(5) Find the image of the half-plane Re(z) > 0 under the Mobius transformation that
maps 0, i, −i to 1, −1, 0, respectively. (Hint: Try to think of a geometrical argument.)

We are given a Möbius transformation that maps the points \( z = 0 \), \( z = i \), and \( z = -i \) to \( w = 1 \), \( w = -1 \), and \( w = 0 \), respectively. We need to find the image of the half-plane \( \text{Re}(z) > 0 \) under this transformation.

### Step 1: General Form of a Möbius Transformation

The general form of a Möbius transformation is:

\[
w = f(z) = \frac{az + b}{cz + d}
\]

This transformation is determined uniquely by three points and their images, so we can use the given conditions to find \( a \), \( b \), \( c \), and \( d \).

### Step 2: Solve for the Möbius Transformation

We know the following mappings:
1. \( f(0) = 1 \)
2. \( f(i) = -1 \)
3. \( f(-i) = 0 \)

Using the general form \( f(z) = \frac{az + b}{cz + d} \), we substitute these points.

#### Condition 1: \( f(0) = 1 \)
\[
f(0) = \frac{a(0) + b}{c(0) + d} = \frac{b}{d} = 1 \implies b = d
\]

#### Condition 2: \( f(i) = -1 \)
\[
f(i) = \frac{a(i) + b}{c(i) + d} = \frac{ai + b}{ci + d} = -1
\]

Substitute \( b = d \):
\[
\frac{ai + d}{ci + d} = -1
\]
Multiply both sides by \( (ci + d) \):
\[
ai + d = -1(ci + d) = -ci - d
\]
This simplifies to:
\[
ai + ci = -2d
\]
\[
i(a + c) = -2d
\]
\[
a + c = -2di
\]
Thus, we have:
\[
a = -2di - c
\]

#### Condition 3: \( f(-i) = 0 \)
\[
f(-i) = \frac{a(-i) + d}{c(-i) + d} = 0
\]
This implies:
\[
a(-i) + d = 0 \implies -ai = d \implies a = di
\]

From this, substitute \( a = di \) into the equation \( a + c = -2di \):
\[
di + c = -2di
\]
This gives:
\[
c = -3di
\]

Thus, the Möbius transformation becomes:
\[
f(z) = \frac{di(z + i)}{-3di(z - i)} = \frac{z + i}{-3(z - i)}
\]

### Step 3: Image of the Half-Plane \( \text{Re}(z) > 0 \)

We now consider the image of the half-plane \( \text{Re}(z) > 0 \) under the transformation \( f(z) = \frac{z + i}{-3(z - i)} \). Since the given transformation is a Möbius transformation, it maps lines and circles in the complex plane to lines and circles.

- The boundary of the half-plane \( \text{Re}(z) = 0 \) is the imaginary axis. We check where the imaginary axis gets mapped by analyzing the transformation.
  
  Substituting \( z = iy \) (with \( y \in \mathbb{R} \)) into the transformation:
  \[
  f(iy) = \frac{iy + i}{-3(iy - i)} = \frac{i(y + 1)}{-3i(y - 1)} = \frac{y + 1}{3(y - 1)}
  \]
  For real values of \( y \), this expression is real. Therefore, the imaginary axis (which is the boundary of the half-plane) is mapped to the real line.

- Since the points \( z = 0 \), \( z = i \), and \( z = -i \) are mapped to \( w = 1 \), \( w = -1 \), and \( w = 0 \), respectively, the half-plane \( \text{Re}(z) > 0 \) (the right half-plane) is mapped to the upper half-plane in the \( w \)-plane.

Thus, the image of the half-plane \( \text{Re}(z) > 0 \) under the Möbius transformation is the upper half-plane \( \text{Im}(w) > 0 \).

### Final Answer:
The image of the half-plane \( \text{Re}(z) > 0 \) under the given Möbius transformation is the upper half-plane \( \text{Im}(w) > 0 \).

Would you like more details on any of the steps, or do you have any other questions?

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Here are five related questions to extend your understanding:
1. What is the general form of a Möbius transformation, and how does it affect the complex plane?
2. How can you find a Möbius transformation given the images of three points?
3. What are the invariants under Möbius transformations, such as circles and lines?
4. How does the upper half-plane relate to the Poincaré disk under a Möbius transformation?
5. Can you find the inverse of the Möbius transformation \( f(z) = \frac{z+i}{-3(z-i)} \)?

**Tip**: Möbius transformations preserve angles and map generalized circles (lines and circles) to generalized circles.

Find the image of the half-plane Re(z) > 0 under the Mobius transformation that maps 0, i, −i to 1, −1, 0, respectively. (Hint: Try to think of a geometrical argument.)

Explore the image of the half-plane Re(z) > 0 under a Mobius transformation that maps 0, i, and -i to 1, -1, and 0, respectively. This problem involves key concepts in complex analysis, including Mobius transformations, and provides a step-by-step solution using geometric arguments.

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