phi) - i\sin(n\phi)) = 2i\sin(n\phi) \] Denominator: \[ (\cos(n\phi) + i\sin(n\phi)) + (\cos(n\phi) - i\sin(n\phi)) = 2\cos(n\phi) \] Calculate the ratio: \[ E = \frac{2i\sin(n\phi)}{2\cos(n\phi)} = i \tan(n\phi) \] Substitute \( \phi = \theta + \alpha \) back: \[ E = i \tan(n(\theta + \alpha)) = i \tan(n\theta + n\alpha) \]
Step 4: Final Answer:

The value is \( i \tan(n\theta + n\alpha) \).