Question

If \( a = e^{i\theta} \), then the expression \( \frac{1+a}{1-a} \) is equal to:

• A. \( \cot\frac{\theta}{2} \)
• B. \( \tan\theta \)
• C. \( i\cot\frac{\theta}{2} \)
• D. \( i\tan\frac{\theta}{2} \)
• E. \( 2\tan\theta \)

Hint:

A useful shortcut: \( \frac{1+e^{i\theta}}{1-e^{i\theta}} = i\cot(\theta/2) \) and conversely \( \frac{e^{i\theta}-1}{e^{i\theta}+1} = i\tan(\theta/2) \). These appear frequently in complex analysis.

Answer 1

The Correct Option is C

Concept: The complex exponential \( e^{i\theta} \) can be expanded using Euler's formula: \( \cos\theta + i\sin\theta \). To simplify fractions involving \( 1 \pm e^{i\theta} \), we use half-angle trigonometric identities to factor out terms and convert the expression into a simpler trigonometric form.

Step 1: Substituting Euler's form and using half-angle identities.
Substitute \( a = \cos\theta + i\sin\theta \): \[ \frac{1+a}{1-a} = \frac{1 + \cos\theta + i\sin\theta}{1 - (\cos\theta + i\sin\theta)} = \frac{(1+\cos\theta) + i\sin\theta}{(1-\cos\theta) - i\sin\theta} \] Using the identities \( 1+\cos\theta = 2\cos^2\frac{\theta}{2} \), \( 1-\cos\theta = 2\sin^2\frac{\theta}{2} \), and \( \sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \): \[ \frac{1+a}{1-a} = \frac{2\cos^2\frac{\theta}{2} + i(2\sin\frac{\theta}{2}\cos\frac{\theta}{2})}{2\sin^2\frac{\theta}{2} - i(2\sin\frac{\theta}{2}\cos\frac{\theta}{2})} \]

Step 2: Factoring and simplifying the complex fraction.
Factor out \( 2\cos\frac{\theta}{2} \) from the numerator and \( 2\sin\frac{\theta}{2} \) from the denominator: \[ \frac{2\cos\frac{\theta}{2} \left( \cos\frac{\theta}{2} + i\sin\frac{\theta}{2} \right)}{2\sin\frac{\theta}{2} \left( \sin\frac{\theta}{2} - i\cos\frac{\theta}{2} \right)} = \cot\frac{\theta}{2} \left( \frac{\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}}{\sin\frac{\theta}{2} - i\cos\frac{\theta}{2}} \right) \] To simplify the term in parentheses, multiply by \( i/i \): \[ \frac{\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}}{-i(i\sin\frac{\theta}{2} + \cos\frac{\theta}{2})} = \frac{1}{-i} = i \] Thus, the final result is \( i\cot\frac{\theta}{2} \).