In this explainer, we will learn how to interpret Möbius transformation in the complex plane.

Möbius transformations form an interesting class of the transformations of the complex plane. They have a number of deep properties and connections to other areas of mathematics, such as group theory, and even have a deep connection to relativity via Lorentz transformations. In this explainer, we will introduce Möbius transformations and consider some of their basic properties including the effect they have on lines and circles in the complex plane.

Before we consider Möbius transformations, we will recap some of the basic transformations of the complex plane.

### Basic Transformations of the Complex Plane

Let , where are constant.

- The transformation represents a translation by the vector .
- The transformation represents a dilation by scale factor and a counterclockwise rotation about the origin by .

These transformations have the property that they map straight lines to straight lines and circles to circles. In this explainer, we will consider a more general class of transformations that map circles and straight lines to circles and straight lines, possibly mapping a straight line to a circle and vice versa.

We will start by considering the effect of the reciprocal map .

### Example 1: Reciprocal Transformation

A transformation which maps the -plane to the -plane is defined by , where .

- Find an equation for the image of under the transformation.
- Find an equation for the image of .
- Find a Cartesian equation for the image of .
- Find a Cartesian equation for the image of .

### Answer

To find an equation representing the image of a particular curve under the transformation , first express in terms of , and then substitute this into the equations defining the particular curve; then we can rearrange to get the expression.

**Part 1**

We begin by expressing in terms of as follows:

We can now substitute this into the equation to get

Using the properties of the modulus, we can rewrite this as

Rearranging the equation, we get

This is the equation of a circle of radius centered at the origin. We can represent the effect of this transformation visually as follows.

**Part 2**

Substituting into yields

Using the properties of the argument, we can rewrite this as

Since , we have

This is the half-line centered at the origin which makes an angle of with the positive real axis. We can represent the effect of this transformation visually as follows.

**Part 3**

Substituting into yields

To find a Cartesian equation, we begin by substituting into this equation as follows:

Multiplying the numerator and denominator by the complex conjugate of the denominator, we can rewrite the fraction as follows:

Hence,

Now we can manipulate this equation to put it into a standard form. To do this, we first multiply through by , which gives

Adding to both sides and dividing through by two gives

We can now complete the square in to get

Hence,

This represents a circle of radius centered at . We can visualize the effect of this transformation on the line as follows.

**Part 4**

Substituting into gives

Rewriting the subject of the modulus as a single fraction we have

Factoring out of the denominator, we have

We can now use the properties of the modulus to rewrite this as

Hence,

This is the equation of a circle. However, to find its Cartesian equation, we need to substitute into the equation as follows:

Squaring both sides yields

We can now use the definition of the modulus to rewrite this as

Expanding the parentheses, we get

Gathering together our like terms, we have

We now divide through by 3, which gives us

Finally, we can complete the square in to get

Hence,

This is the equation of a circle of radius centered at . We can visually represent the effect of this transformation on the circle as follows.

The previous example showed that the transformation maps some lines to circles and some circles to circles. By looking carefully at the previous example, we can piece together what the transformation is doing. The first part generalizes to tell us that the transformation maps a circle of radius centered at the origin to a circle of radius centered at the origin. The second part shows us that a ray from the origin gets mapped to another ray which has been reflected in the real axis. The third part shows us that lines which do not pass through the origin get mapped to circles which pass through the origin. Finally, the fourth part shows us that circles which do not pass through the origin get mapped to other circles. Putting these facts together and recalling some geometry, we can see that the transformation is a combination of inversion in the unit circle and reflection in the real axis.

We now turn our attention to the general definition of Möbius transformations.

Möbius transformations are the transformations we get by composing the following three basic types of transformation in the complex plane:

- Translations: ,
- Rotations and dilations: ,
- Inversion with reflection in the real axis: .

Formally speaking, we define Möbius Transformations as follows.

### Möbius Transformations

A transformation of the form where and , is called a Möbius transformation.

In the definition of a Möbius transformation, we gave the restriction that . In the next example, we will explore why we impose this restriction.

### Example 2: Degenerate Case

Given that , simplify the expression where .

### Answer

To use the property that , we can multiply the expression by to get

We can use the property and substitute for as follows:

Given that , we can cancel the expression from the numerator and denominator to get

As we saw in the previous example, we include the restriction that to avoid the trivial case where the function reduces to a constant function. We now consider what happens if we conform to Möbius transformations.

### Example 3: Composition of Möbius Transformations

Consider the two Möbius transformations and . Write an expression for the composition in terms of , , , , , , , and and determine whether the resulting transformation is a Möbius transformation.

### Answer

To solve a problem like this, we simply need to substitute one equation into the other and work through the algebra. Hence, substituting into the formula defining , we have

We can simplify this expression by multiplying the numerator and denominator by as follows:

We now gather the terms and constant terms in the numerator and denominator to get

To confirm that this is a Möbius transformation, we need to check that . To do this, we expand the parentheses as follows:

Since and are Möbius transformations, we know that and ; therefore, , which implies that is a Möbius transformation.

The last example showed us the relationship between the four complex coefficients that define a Möbius transformation under composition. A careful examination of the relationship highlights an interesting connection between Möbius transformations and matrices. In particular, if we write the coefficients of in the matrix and the coefficients of in another matrix the coefficients of are the elements of the matrix . Furthermore, the condition that is simply the condition that the matrix is not singular. Although this relationship between matrices and Möbius transformations is interesting to explore in more depth, we will not explore this further here. Instead we will turn our attention to the effect of Möbius transformations on lines and circles in the complex plane.

### Example 4: Properties of Möbius Transformations

A transformation which maps the -plane to the -plane is given by , where .

- Find the Cartesian equation of the image of under the transformation.
- Find the Cartesian equation of the image of under the transformation.

### Answer

**Part 1**

To find the equations of the image of , we first express in terms of . Starting from the definition of the transformation we multiply through by which gives

Gathering the terms of the left-hand side, we have

Dividing both sides of the equation by gives

We can now substitute this into the equation , to find an equation for its image in the -plane. Hence,

Factoring out from the numerator, we have

Using the properties of the modulus, we can rewrite this as

Rearranging and using the fact that , we have

This is the equation of a line; in particular, it is the perpendicular bisector of the line segment between and . We can write the Cartesian equation for this line as . It is also possible to derive this equation by substituting into the equation as we will show. Starting from the equation we can square both sides to get

Using the definition of the modulus, we can rewrite this as

Canceling the terms and expanding the parentheses, we have

Canceling and rearranging, we get the equation for the line as . We can represent the effect of this Möbius transformation on this circle visually as follows.

**Part 2**

We can use the calculations we made in the previous part of the question to simplify finding the equation of the image of . In particular, we can use the left-hand side of equation (1), since this is equal to , and set it equal to 1 as follows:

Multiplying by and simplifying, we have

This is the equation of a circle. To find its Cartesian equation, we substitute into the equation to get

Squaring both sides of the equation yields

Using the definition of the modulus, we can rewrite this as

We now expand the parentheses to get

Rearranging, we have

Dividing through by 8 gives

We can now complete the square in to get

Hence, which represents a circle of radius centered at . We can represent the effect of this Möbius transformation on this circle visually as follows.

As we can see, much like the transformation , Möbius transformations map circles to both lines and circles. In the final example, we will consider what happens to lines that pass through the point where the denominator vanishes—we often refer to this point as a poll of the transformation.

### Example 5: Properties of Möbius Transformations

A transformation, , which maps the -plane to the -plane is given by , where .

- Find a Cartesian equation for the image of the imaginary axis under the transformation .
- Hence, find the image of the region under the transformation .

### Answer

**Part 1**

We begin by expressing in terms of . Starting from the equation defining the transformation we multiply both sides by which give us

Gathering the terms on the left-hand side, we have

Hence, by dividing by , we have

Since we would like to consider the image of the imaginary axis (), we would need to find an expression for the real part of . To find this, we substitute in as follows: Gathering the real and imaginary parts in the numerator and denominator, we have

Multiplying the numerator and denominator by the complex conjugate of the denominator, we get

Expanding the parentheses in the numerator, we have

Hence,

We can now set the real part to be equal to zero to get

Multiplying by yields

Finally we can expand these parentheses to get

Hence, which is equivalent to the equation

Therefore, we can represent the effect of the transformation on the imaginary axis as follows.