How Can I Create a Mobius Transformation?

[Stephen88]

I want to understand how to make a Mobius Transformation.If someone can help me with an example that will be great.
Let's say we have f(0) = i, f(1) = 1, f(−1) = −1 for instance ...how should I proceed in finding one?Thank you

 

[Fernando Revilla]

Write $f(z)=\dfrac{az+b}{cz+d}$ , then $f(0)=i\Leftrightarrow b=di\;,\;\ldots$ etc , and solve the system on the unknowns $a,b,c,d$ .

 

[Stephen88]

I saw an example f(z)=(z-z1)(z2-z3)/(z-z3)(z2-z1)...and I've used that and I got 2z/(z+1) and for f(w) I got 2(w-i)/(w+1)(1-i)...then by computing f(z) o f^(-1)(w)=mobius transformation...but I don't know how to get f^(-1)(w).

 

[Fernando Revilla]

I saw an example f(z)=(z-z1)(z2-z3)/(z-z3)(z2-z1)...and I've used that and I got 2z/(z+1) and for f(w) I got 2(w-i)/(w+1)(1-i)...then by computing f(z) o f^(-1)(w)=mobius transformation...but I don't know how to get f^(-1)(w).

If you prefer this method, the result is: there exists a Möbius transformation $w$ such that $w(z_1)=w_1,w(z_2)=w_2,w(z_3)=w_3$ . This transformation is defined by $\dfrac{(w-w_1)(w_2-w_3)}{(w-w_3)(w_2-w_1)}=\dfrac{(z-z_1)(z_2-z_3)}{(z-z_3)(z_2-z_1)}$ . As you say, $\dfrac{2(w-i)}{(w+1)(1-i)}=\dfrac{2z}{z+1}$ . Now, you can easily express $w=\ldots$ as a function of $z$ .

 

[blue_raver22]

A Mobius transformation is a type of transformation in mathematics that maps a complex plane onto itself. It is defined by the formula f(z) = (az + b) / (cz + d), where a, b, c, and d are complex numbers and z is a complex variable. This transformation is also known as a linear fractional transformation.

To understand how to make a Mobius transformation, it is important to first understand its properties. One of the key properties of a Mobius transformation is that it preserves circles. This means that if a circle is mapped onto the complex plane using a Mobius transformation, the resulting image will also be a circle.

To find a specific example of a Mobius transformation, we can use the given conditions f(0) = i, f(1) = 1, f(-1) = -1. These conditions can be represented as three equations:

(i) a(0) + b = i
(ii) a(1) + b = 1
(iii) a(-1) + b = -1

Solving these equations simultaneously, we get a = 1, b = i. Substituting these values into the formula for a Mobius transformation, we get:

f(z) = (z + i) / (z + 1)

This is an example of a Mobius transformation that satisfies the given conditions. You can verify this by plugging in different values for z and observing the resulting transformation.

In general, to find a Mobius transformation that satisfies certain conditions, you can follow a similar approach by setting up equations and solving for the unknown variables. It is also helpful to keep in mind the key properties of Mobius transformations, such as preserving circles and having a linear fractional form. Additionally, there are many online resources and software programs available that can help you generate and visualize Mobius transformations. I hope this helps you understand how to make a Mobius transformation.