Question

If \(z\) is a complex number of unit modulus and argument \(\theta\), then the real part of \(\dfrac{z(1-\bar{z})}{\bar{z}(1+z)}\) is :

• A. \(1+\cos\dfrac{\theta}{2}\)
• B. \(1-\sin\dfrac{\theta}{2}\)
• C. \(-2\sin^{2}\dfrac{\theta}{2}\)
• D. \(2\cos^{2}\dfrac{\theta}{2}\)

Hint:

Use \(\bar z = 1/z\) for \(|z|=1\) and write \(z=e^{i\theta}\); the expression reduces to \(z\cdot i\tan(\theta/2)\).

Answer 1

The Correct Option is C

Since \(|z|=1\), \(\bar z = \dfrac1z\), so \(\dfrac{z}{\bar z}=z^2\) and \(1-\bar z = \dfrac{z-1}{z}\).
The expression becomes \(z^2\cdot\dfrac{z-1}{z(1+z)} = \dfrac{z(z-1)}{z+1}\).
With \(z=e^{i\theta}\): \(z-1 = e^{i\theta/2}\cdot 2i\sin\dfrac\theta2\) and \(z+1 = e^{i\theta/2}\cdot 2\cos\dfrac\theta2\), so \(\dfrac{z-1}{z+1}=i\tan\dfrac\theta2\).
Hence the expression equals \(i\tan\dfrac\theta2(\cos\theta + i\sin\theta)\).
Real part \(= -\tan\dfrac\theta2\sin\theta = -\tan\dfrac\theta2\cdot 2\sin\dfrac\theta2\cos\dfrac\theta2 = -2\sin^2\dfrac\theta2\).
Answer: (3) \(-2\sin^{2}\dfrac{\theta}{2}\)