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Lesson Explainer: MΓΆbius Transformations Mathematics

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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 𝑧=π‘₯+π‘¦π‘–οŠ§, where π‘₯,π‘¦βˆˆβ„ are constant.

  1. The transformation 𝑇(𝑧)=𝑧+π‘§οŠ§οŠ§ represents a translation by the vector ο€»π‘₯𝑦.
  2. The transformation 𝑇(𝑧)=π‘§π‘§οŠ¨οŠ§ 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 π‘‡βˆΆπ‘§β†¦1𝑧.

Example 1: Reciprocal Transformation

A transformation which maps the 𝑧-plane to the 𝑀-plane is defined by π‘‡βˆΆπ‘§β†¦1𝑧, where 𝑧≠0.

  1. Find an equation for the image of |𝑧|=2 under the transformation.
  2. Find an equation for the image of arg(𝑧)=3πœ‹4.
  3. Find a Cartesian equation for the image of Im(𝑧)=2.
  4. Find a Cartesian equation for the image of |π‘§βˆ’π‘–|=12.

Answer

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

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

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

Rearranging the equation, we get |𝑀|=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=1𝑀 into arg(𝑧)=3πœ‹4 yields argο€Ό1π‘€οˆ=3πœ‹4.

Using the properties of the argument, we can rewrite this as argarg(1)βˆ’(𝑀)=3πœ‹4.

Since arg(1)=0, we have arg(𝑀)=βˆ’3πœ‹4.

This is the half-line centered at the origin which makes an angle of βˆ’3πœ‹4 with the positive real axis. We can represent the effect of this transformation visually as follows.

Part 3

Substituting 𝑧=1𝑀 into Im(𝑧)=2 yields Imο€Ό1π‘€οˆ=2.

To find a Cartesian equation, we begin by substituting 𝑀=𝑒+𝑣𝑖 into this equation as follows: Imο€Ό1𝑒+π‘£π‘–οˆ=2.

Multiplying the numerator and denominator by the complex conjugate of the denominator, we can rewrite the fraction as follows: Imο€½π‘’βˆ’π‘£π‘–π‘’+𝑣=2.

Hence, βˆ’π‘£π‘’+𝑣=2.

Now we can manipulate this equation to put it into a standard form. To do this, we first multiply through by 𝑒+π‘£οŠ¨οŠ¨, which gives βˆ’π‘£=2𝑒+2𝑣.

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

We can now complete the square in 𝑣 to get 𝑒+𝑣+14οˆβˆ’14=0.

Hence, 𝑒+𝑣+14=14.

This represents a circle of radius 14 centered at ο€Ό0,βˆ’14. We can visualize the effect of this transformation on the line Im(𝑧)=2 as follows.

Part 4

Substituting 𝑧=1𝑀 into |π‘§βˆ’π‘–|=12 gives |||1π‘€βˆ’π‘–|||=12.

Rewriting the subject of the modulus as a single fraction we have |||1βˆ’π‘–π‘€π‘€|||=12.

Factoring βˆ’π‘– out of the denominator, we have |||βˆ’π‘–(𝑀+𝑖)𝑀|||=12.

We can now use the properties of the modulus to rewrite this as |βˆ’π‘–||𝑀+𝑖||𝑀|=12.

Hence, 2|𝑀+𝑖|=|𝑀|.

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

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

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

Expanding the parentheses, we get 4𝑒+4𝑣+2𝑣+1=𝑒+𝑣.

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

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

Finally, we can complete the square in 𝑣 to get 𝑒+𝑣+43οˆβˆ’169+43=0.

Hence, 𝑒+𝑣+43=49.

This is the equation of a circle of radius 23 centered at ο€Ό0,βˆ’43. We can visually represent the effect of this transformation on the circle |π‘§βˆ’π‘–|=12 as follows.

The previous example showed that the transformation 𝑀=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 π‘Ÿ centered at the origin to a circle of radius 1π‘Ÿ 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:

  1. Translations: 𝑇(𝑧)=𝑧+π‘§οŠ§οŠ§,
  2. Rotations and dilations: 𝑇(𝑧)=π‘Žπ‘§οŠ¨,
  3. Inversion with reflection in the real axis: 𝑇(𝑧)=1π‘§οŠ©.

Formally speaking, we define MΓΆbius Transformations as follows.

MΓΆbius Transformations

A transformation 𝑇 of the form 𝑇(𝑧)=π‘Žπ‘§+𝑏𝑐𝑧+𝑑, where π‘Ž,𝑐,𝑏,π‘‘βˆˆβ„‚ and π‘Žπ‘‘βˆ’π‘π‘β‰ 0, is called a MΓΆbius transformation.

In the definition of a MΓΆbius transformation, we gave the restriction that π‘Žπ‘‘βˆ’π‘π‘β‰ 0. In the next example, we will explore why we impose this restriction.

Example 2: Degenerate Case

Given that π‘Žπ‘‘βˆ’π‘π‘=0, simplify the expression π‘Žπ‘§+𝑏𝑐𝑧+𝑑, where π‘§β‰ βˆ’π‘‘π‘.

Answer

To use the property that π‘Žπ‘‘βˆ’π‘π‘=0, we can multiply the expression by 𝑑𝑑 to get π‘Žπ‘§+𝑏𝑐𝑧+𝑑=π‘Žπ‘‘π‘§+𝑏𝑑𝑐𝑑𝑧+𝑑.

We can use the property π‘Žπ‘‘βˆ’π‘π‘=0 and substitute π‘Žπ‘‘ for 𝑏𝑐 as follows: π‘Žπ‘§+𝑏𝑐𝑧+𝑑=𝑏𝑐𝑧+𝑏𝑑𝑐𝑑𝑧+𝑑=𝑏(𝑐𝑧+𝑑)𝑑(𝑐𝑧+𝑑).

Given that π‘§β‰ βˆ’π‘‘π‘, we can cancel the expression 𝑐𝑧+𝑑 from the numerator and denominator to get π‘Žπ‘§+𝑏𝑐𝑧+𝑑=𝑏𝑑.

As we saw in the previous example, we include the restriction that π‘Žπ‘‘βˆ’π‘π‘β‰ 0 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 𝑇(𝑧)=π‘Žπ‘§+𝑏𝑐𝑧+π‘‘οŠ§ and 𝑇(𝑧)=𝛼𝑧+𝛽𝛾𝑧+π›ΏοŠ¨. Write an expression for the composition π‘‡βˆ˜π‘‡οŠ§οŠ¨ in terms of π‘Ž, 𝑏, 𝑐, 𝑑, 𝛼, 𝛽, 𝛾, and 𝛿 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 𝑇(𝑧) into the formula defining π‘‡οŠ§, we have π‘‡βˆ˜π‘‡(𝑧)=π‘Žο€Όοˆ+π‘π‘ο€Όοˆ+𝑑.οŠ§οŠ¨ο΅ο™οŠ°οΆο·ο™οŠ°οΈο΅ο™οŠ°οΆο·ο™οŠ°οΈ

We can simplify this expression by multiplying the numerator and denominator by 𝛾𝑧+𝛿 as follows: π‘‡βˆ˜π‘‡(𝑧)=π‘Ž(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 π‘‡βˆ˜π‘‡(𝑧)=(π‘Žπ›Ό+𝑏𝛾)𝑧+(π‘Žπ›½+𝑏𝛿)(𝑐𝛼+𝑑𝛾)𝑧+(𝑐𝛽+𝑑𝛿).

To confirm that this is a MΓΆbius transformation, we need to check that Ξ”=(π‘Žπ›Ό+𝑏𝛾)(𝑐𝛽+𝑑𝛿)βˆ’(π‘Žπ›½+𝑏𝛿)(𝑐𝛼+𝑑𝛾)β‰ 0. To do this, we expand the parentheses as follows: Ξ”=π‘Žπ‘π›Όπ›½+π‘Žπ‘‘π›Όπ›Ώ+𝑏𝑐𝛽𝛾+π‘π‘‘π›Ύπ›Ώβˆ’(π‘Žπ‘π›Όπ›½+π‘Žπ‘‘π›Όπ›Ύ+𝑏𝑐𝛼𝛿+𝑏𝑑𝛿𝛾)=π‘Žπ‘‘(π›Όπ›Ώβˆ’π›Όπ›Ύ)+𝑏𝑐(π›½π›Ύβˆ’π›Όπ›Ώ)=(π‘Žπ‘‘βˆ’π‘π‘)(π›Όπ›Ώβˆ’π›Όπ›Ύ).

Since π‘‡οŠ§ and π‘‡οŠ¨ are MΓΆbius transformations, we know that π‘Žπ‘‘βˆ’π‘π‘β‰ 0 and π›Όπ›Ώβˆ’π›Όπ›Ύβ‰ 0; therefore, Ξ”β‰ 0, which implies that π‘‡βˆ˜π‘‡οŠ§οŠ¨ 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 π‘‡οŠ§ in the matrix 𝑀=ο€Όπ‘Žπ‘π‘π‘‘οˆ, and the coefficients of π‘‡οŠ¨ in another matrix 𝑀=ο€Ύπ›Όπ›½π›Ύπ›ΏοŠ, the coefficients of π‘‡βˆ˜π‘‡οŠ§οŠ¨ are the elements of the matrix π‘€π‘€οŠ§οŠ¨. Furthermore, the condition that π‘Žπ‘‘βˆ’π‘π‘β‰ 0 is simply the condition that the matrix π‘€οŠ§ 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 𝑧-plane to the 𝑀-plane is given by 𝑀=𝑖𝑧+6π‘§βˆ’3𝑖, where 𝑧≠3𝑖.

  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 𝑀. Starting from the definition of the transformation 𝑀=𝑖𝑧+6π‘§βˆ’3𝑖, we multiply through by π‘§βˆ’3𝑖 which gives 𝑀(π‘§βˆ’3𝑖)=𝑖𝑧+6.

Gathering the 𝑧 terms of the left-hand side, we have π‘€π‘§βˆ’π‘–π‘§=6+3𝑖𝑀.

Dividing both sides of the equation by π‘€βˆ’π‘– gives 𝑧=6+3π‘–π‘€π‘€βˆ’π‘–.

We can now substitute this into the equation |𝑧|=3, to find an equation for its image in the 𝑀-plane. Hence, |||6+3π‘–π‘€π‘€βˆ’π‘–|||=3.

Factoring out 3𝑖 from the numerator, we have |||3𝑖(π‘€βˆ’2𝑖)π‘€βˆ’π‘–|||=3.

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

|3𝑖||π‘€βˆ’2𝑖||π‘€βˆ’π‘–|=3.(1)

Rearranging and using the fact that |3𝑖|=3, we have |π‘€βˆ’2𝑖|=|π‘€βˆ’π‘–|.

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 𝑀=𝑒+𝑣𝑖 into the equation |π‘€βˆ’2𝑖|=|π‘€βˆ’π‘–| as we will show. Starting from the equation |𝑒+π‘£π‘–βˆ’2𝑖|=|𝑒+π‘£π‘–βˆ’π‘–|, we can square both sides to get |𝑒+(π‘£βˆ’2)𝑖|=|𝑒+(π‘£βˆ’1)𝑖|.

Using the definition of the modulus, we can rewrite this as 𝑒+(π‘£βˆ’2)=𝑒+(π‘£βˆ’1).

Canceling the π‘’οŠ¨ terms and expanding the parentheses, we have π‘£βˆ’4𝑣+4=π‘£βˆ’2𝑣+1.

Canceling π‘£οŠ¨ and rearranging, we get the equation for the line as 𝑣=32. 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 |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||π‘€βˆ’2𝑖||π‘€βˆ’π‘–|=1.

Multiplying by |π‘€βˆ’π‘–| and simplifying, we have 3|π‘€βˆ’2𝑖|=|π‘€βˆ’π‘–|.

This is the equation of a circle. To find its Cartesian equation, we substitute 𝑀=𝑒+𝑖𝑣 into the equation to get 3|𝑒+π‘–π‘£βˆ’2𝑖|=|𝑒+π‘–π‘£βˆ’π‘–|.

Squaring both sides of the equation yields 9|𝑒+(π‘£βˆ’2)𝑖|=|𝑒+(π‘£βˆ’1)𝑖|.

Using the definition of the modulus, we can rewrite this as9𝑒+(π‘£βˆ’2)=𝑒+(π‘£βˆ’1).

We now expand the parentheses to get 9𝑒+9ο€Ήπ‘£βˆ’4𝑣+4=𝑒+π‘£βˆ’2𝑣+1.

Rearranging, we have 8𝑒+8π‘£βˆ’34𝑣+35=0.

Dividing through by 8 gives 𝑒+π‘£βˆ’174𝑣+358=0.

We can now complete the square in 𝑣 to get 𝑒+ο€Όπ‘£βˆ’178οˆβˆ’28964+358=0.

Hence, 𝑒+ο€Όπ‘£βˆ’178=964, which represents a circle of radius 38 centered at ο€Ό0,178. We can represent the effect of this MΓΆbius transformation on this circle visually as follows.

As we can see, much like the transformation 𝑀=1𝑧, 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 vanishesβ€”we often refer to this point as a poll of the transformation.

Example 5: Properties of MΓΆbius Transformations

A transformation, 𝑇, which maps the 𝑧-plane to the 𝑀-plane is given by 𝑇(𝑧)=(2+𝑖)𝑧+4π‘§βˆ’π‘–, where 𝑧≠𝑖.

  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 𝑀. Starting from the equation defining the transformation 𝑀=(2+𝑖)𝑧+4π‘§βˆ’π‘–, we multiply both sides by π‘§βˆ’π‘– which give us π‘€π‘§βˆ’π‘€π‘–=(2+𝑖)𝑧+4.

Gathering the 𝑧 terms on the left-hand side, we have 𝑧(π‘€βˆ’(2+𝑖))=𝑖𝑀+4.

Hence, by dividing by π‘€βˆ’(2+𝑖), we have 𝑧=𝑖𝑀+4π‘€βˆ’(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 𝑖𝑀+4π‘€βˆ’(2+𝑖). To find this, we substitute in 𝑀=𝑒+𝑣𝑖 as follows: 𝑧=𝑖(𝑒+𝑣𝑖)+4𝑒+π‘£π‘–βˆ’2βˆ’π‘–. Gathering the real and imaginary parts in the numerator and denominator, we have 𝑧=4βˆ’π‘£+π‘’π‘–π‘’βˆ’2+(π‘£βˆ’1)𝑖.

Multiplying the numerator and denominator by the complex conjugate of the denominator, we get 𝑧=((4βˆ’π‘£)+𝑒𝑖)((π‘’βˆ’2)βˆ’(π‘£βˆ’1)𝑖)(π‘’βˆ’2)+(π‘£βˆ’1).

Expanding the parentheses in the numerator, we have 𝑧=(4βˆ’π‘£)(π‘’βˆ’2)βˆ’(4βˆ’π‘£)(π‘£βˆ’1)𝑖+𝑒(π‘’βˆ’2)𝑖+𝑒(π‘£βˆ’1)(π‘’βˆ’2)+(π‘£βˆ’1).

Hence, 𝑧=(4βˆ’π‘£)(π‘’βˆ’2)+𝑒(π‘£βˆ’1)βˆ’((4βˆ’π‘£)(π‘£βˆ’1)βˆ’π‘’(π‘’βˆ’2))𝑖(π‘’βˆ’2)+(π‘£βˆ’1).

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

0=(4βˆ’π‘£)(π‘’βˆ’2)+𝑒(π‘£βˆ’1)(π‘’βˆ’2)+(π‘£βˆ’1).(2)

Multiplying by (π‘’βˆ’2)+(π‘£βˆ’1) yields (4βˆ’π‘£)(π‘’βˆ’2)+𝑒(π‘£βˆ’1)=0.

Finally we can expand these parentheses to get 4π‘’βˆ’8βˆ’π‘£π‘’+2𝑣+π‘’π‘£βˆ’π‘’=0.

Hence, 3π‘’βˆ’8+2𝑣=0, which is equivalent to the equation 𝑣=4βˆ’32𝑒.

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 Re(𝑧)>0. 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 0<(4βˆ’π‘£)(π‘’βˆ’2)+𝑒(π‘£βˆ’1)(π‘’βˆ’2)+(π‘£βˆ’1).

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, 0<(4βˆ’π‘£)(π‘’βˆ’2)+𝑒(π‘£βˆ’1).

Working through the same algebra as in the last part, we get 0<3π‘’βˆ’8+2𝑣, which we can rearrange to 𝑣>4βˆ’32𝑒.

We can represent this on the 𝑀-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 𝑇 of the form 𝑇(𝑧)=π‘Žπ‘§+𝑏𝑐𝑧+𝑑, where π‘Ž,𝑐,𝑏,π‘‘βˆˆβ„‚ and π‘Žπ‘‘βˆ’π‘π‘β‰ 0, 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.

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