Find Linear Fractional Transformations for Points -1, 0, -1 to j, $\infty$, 1

[lolproe]

Homework Statement

find the linear fractional transformations (bilinear transformations) which map the ponts:

$$z_{1} = -1, z_{2} = 0, z_{3} = -1$$ into $$w_{1} = j, w_{2} = \infty, w_{3} = 1$$

Homework Equations

N/A

The Attempt at a Solution

I really don't have anything. For every question of this type I've tried before, I used the shortcut of setting
$$\frac{(w - w_{1})}{(w - w_{2})} \frac{(w_{3} - w_{2})}{(w_{3} - w_{1})} = \frac{(z - z_{1})}{(z - z_{2})} \frac{(z_{3} - z_{2})}{(z_{3} - z_{1})}$$
So the equation goes to zero (or one) when w or z are equal to one of the given points. I just don't know how to translate that into this example. I can't just throw infinity into the equation as a constant so I really don't know what to do with it. All I can figure out is that I suppose my final expression for w will probably end up with the denominator as a multiple of z alone, seeing as 0 maps to infinity. But other than that I have no idea how to start this.

My notes for this class don't have any examples where there is a point at infinity, and there aren't any examples where the above formula can't be used. They mention that you can brute-force for all the coefficients in your final expression in w, but that's all the information given about that method so I don't know if that would work.

Thanks for any help

[edit]Just for fun, I ended up throwing infinity in as w₂ in the equation and everything seemed to work out. I ended up with $w = \frac{(1+j)z + (j-1)}{2z}$ and that seems to map all the points properly, at least if $\frac{j - 1}{0} = \infty$. My main problem with this is that I had to algebraically say that $\frac{-\infty}{-\infty} = 1$ when I first subbed it in, and something about that rubs me the wrong way. The whole substituting infinity thing in general seems wrong to me, but saying that $(1 - \infty) = (w - \infty) = -\infty$ just seems a bit sloppy to me and I'm not sure if I can say that.

 

[mighty2000]

it's important to be precise in your calculations and not make assumptions or shortcuts that may lead to incorrect results. In this case, substituting infinity into the equation may not be the most accurate approach.

Instead, you can approach this problem by using the properties of linear fractional transformations. In this case, we have three points in the z-plane and three corresponding points in the w-plane. This means we can write a linear fractional transformation in the form:

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

where a, b, c, and d are constants that we need to solve for. To do this, we can use the fact that the cross-ratio of four points is preserved under linear fractional transformations. In other words, for any four points z1, z2, z3, z4 in the z-plane and their corresponding points w1, w2, w3, w4 in the w-plane, we have:

\frac{(w1 - w2)(w3 - w4)}{(w1 - w3)(w2 - w4)} = \frac{(z1 - z2)(z3 - z4)}{(z1 - z3)(z2 - z4)}

Using this property, we can set up a system of equations to solve for a, b, c, and d. In this case, we have:

\frac{(j - \infty)(1 - 1)}{(j - 1)(\infty - 1)} = \frac{(-1 - 0)(-1 - \infty)}{(-1 - \infty)(0 - \infty)}

Simplifying this equation, we get:

\frac{1}{j-1} = -\frac{1}{\infty}

Since \infty is not a well-defined number, we can't use it in our calculations. Instead, we can use the fact that as z approaches \infty, the value of \frac{1}{z} approaches 0. This means we can rewrite our equation as:

\frac{1}{j-1} = 0

Solving for j, we get:

j = 1

Now we can use this value of j in our other equations to solve for a, b, c, and d. Plugging in the values for z1, z2, z3, and w1, we get