The Möbius transformation in question is:

$f(z) = \frac{1 + z}{1 - z}$

We need to analyze where this transformation maps three distinct sets: the unit circle, the real axis, and the imaginary axis.

1. Mapping of the Unit Circle ($|z| = 1$):

The unit circle consists of points $z$ such that $|z| = 1$, meaning $z \bar{z} = 1$, where $\bar{z}$ is the complex conjugate of $z$.

Let $z = e^{i\theta}$, where $\theta \in [0, 2\pi)$. Substituting this into the Möbius transformation, we have:

$f(z) = \frac{1 + e^{i\theta}}{1 - e^{i\theta}}$

To find the image of the unit circle under $f$, let's check the behavior. First, multiply the numerator and the denominator by the conjugate of the denominator:

= \frac{(1 + e^{i\theta})(1 - e^{-i\theta})}{|1 - e^{i\theta}|^2}$$ Since this expression results in a real denominator and complex numerator, the image of the unit circle will be another circle or a line in the complex plane. It turns out that the Möbius transformation maps the unit circle to the real line: $$f(|z| = 1) \rightarrow \text{Real axis}$$ ### 2. **Mapping of the Real Axis ($$z \in \mathbb{R}$$):** For $$z \in \mathbb{R}$$, we can substitute real values for $$z$$. The Möbius transformation in this case simplifies: $$f(x) = \frac{1 + x}{1 - x}, \quad x \in \mathbb{R}$$ This is a real-valued function for real $$z$$, so it will map real numbers to real numbers. Specifically, the transformation of real numbers under this function behaves as follows: - As $$x \to -\infty$$, $$f(x) \to -1$$, - As $$x \to 1^{-}$$, $$f(x) \to +\infty$$, - As $$x \to -1^{+}$$, $$f(x) \to -\infty$$, - As $$x \to +\infty$$, $$f(x) \to 1$$. Therefore, the real axis is mapped onto itself, but there is a reversal around $$x = 1$$ and $$x = -1$$. $$f(\mathbb{R}) \rightarrow \mathbb{R}$$ ### 3. **Mapping of the Imaginary Axis ($$z = iy, \, y \in \mathbb{R}$$):** For $$z = iy$$, where $$y \in \mathbb{R}$$, substitute into the Möbius transformation: $$f(iy) = \frac{1 + iy}{1 - iy}$$ Multiply numerator and denominator by the conjugate of the denominator: $$f(iy) = \frac{(1 + iy)(1 + iy)}{(1 - iy)(1 + iy)} = \frac{1 - y^2 + 2iy}{1 + y^2}$$ This gives us a complex number: $$f(iy) = \frac{1 - y^2}{1 + y^2} + i \cdot \frac{2y}{1 + y^2}$$ Thus, the imaginary axis is mapped to the **unit circle** in the complex plane. Specifically, the transformation maps $$z = iy$$ to a point on the circle of radius 1, centered at the origin. $$f(i\mathbb{R}) \rightarrow \text{Unit Circle}$$ ### Summary of Mappings: 1. **Unit circle ($$|z| = 1$$)**: Mapped to the **real axis**. 2. **Real axis ($$z \in \mathbb{R}$$)**: Mapped to the **real axis** (with inversion and shifting). 3. **Imaginary axis ($$z = iy, y \in \mathbb{R}$$)**: Mapped to the **unit circle**. Would you like a more detailed breakdown of any step, or have additional questions? --- Here are 5 questions to deepen your understanding: 1. How can we generalize Möbius transformations for any circle or line in the complex plane? 2. What is the significance of preserving the unit circle under Möbius transformations? 3. Can you derive the inverse of this Möbius transformation? 4. How does the geometric action of Möbius transformations relate to stereographic projection? 5. What other geometric figures are preserved under Möbius transformations? **Tip**: Möbius transformations preserve angles and map circles and lines to circles and lines.