In this explainer, we will learn how to interpret MΓΆbius transformation in the complex plane.
MΓΆbius 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 MΓΆbius transformations and consider some of their basic properties including the effect they have on lines and circles in the complex plane.
Before we consider MΓΆbius 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 MΓΆbius transformations.
MΓΆbius 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 MΓΆbius Transformations as follows.
MΓΆbius Transformations
A transformation of the form where and , is called a MΓΆbius transformation.
In the definition of a MΓΆbius 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 MΓΆbius transformations.
Example 3: Composition of MΓΆbius Transformations
Consider the two MΓΆbius transformations and . Write an expression for the composition in terms of , , , , , , , and and determine whether the resulting transformation is a MΓΆbius 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 MΓΆbius transformation, we need to check that . To do this, we expand the parentheses as follows:
Since and are MΓΆbius transformations, we know that and ; therefore, , which implies that is a MΓΆbius transformation.
The last example showed us the relationship between the four complex coefficients that define a MΓΆbius transformation under composition. A careful examination of the relationship highlights an interesting connection between MΓΆbius 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 MΓΆbius 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 MΓΆbius transformations on lines and circles in the complex plane.
Example 4: Properties of MΓΆbius 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 MΓΆbius 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 MΓΆbius transformation on this circle visually as follows.
As we can see, much like the transformation , MΓΆbius 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 vanishesβwe often refer to this point as a poll of the transformation.
Example 5: Properties of MΓΆbius 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.
Part 2
We would like to find the image of the region defined by . We have already found the equation of the boundary line in the previous part, we just need to consider the direction of the inequality. Looking back to equation (2), this is the point where we used the fact that the real part was equal to zero. Instead, we can use the fact that the real part is greater than or equal to 0. Hence, replacing the equality sign with an inequality sign, we get
Since the denominator is positive, we can multiply both sides of the equation by the denominator and it will not affect the direction of the inequality. Hence,
Working through the same algebra as in the last part, we get which we can rearrange to
We can represent this on the -plane as follows.
Key Points
- MΓΆbius transformations are a special type of transformations of the complex plane that may transform lines and circles to other circles and lines.
- A transformation of the form where and , is called a MΓΆbius transformation.
- Composition of MΓΆbius transformations results in another MΓΆbius transformation.
- There are some interesting connections between MΓΆbius transformations and other areas of mathematics.