Explainer: Möbius Transformations

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 𝑧=𝑥+𝑦𝑖, where 𝑥,𝑦 are constant.

  1. The transformation 𝑇(𝑧)=𝑧+𝑧 represents a translation by the vector 𝑥𝑦.
  2. The transformation 𝑇(𝑧)=𝑧𝑧 represents a dilation by scale factor |𝑧| and a counterclockwise rotation about the origin by arg(𝑧).

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 𝑇𝑧1𝑧.

Example 1: Reciprocal Transformation

A transformation which maps the 𝑧-plane to the 𝑤-plane is defined by 𝑇𝑧1𝑧, where 𝑧0.

  1. Find an equation for the image of |𝑧|=2 under the transformation.
  2. Find an equation for the image of arg(𝑧)=3𝜋4.
  3. Find a Cartesian equation for the image of Im(𝑧)=2.
  4. Find a Cartesian equation for the image of |𝑧𝑖|=12.

Answer

To find an equation representing the image of a particular curve under the transformation 𝑇, first express 𝑧 in terms of 𝑤, 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 𝑧 in terms of 𝑤 as follows: 𝑧=1𝑤.

We can now substitute this into the equation |𝑧|=2 to get |||1𝑤|||=2.

Using the properties of the modulus, we can rewrite this as 1|𝑤|=2.

Rearranging the equation, we get |𝑤|=12.

This is the equation of a circle of radius 12 centered at the origin. We can represent the effect of this transformation visually as follows.

Part 2

Substituting 𝑧=1𝑤 into arg(𝑧)=3𝜋4 yields arg1𝑤=3𝜋4.

Using the properties of the argument, we can rewrite this as argarg(1)(𝑤)=3𝜋4.

Since arg(1)=0, we have arg(𝑤)=3𝜋4.

This is the half-line centered at the origin which makes an angle of 3𝜋4 with the positive real axis. We can represent the effect of this transformation visually as follows.

Part 3

Substituting 𝑧=1𝑤 into Im(𝑧)=2 yields Im1𝑤=2.

To find a Cartesian equation, we begin by substituting 𝑤=𝑢+𝑣𝑖 into this equation as follows: Im1𝑢+𝑣𝑖=2.

Multiplying the numerator and denominator by the complex conjugate of the denominator, we can rewrite the fraction as follows: Im𝑢𝑣𝑖𝑢+𝑣=2.

Hence, 𝑣𝑢+𝑣=2.

Now we can manipulate this equation to put it into a standard form. To do this, we first multiply through by 𝑢+𝑣, which gives 𝑣=2𝑢+2𝑣.

Adding 𝑣 to both sides and dividing through by two gives 𝑢+𝑣+12𝑣=0.

We can now complete the square in 𝑣 to get 𝑢+𝑣+1414=0.

Hence, 𝑢+𝑣+14=14.

This represents a circle of radius 14 centered at 0,14. We can visualize the effect of this transformation on the line Im(𝑧)=2 as follows.

Part 4

Substituting 𝑧=1𝑤 into |𝑧𝑖|=12 gives |||1𝑤𝑖|||=12.

Rewriting the subject of the modulus as a single fraction we have |||1𝑖𝑤𝑤|||=12.

Factoring 𝑖 out of the denominator, we have |||𝑖(𝑤+𝑖)𝑤|||=12.

We can now use the properties of the modulus to rewrite this as |𝑖||𝑤+𝑖||𝑤|=12.

Hence, 2|𝑤+𝑖|=|𝑤|.

This is the equation of a circle. However, to find its Cartesian equation, we need to substitute 𝑤=𝑢+𝑣𝑖 into the equation as follows: 2|𝑢+𝑣𝑖+𝑖|=|𝑢+𝑣𝑖|.

Squaring both sides yields 4|𝑢+(𝑣+1)𝑖|=|𝑢+𝑣𝑖|.

We can now use the definition of the modulus to rewrite this as 4𝑢+(𝑣+1)=𝑢+𝑣.

Expanding the parentheses, we get 4𝑢+4𝑣+2𝑣+1=𝑢+𝑣.

Gathering together our like terms, we have 3𝑢+3𝑣+8𝑣+4=0.

We now divide through by 3, which gives us 𝑢+𝑣+83𝑣+43=0.

Finally, we can complete the square in 𝑣 to get 𝑢+𝑣+43169+43=0.

Hence, 𝑢+𝑣+43=49.

This is the equation of a circle of radius 23 centered at 0,43. We can visually represent the effect of this transformation on the circle |𝑧𝑖|=12 as follows.

The previous example showed that the transformation 𝑤=1𝑧 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 𝑟 centered at the origin to a circle of radius 1𝑟 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:

  1. Translations: 𝑇(𝑧)=𝑧+𝑧,
  2. Rotations and dilations: 𝑇(𝑧)=𝑎𝑧,
  3. Inversion with reflection in the real axis: 𝑇(𝑧)=1𝑧.

Formally speaking, we define Möbius Transformations as follows.

Möbius Transformations

A transformation 𝑇 of the form 𝑇(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑, where 𝑎,𝑐,𝑏,𝑑 and 𝑎𝑑𝑏𝑐0, is called a Möbius transformation.

In the definition of a Möbius transformation, we gave the restriction that 𝑎𝑑𝑏𝑐0. In the next example, we will explore why we impose this restriction.

Example 2: Degenerate Case

Given that 𝑎𝑑𝑏𝑐=0, simplify the expression 𝑎𝑧+𝑏𝑐𝑧+𝑑, where 𝑧𝑑𝑐.

Answer

To use the property that 𝑎𝑑𝑏𝑐=0, we can multiply the expression by 𝑑𝑑 to get 𝑎𝑧+𝑏𝑐𝑧+𝑑=𝑎𝑑𝑧+𝑏𝑑𝑐𝑑𝑧+𝑑.

We can use the property 𝑎𝑑𝑏𝑐=0 and substitute 𝑎𝑑 for 𝑏𝑐 as follows: 𝑎𝑧+𝑏𝑐𝑧+𝑑=𝑏𝑐𝑧+𝑏𝑑𝑐𝑑𝑧+𝑑=𝑏(𝑐𝑧+𝑑)𝑑(𝑐𝑧+𝑑).

Given that 𝑧𝑑𝑐, we can cancel the expression 𝑐𝑧+𝑑 from the numerator and denominator to get 𝑎𝑧+𝑏𝑐𝑧+𝑑=𝑏𝑑.

As we saw in the previous example, we include the restriction that 𝑎𝑑𝑏𝑐0 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 𝑇(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑 and 𝑇(𝑧)=𝛼𝑧+𝛽𝛾𝑧+𝛿. Write an expression for the composition 𝑇𝑇 in terms of 𝑎, 𝑏, 𝑐, 𝑑, 𝛼, 𝛽, 𝛾, and 𝛿 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 𝑇(𝑧) into the formula defining 𝑇, we have 𝑇𝑇(𝑧)=𝑎+𝑏𝑐+𝑑.

We can simplify this expression by multiplying the numerator and denominator by 𝛾𝑧+𝛿 as follows: 𝑇𝑇(𝑧)=𝑎(𝛼𝑧+𝛽)+𝑏(𝛾𝑧+𝛿)𝑐(𝛼𝑧+𝛽)+𝑑(𝛾𝑧+𝛿).

We now gather the 𝑧 terms and constant terms in the numerator and denominator to get 𝑇𝑇(𝑧)=(𝑎𝛼+𝑏𝛾)𝑧+(𝑎𝛽+𝑏𝛿)(𝑐𝛼+𝑑𝛾)𝑧+(𝑐𝛽+𝑑𝛿).

To confirm that this is a Möbius transformation, we need to check that Δ=(𝑎𝛼+𝑏𝛾)(𝑐𝛽+𝑑𝛿)(𝑎𝛽+𝑏𝛿)(𝑐𝛼+𝑑𝛾)0. To do this, we expand the parentheses as follows: Δ=𝑎𝑐𝛼𝛽+𝑎𝑑𝛼𝛿+𝑏𝑐𝛽𝛾+𝑏𝑑𝛾𝛿(𝑎𝑐𝛼𝛽+𝑎𝑑𝛼𝛾+𝑏𝑐𝛼𝛿+𝑏𝑑𝛿𝛾)=𝑎𝑑(𝛼𝛿𝛼𝛾)+𝑏𝑐(𝛽𝛾𝛼𝛿)=(𝑎𝑑𝑏𝑐)(𝛼𝛿𝛼𝛾).

Since 𝑇 and 𝑇 are Möbius transformations, we know that 𝑎𝑑𝑏𝑐0 and 𝛼𝛿𝛼𝛾0; therefore, Δ0, which implies that 𝑇𝑇 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 𝑇 in the matrix 𝑀=𝑎𝑏𝑐𝑑, and the coefficients of 𝑇 in another matrix 𝑀=𝛼𝛽𝛾𝛿, the coefficients of 𝑇𝑇 are the elements of the matrix 𝑀𝑀. Furthermore, the condition that 𝑎𝑑𝑏𝑐0 is simply the condition that the matrix 𝑀 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 𝑧-plane to the 𝑤-plane is given by 𝑤=𝑖𝑧+6𝑧3𝑖, where 𝑧3𝑖.

  1. Find the Cartesian equation of the image of |𝑧|=3 under the transformation.
  2. Find the Cartesian equation of the image of |𝑧|=1 under the transformation.

Answer

Part 1

To find the equations of the image of |𝑧|=3, we first express 𝑧 in terms of 𝑤. Starting from the definition of the transformation 𝑤=𝑖𝑧+6𝑧3𝑖, we multiply through by 𝑧3𝑖 which gives 𝑤(𝑧3𝑖)=𝑖𝑧+6.

Gathering the 𝑧 terms of the left-hand side, we have 𝑤𝑧𝑖𝑧=6+3𝑖𝑤.

Dividing both sides of the equation by 𝑤𝑖 gives 𝑧=6+3𝑖𝑤𝑤𝑖.

We can now substitute this into the equation |𝑧|=3, to find an equation for its image in the 𝑤-plane. Hence, |||6+3𝑖𝑤𝑤𝑖|||=3.

Factoring out 3𝑖 from the numerator, we have |||3𝑖(𝑤2𝑖)𝑤𝑖|||=3.

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

|3𝑖||𝑤2𝑖||𝑤𝑖|=3.(1)

Rearranging and using the fact that |3𝑖|=3, we have |𝑤2𝑖|=|𝑤𝑖|.

This is the equation of a line; in particular, it is the perpendicular bisector of the line segment between 𝑖 and 2𝑖. We can write the Cartesian equation for this line as 𝑣=32. It is also possible to derive this equation by substituting 𝑤=𝑢+𝑣𝑖 into the equation |𝑤2𝑖|=|𝑤𝑖| as we will show. Starting from the equation |𝑢+𝑣𝑖2𝑖|=|𝑢+𝑣𝑖𝑖|, we can square both sides to get |𝑢+(𝑣2)𝑖|=|𝑢+(𝑣1)𝑖|.

Using the definition of the modulus, we can rewrite this as 𝑢+(𝑣2)=𝑢+(𝑣1).

Canceling the 𝑢 terms and expanding the parentheses, we have 𝑣4𝑣+4=𝑣2𝑣+1.

Canceling 𝑣 and rearranging, we get the equation for the line as 𝑣=32. 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 |𝑧|=1. In particular, we can use the left-hand side of equation (1), since this is equal to |𝑧|, and set it equal to 1 as follows: |3𝑖||𝑤2𝑖||𝑤𝑖|=1.

Multiplying by |𝑤𝑖| and simplifying, we have 3|𝑤2𝑖|=|𝑤𝑖|.

This is the equation of a circle. To find its Cartesian equation, we substitute 𝑤=𝑢+𝑖𝑣 into the equation to get 3|𝑢+𝑖𝑣2𝑖|=|𝑢+𝑖𝑣𝑖|.

Squaring both sides of the equation yields 9|𝑢+(𝑣2)𝑖|=|𝑢+(𝑣1)𝑖|.

Using the definition of the modulus, we can rewrite this as9𝑢+(𝑣2)=𝑢+(𝑣1).

We now expand the parentheses to get 9𝑢+9𝑣4𝑣+4=𝑢+𝑣2𝑣+1.

Rearranging, we have 8𝑢+8𝑣34𝑣+35=0.

Dividing through by 8 gives 𝑢+𝑣174𝑣+358=0.

We can now complete the square in 𝑣 to get 𝑢+𝑣17828964+358=0.

Hence, 𝑢+𝑣178=964, which represents a circle of radius 38 centered at 0,178. We can represent the effect of this Möbius transformation on this circle visually as follows.

As we can see, much like the transformation 𝑤=1𝑧, 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, 𝑇, which maps the 𝑧-plane to the 𝑤-plane is given by 𝑇(𝑧)=(2+𝑖)𝑧+4𝑧𝑖, where 𝑧𝑖.

  1. Find a Cartesian equation for the image of the imaginary axis under the transformation 𝑇.
  2. Hence, find the image of the region Im(𝑧)>0 under the transformation 𝑇.

Answer

Part 1

We begin by expressing 𝑧 in terms of 𝑤. Starting from the equation defining the transformation 𝑤=(2+𝑖)𝑧+4𝑧𝑖, we multiply both sides by 𝑧𝑖 which give us 𝑤𝑧𝑤𝑖=(2+𝑖)𝑧+4.

Gathering the 𝑧 terms on the left-hand side, we have 𝑧(𝑤(2+𝑖))=𝑖𝑤+4.

Hence, by dividing by 𝑤(2+𝑖), we have 𝑧=𝑖𝑤+4𝑤(2+𝑖).

Since we would like to consider the image of the imaginary axis (Re(𝑧)=0), we would need to find an expression for the real part of 𝑖𝑤+4𝑤(2+𝑖). To find this, we substitute in 𝑤=𝑢+𝑣𝑖 as follows: 𝑧=𝑖(𝑢+𝑣𝑖)+4𝑢+𝑣𝑖2𝑖. Gathering the real and imaginary parts in the numerator and denominator, we have 𝑧=4𝑣+𝑢𝑖𝑢2+(𝑣1)𝑖.

Multiplying the numerator and denominator by the complex conjugate of the denominator, we get 𝑧=((4𝑣)+𝑢𝑖)((𝑢2)(𝑣1)𝑖)(𝑢2)+(𝑣1).

Expanding the parentheses in the numerator, we have 𝑧=(4𝑣)(𝑢2)(4𝑣)(𝑣1)𝑖+𝑢(𝑢2)𝑖+𝑢(𝑣1)(𝑢2)+(𝑣1).

Hence, 𝑧=(4𝑣)(𝑢2)+𝑢(𝑣1)((4𝑣)(𝑣1)𝑢(𝑢2))𝑖(𝑢2)+(𝑣1).

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

0=(4𝑣)(𝑢2)+𝑢(𝑣1)(𝑢2)+(𝑣1).(2)

Multiplying by (𝑢2)+(𝑣1) yields (4𝑣)(𝑢2)+𝑢(𝑣1)=0.

Finally we can expand these parentheses to get 4𝑢8𝑣𝑢+2𝑣+𝑢𝑣𝑢=0.

Hence, 3𝑢8+2𝑣=0, which is equivalent to the equation 𝑣=432𝑢.

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