To find the bilinear (or Möbius) transformation that maps the points $z = -i, 1, i$ onto $w = -1, 0, 1$, respectively, we use the general form of a bilinear transformation:

$w = \frac{az + b}{cz + d}$

The bilinear transformation is determined by four constants $a$, $b$, $c$, and $d$. To find these, we can substitute the corresponding points $z$ and $w$ into the equation and solve the system of equations.

Step 1: Substitute the points into the transformation equation

For each point, substitute $z$ and $w$ into $w = \frac{az + b}{cz + d}$:

1. For $z = -i$, $w = -1$: $-1 = \frac{a(-i) + b}{c(-i) + d}$

2. For $z = 1$, $w = 0$: $0 = \frac{a(1) + b}{c(1) + d}$

3. For $z = i$, $w = 1$: $1 = \frac{a(i) + b}{c(i) + d}$

Step 2: Simplify each equation

1. $-1 = \frac{-ai + b}{-ci + d}$ $-1(-ci + d) = -ai + b$ Expanding:

   \quad \text{(Equation 1)}$$

2. $0 = \frac{a + b}{c + d}$ leads to: $a + b = 0 \quad \text{(Equation 2)}$

3. $1 = \frac{ai + b}{ci + d}$ $1(ci + d) = ai + b$ Expanding:

   \quad \text{(Equation 3)}$$

Step 3: Solve the system of equations

From Equation 2, we know that $a + b = 0$, so $b = -a$.

Substitute $b = -a$ into Equations 1 and 3:

• Equation 1: $ci - d = -ai - a$ Equating real and imaginary parts: $c = -a, \quad -d = -a \implies d = a$

• Equation 3: $ci + d = ai - a$ Equating real and imaginary parts: $c = a, \quad d = -a$

Step 4: Write the final transformation

Now we know that: $c = -a, \quad d = a, \quad b = -a$

Thus, the bilinear transformation becomes: $w = \frac{az - a}{-az + a}$

Factor out $a$ from the numerator and denominator: $w = \frac{a(z - 1)}{a(1 - z)} = \frac{z - 1}{1 - z}$

Thus, the bilinear transformation is:

$w = \frac{z - 1}{1 - z}$

Final Answer:

The bilinear transformation that maps $z = -i, 1, i$ onto $w = -1, 0, 1$ is:

$w = \frac{z - 1}{1 - z}$

Would you like any further details or clarification?

Here are 5 related questions:

1. What are the general properties of bilinear transformations?
2. How can we verify if a transformation is conformal?
3. What happens to circles under bilinear transformations?
4. How can we derive the inverse of a bilinear transformation?
5. How can bilinear transformations be used in complex function theory?

Tip: Bilinear transformations always map circles and straight lines in the complex plane to other circles or straight lines!