Question

If \(n\) is an integer and \(Z = \cos \theta + i \sin \theta\), \(\theta \neq (2n + 1)\) \(\frac{\pi}{2}\), then: \(\frac{1 + Z^{2n}}{1 - Z^{2n}}\) = ?

• A. \( i \tan n\theta \)
• B. \( i \cot n\theta \)
• C. \( -i \tan n\theta \)
• D. \( -i \cot n\theta \)

Hint:

When working with complex exponentials, apply De Moivre's theorem to express the powers of complex numbers in terms of sines and cosines. Use the conjugate to simplify the fraction.

Answer 1

The Correct Option is B

We are given \( Z = \cos \theta + i \sin \theta \), which is the polar form of a complex number, and we are asked to evaluate: \[ \frac{1 + Z^{2n}}{1 - Z^{2n}}. \] Using De Moivre's theorem, we know: \[ Z^{2n} = \cos(2n\theta) + i \sin(2n\theta). \] Step 1: Substituting for \( Z^{2n} \) Substituting into the expression: \[ \frac{1 + \cos(2n\theta) + i \sin(2n\theta)}{1 - \cos(2n\theta) - i \sin(2n\theta)}. \] Step 2: Simplifying the expression Multiply the numerator and the denominator by the conjugate of the denominator: \[ \frac{(1 + \cos(2n\theta) + i \sin(2n\theta)) (1 - \cos(2n\theta) + i \sin(2n\theta))}{(1 - \cos(2n\theta) - i \sin(2n\theta)) (1 - \cos(2n\theta) + i \sin(2n\theta))}. \] Simplifying the denominator: \[ (1 - \cos(2n\theta))^2 + \sin^2(2n\theta) = 2(1 - \cos(2n\theta)) = 4\sin^2(n\theta). \] The numerator simplifies to: \[ i \cot n\theta. \] Thus, the value of the given expression is \( i \cot n\theta \).

Answer 2

Given: Let \( n \) be an integer and \[ Z = \cos \theta + i \sin \theta, \] with \( \theta \neq (2n + 1) \frac{\pi}{2} \). Find the value of: \[ \frac{1 + Z^{2n}}{1 - Z^{2n}}. \]

Step 1: Express \( Z^{2n} \) using Euler’s formula Recall that \[ Z = e^{i\theta} \implies Z^{2n} = e^{i 2n \theta} = \cos 2n\theta + i \sin 2n\theta. \]

Step 2: Simplify the expression \[ \frac{1 + Z^{2n}}{1 - Z^{2n}} = \frac{1 + \cos 2n\theta + i \sin 2n\theta}{1 - \cos 2n\theta - i \sin 2n\theta}. \] Multiply numerator and denominator by the conjugate of the denominator: \[ \frac{(1 + \cos 2n\theta) + i \sin 2n\theta}{(1 - \cos 2n\theta) - i \sin 2n\theta} \times \frac{(1 - \cos 2n\theta) + i \sin 2n\theta}{(1 - \cos 2n\theta) + i \sin 2n\theta}. \]

Step 3: Calculate numerator and denominator separately Denominator: \[ (1 - \cos 2n\theta)^2 + (\sin 2n\theta)^2 = 1 - 2\cos 2n\theta + \cos^2 2n\theta + \sin^2 2n\theta = 2 - 2\cos 2n\theta = 4 \sin^2 n\theta, \] using the identity \( 1 - \cos 2A = 2 \sin^2 A \). Numerator: \[ [(1 + \cos 2n\theta)(1 - \cos 2n\theta) - \sin^2 2n\theta] + i[(1 + \cos 2n\theta) \sin 2n\theta + \sin 2n\theta (1 - \cos 2n\theta)]. \] Simplify the real part: \[ (1 - \cos^2 2n\theta) - \sin^2 2n\theta = \sin^2 2n\theta - \sin^2 2n\theta = 0. \] Simplify the imaginary part: \[ \sin 2n\theta (1 + \cos 2n\theta + 1 - \cos 2n\theta) = \sin 2n\theta \times 2 = 2 \sin 2n\theta. \] Therefore, numerator is: \[ i \times 2 \sin 2n\theta. \]

Step 4: Final expression \[ \frac{1 + Z^{2n}}{1 - Z^{2n}} = \frac{i \times 2 \sin 2n\theta}{4 \sin^2 n\theta} = \frac{i \sin 2n\theta}{2 \sin^2 n\theta}. \] Using the identity \( \sin 2A = 2 \sin A \cos A \), \[ = \frac{i \times 2 \sin n\theta \cos n\theta}{2 \sin^2 n\theta} = i \frac{\cos n\theta}{\sin n\theta} = i \cot n\theta. \]

Final answer: \[ \boxed{i \cot n\theta}. \]