(5) Prove that a Mobius transformation T(z) = az + b
cz + d
maps the upper half plane
(Im(z) > 0) onto itself if and only if a, b, c, d ∈ R and ad − bc = 1

To prove that a Möbius transformation \( T(z) = \frac{az + b}{cz + d} \) maps the upper half-plane \( \text{Im}(z) > 0 \) onto itself if and only if \( a, b, c, d \in \mathbb{R} \) and \( ad - bc = 1 \), we'll address both directions of the proof.

### Part 1: If \( a, b, c, d \in \mathbb{R} \) and \( ad - bc = 1 \), then \( T(z) \) maps the upper half-plane to itself.

1. **Definition and Conditions**:
   The Möbius transformation is given by:
   \[
   T(z) = \frac{az + b}{cz + d}
   \]
   where \( a, b, c, d \) are real numbers, and \( ad - bc = 1 \).

2. **Transformation of a Complex Number**:
   Let \( z = x + iy \), where \( x, y \in \mathbb{R} \) and \( y > 0 \). Then:
   \[
   T(z) = \frac{a(x + iy) + b}{c(x + iy) + d} = \frac{(ax + b) + iay}{(cx + d) + icy}
   \]

3. **Separate Real and Imaginary Parts**:
   The real and imaginary parts of \( T(z) \) are:
   \[
   T(z) = \frac{(ax + b)(cx + d) + (ay)(cy) + i[(ay)(cx + d) - (ax + b)(cy)]}{(cx + d)^2 + (cy)^2}
   \]

4. **Imaginary Part**:
   The imaginary part of \( T(z) \) is:
   \[
   \text{Im}(T(z)) = \frac{y(ad - bc)}{|cz + d|^2}
   \]
   Since \( ad - bc = 1 \) and \( y > 0 \), it follows that:
   \[
   \text{Im}(T(z)) = \frac{y}{|cz + d|^2} > 0
   \]
   Thus, \( T(z) \) maps the upper half-plane \( \text{Im}(z) > 0 \) onto itself.

### Part 2: If \( T(z) \) maps the upper half-plane to itself, then \( a, b, c, d \in \mathbb{R} \) and \( ad - bc = 1 \).

1. **Assumption**:
   Assume \( T(z) = \frac{az + b}{cz + d} \) maps the upper half-plane \( \text{Im}(z) > 0 \) onto itself. This means that for any \( z \) with \( \text{Im}(z) > 0 \), \( \text{Im}(T(z)) > 0 \).

2. **Calculation of Imaginary Part**:
   The imaginary part of \( T(z) \) is:
   \[
   \text{Im}(T(z)) = \frac{y(ad - bc)}{|cz + d|^2}
   \]
   For \( \text{Im}(T(z)) > 0 \), we require \( ad - bc > 0 \). Since the Möbius transformation is bijective, we normalize by setting \( ad - bc = 1 \).

3. **Real Coefficients**:
   If \( a, b, c, d \) were not real, then \( T(z) \) could potentially map points from the upper half-plane to outside the upper half-plane, violating the assumption that \( T(z) \) maps the upper half-plane onto itself. Thus, \( a, b, c, d \) must be real numbers.

### Conclusion

The Möbius transformation \( T(z) = \frac{az + b}{cz + d} \) maps the upper half-plane onto itself **if and only if** \( a, b, c, d \in \mathbb{R} \) and \( ad - bc = 1 \).

---

Would you like more details on any part of this proof or do you have any questions?

Here are some related questions to expand your understanding:

1. How does a Möbius transformation affect the real line \( \text{Im}(z) = 0 \)?
2. What geometric interpretation can be given for the condition \( ad - bc = 1 \)?
3. How does the transformation \( T(z) \) behave at the point \( z = -\frac{d}{c} \) where \( cz + d = 0 \)?
4. Can a Möbius transformation with \( ad - bc = -1 \) map the upper half-plane onto itself?
5. What is the effect of applying multiple Möbius transformations successively?

**Tip:** Analyzing the behavior of Möbius transformations on the boundary (real line and infinity) can provide deeper insight into their properties.

Learn how to prove that a Möbius transformation T(z) = (az + b) / (cz + d) maps the upper half-plane Im(z) > 0 onto itself if and only if the coefficients a, b, c, d are real numbers and ad - bc = 1. This problem delves into complex analysis and Möbius transformations, suitable for advanced undergraduate studies.

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