To solve the given problem of finding the bilinear transformation that maps $z = 1, i, -1$ onto $w = 1, 0, -1$, and to find the fixed points of this transformation, we will follow these steps:

Step 1: Understanding Bilinear Transformations

A bilinear transformation (or Möbius transformation) has the form: $w = \frac{az + b}{cz + d}$ where $a, b, c, d$ are complex constants satisfying $ad - bc \neq 0$.

Step 2: Setting Up the Equations

Given the points: $z_1 = 1 \rightarrow w_1 = 1$ $z_2 = i \rightarrow w_2 = 0$ $z_3 = -1 \rightarrow w_3 = -1$

We can substitute these points into the bilinear transformation equation to obtain a system of equations.

Step 3: Forming the System of Equations

For $z_1 = 1$ and $w_1 = 1$: $1 = \frac{a(1) + b}{c(1) + d}$ $1 = \frac{a + b}{c + d}$ $a + b = c + d \quad \text{(Equation 1)}$

For $z_2 = i$ and $w_2 = 0$: $0 = \frac{a(i) + b}{c(i) + d}$ $0 = \frac{ai + b}{ci + d}$ $ai + b = 0$ $b = -ai \quad \text{(Equation 2)}$

For $z_3 = -1$ and $w_3 = -1$: $-1 = \frac{a(-1) + b}{c(-1) + d}$ $-1 = \frac{-a + b}{-c + d}$ $-1 = \frac{-a + b}{-c + d}$ $-a + b = c - d \quad \text{(Equation 3)}$

Step 4: Solving the System of Equations

From Equation 2: $b = -ai$

Substitute $b$ into Equation 1: $a - ai = c + d$

Substitute $b$ into Equation 3: $-a - ai = c - d$

Now, we have: $a(1 - i) = c + d \quad \text{(Equation 4)}$ $a(-1 - i) = c - d \quad \text{(Equation 5)}$

Add Equations 4 and 5: $a(1 - i) + a(-1 - i) = (c + d) + (c - d)$ $a(1 - i - 1 - i) = 2c$ $a(-2i) = 2c$ $c = -ai$

Subtract Equation 5 from Equation 4: $a(1 - i) - a(-1 - i) = (c + d) - (c - d)$ $a(1 - i + 1 + i) = 2d$ $a(2) = 2d$ $d = a$

Thus, we have: $c = -ai$ $d = a$

Step 5: Constructing the Transformation

Substitute $c$ and $d$ back into the bilinear transformation form: $w = \frac{az + b}{cz + d} = \frac{az - ai}{-aiz + a} = \frac{a(z - i)}{a(1 - iz)} = \frac{z - i}{1 - iz}$

Step 6: Finding the Fixed Points

Fixed points occur when $w = z$: $z = \frac{z - i}{1 - iz}$ $z(1 - iz) = z - i$ $z - iz^2 = z - i$ $-iz^2 = -i$ $z^2 = 1$ $z = \pm 1$

Final Answer

The bilinear transformation that maps $z = 1, i, -1$ onto $w = 1, 0, -1$ is: $w = \frac{z - i}{1 - iz}$

The fixed points of this transformation are: $z = \pm 1$

Would you like more details or have any questions? Here are some related questions for further exploration:

1. How do you verify if a bilinear transformation is bijective?
2. What are the properties of Möbius transformations?
3. Can a bilinear transformation map a circle to a line? How?
4. How do you find the inverse of a bilinear transformation?
5. What is the significance of fixed points in complex analysis?
6. How can bilinear transformations be used in practical applications?
7. What is the geometric interpretation of the bilinear transformation found?
8. How do bilinear transformations relate to the cross-ratio in projective geometry?

Tip: Always ensure the coefficients $a, b, c, d$ of a bilinear transformation satisfy the condition $ad - bc \neq 0$ to ensure it's a valid transformation.