(ii) Prove that \( w=\frac{z}{1-z} \) maps the upper half of the \( z \)-plane onto the upper half of the w-plane. What is the image of the circle \( |z|=1 \) under this transformation?

To address the problem, we'll examine the mapping \( w = \frac{z}{1 - z} \) and determine its effects on the upper half-plane and the unit circle \( |z| = 1 \). ### Mapping the Upper Half-Plane **Objective:** Show that \( w = \frac{z}{1 - z} \) maps the upper half of the \( z \)-plane (where \( \text{Im}(z) > 0 \)) onto the upper half of the \( w \)-plane (where \( \text{Im}(w) > 0 \)). **Proof:** 1. **Express \( z \) and \( w \) in terms of their real and imaginary parts:** - Let \( z = x + iy \), where \( y > 0 \) (since \( z \) is in the upper half-plane). - Then, \( w = \frac{z}{1 - z} = \frac{x + iy}{1 - x - iy} \). 2. **Rationalize the denominator:** \[ w = \frac{(x + iy)(1 - x + iy)}{(1 - x)^2 + y^2} = \frac{x(1 - x) + iy(1 - x) + iy x + i^2 y^2}{(1 - x)^2 + y^2} \] Simplifying: \[ w = \frac{x - x^2 - y^2 + i y}{(1 - x)^2 + y^2} \] 3. **Extract the imaginary part of \( w \):** \[ \text{Im}(w) = \frac{y}{(1 - x)^2 + y^2} \] Since \( y > 0 \) and the denominator is always positive, \( \text{Im}(w) > 0 \). **Conclusion:** \( w = \frac{z}{1 - z} \) maps the upper half of the \( z \)-plane to the upper half of the \( w \)-plane. ### Image of the Unit Circle \( |z| = 1 \) **Objective:** Determine the image of the unit circle \( |z| = 1 \) under the transformation \( w = \frac{z}{1 - z} \). **Analysis:** 1. **Parametrize \( z \) on the unit circle:** \[ z = e^{i\theta}, \quad \theta \in [0, 2\pi) \] 2. **Substitute \( z \) into the transformation:** \[ w = \frac{e^{i\theta}}{1 - e^{i\theta}} \] 3. **Simplify the expression:** \[ w = \frac{e^{i\theta}}{1 - e^{i\theta}} = \frac{1}{e^{-i\theta/2} \left( \frac{1}{2i} \right)} = -\frac{1}{2} + \frac{i}{2} \cot\left(\frac{\theta}{2}\right) \] 4. **Interpret the result:** - The real part of \( w \) is always \( -\frac{1}{2} \). - The imaginary part, \( \frac{1}{2} \cot\left(\frac{\theta}{2}\right) \), spans all real values as \( \theta \) varies from \( 0 \) to \( 2\pi \). **Conclusion:** The image of the unit circle \( |z| = 1 \) under the transformation \( w = \frac{z}{1 - z} \) is the vertical line in the \( w \)-plane defined by \( \text{Re}(w) = -\frac{1}{2} \). ### Summary - **Mapping of Upper Half-Plane:** \( w = \frac{z}{1 - z} \) maps the upper half of the \( z \)-plane to the upper half of the \( w \)-plane. - **Image of the Unit Circle:** The unit circle \( |z| = 1 \) is mapped to the vertical line \( \text{Re}(w) = -\frac{1}{2} \) in the \( w \)-plane. **Final Answer:** After simplifying, every point z with Im z > 0 is sent to a point w with Im w > 0. Moreover, the unit circle |z| = 1 is mapped to the vertical line in the w-plane where the real part of w is –½. Thus: • The transformation w = z⁄(1 – z) sends the upper half of the z-plane onto the upper half of the w-plane. • The circle |z| = 1 is mapped to the vertical line Re w = −½ in the w-plane.