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`Complex Function Mobius Transformations Properties Examples Facts About Mobius Transformations Summary Further Facts Examples Fact Images of Circles and Lines`

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# Complex Function

## Mobius Transformations

A Mobius transformation, also called fractional linear transformation, is a function of the form

`𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑`

where 𝑎, 𝑏, 𝑐, 𝑑∈ℂ such that 𝑎𝑑−𝑏𝑐≠0.

Notes:

• As 𝑧→∞, 𝑓(𝑧)→𝑎𝑐 if 𝑐≠0 and 𝑓(𝑧)→∞ if 𝑐=0.

Thus allow 𝑧=∞ and define 𝑓(∞)=𝑎𝑐 if 𝑐≠0 and 𝑓(∞)=∞ if 𝑐=0.

• Similarly, 𝑓(𝑑𝑐)=∞, if 𝑐≠0.
• Thus regard 𝑓 as a mapping from the extended complex plane =ℂ∪{∞} to the extended plane

## Properties

Properties of 𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑 where 𝑎, 𝑏, 𝑐, 𝑑∈ℂ and 𝑎𝑑−𝑏𝑐≠0.

• 𝑓′(𝑧)=(𝑐𝑧+𝑑)𝑎−(𝑎𝑧+𝑏)𝑐(𝑐𝑧+𝑑)2=𝑎𝑑−𝑏𝑐(𝑐𝑧+𝑑)2.

The condition 𝑎𝑑−𝑏𝑐≠0 thus simply guarantees that 𝑓 is non-constant.

• If multiply each parameter 𝑎, 𝑏, 𝑐, 𝑑 by a constant 𝑘≠0, and obtain the same mapping, so a given mapping does not uniquely determine 𝑎, 𝑏, 𝑐, 𝑑.
• Mobius transformation is one-to-one and onto from to . To prove this, pick 𝑤∈ and observe

```𝑓(𝑧)=𝑤⇔𝑎𝑧+𝑏=𝑤(𝑐𝑧+𝑑)  ⇔𝑧(𝑎−𝑤𝑐)=𝑤𝑑−𝑏  ⇔𝑧=𝑤𝑑−𝑏−𝑤𝑐+𝑎 ``` For each 𝑤∈ there is one and only one 𝑧∈ such that 𝑓(𝑧)=𝑤.

Mobius transformations are thus conformal mapping from to . Mobius transformations are the only conformal mappings from to .

## Examples

• 𝑐=0, 𝑑=1: Then 𝑓(𝑧)=𝑎𝑧+𝑏 (𝑎≠0). These are also called affine transformations. They map to and therefore map to . They are hence conformal mappings from to (and in fact, these are the only conformal mappings from to ).
• 𝑓(𝑧)=𝑎𝑧 (i.e. 𝑏=0) is a rotation and dilation.
• 𝑓(𝑧)=𝑧+𝑏 (i.e. 𝑎=1) is a translation
• 𝑎=0, 𝑏=1, 𝑐=1, 𝑑=0: Then 𝑓(𝑧)=1𝑧. This is an inversion.
• If 𝑧=𝑟ℯ𝑖𝜃 then 1𝑧=1𝑟−𝑖𝜃.
• 𝑓 interchanges outside and inside of the unit circle.
• A circle, centered at 0, is clearly mapped to a circle, centered at 0, of reciprocal radius.
• The images of circles under 𝑓, where the inversion 𝑓(𝑧)=1/𝑧
• Let 𝐾={𝑧:|𝑧−3|=1} be the circle of radius 1, centered at 3. the image of 𝑓(𝐾).

``` 𝑤∈𝑓(𝐾)⇔1𝑤∈𝐾⇔1𝑤−3=1  ⇔|1−3𝑤|2=|𝑤|2  ⇔1−3𝑤−3𝑤+9|𝑤|2=|𝑤|2  ⇔8|𝑤|2−3𝑤−3𝑤=−1  ⇔(𝑤−38)(𝑤−38)= 964−18  ⇔𝑤−38=18 ```

Thus the image of the circle 𝐾={𝑧:|𝑧−3|=1} under 𝑓(𝑧)=1𝑧 is another circle, namely the circle of radius 18, centered at 38

• Let 𝐾={𝑧:|𝑧−1|=1} be the circle of radius 1, centered at 1. the image of 𝑓(𝐾).

``` 𝑤∈𝑓(𝐾)⇔1𝑤∈𝐾⇔1𝑤−1=1  ⇔|1−𝑤|2=|𝑤|2  ⇔1−𝑤−𝑤+|𝑤|2=|𝑤|2  ⇔𝑤+𝑤=1  ⇔Re𝑤=12 ```

Thus the image of the circle 𝐾={𝑧:|𝑧−1|=1} is the vertical line 𝑓(𝐾)={𝑤:Re𝑤=12}

• Since 𝑓(𝑓(𝑧))=𝑓(1𝑧)=𝑧, finding that
• 𝑓 maps the line {𝑧:Re𝑧=12} to the circle {𝑧:|𝑧−1|=1}
• 𝑓 maps the circle {𝑧:|𝑧−38|=18} to the circle {𝑧:|𝑧-3|=1}
• Let now 𝐿 be the line {𝑧:𝑧=𝑡+𝑖𝑡, -∞<𝑡<∞}. If 𝑧=𝑡+𝑖𝑡, then

```𝑓(𝑧)=1𝑧=𝑧|𝑧|2= 𝑡-𝑖𝑡2𝑡2=12𝑡-𝑖12𝑡= 𝑠-𝑖𝑠 ```

Thus the image of the line {𝑡+𝑖𝑡:𝑡∈ℝ} is the line {𝑠-𝑖𝑠:𝑠∈ℝ}.

Images of lines and circles seem to be lines or circles.

By fact. Every Mobius transformation maps circles and lines to circles or lines.

Note: A line could be viewed as a "circle through infinity".

By fact. Given three distinct points 𝑧1, 𝑧2, 𝑧3, there exists a unique Mobius transformation 𝑓 such that 𝑓(𝑧1)=0, 𝑓(𝑧2)=1, and 𝑓(𝑧3)=∞.

This Mobius transformation is

```𝑓(𝑧)=𝑧-𝑧1𝑧-𝑧3⋅ 𝑧2-𝑧3𝑧2-𝑧1 ```

## Summary

`𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑` with 𝑎, 𝑏, 𝑐, 𝑑∈ℂand 𝑎𝑑−𝑏𝑐≠0 is called a Mobius transformation..

• 𝑓 maps =ℂ∪{∞} to .
• 𝑓 is a conformal map from to
• If 𝑐=0, 𝑑=1:𝑓(𝑧)=𝑎𝑧+𝑏 is a conformal map from to
• Mobius transformations map circles and lines to circles or lines.
• For distinct 𝑧1,𝑧2,𝑧3, the Mobius transformation ```𝑓(𝑧)=𝑧−𝑧1𝑧−𝑧3⋅ 𝑧2−𝑧3𝑧2−𝑧1 ``` maps 𝑧1,𝑧2,𝑧3 to 0, 1, ∞, respectively.

## Further Facts

• The composition of two Mobius transformations is a Mobius transformation, and so is the inverse.

```𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑, 𝑔(𝑧)=ℎ𝑧+𝑖𝑗𝑧+𝑘 𝑓(𝑔(𝑧)=𝑎(𝑔(𝑧)+𝑏𝑐(𝑔(𝑧)+𝑑=𝑎(ℎ𝑧+𝑖)+𝑏(𝑗𝑧+𝑘)𝑐(ℎ𝑧+𝑖)+𝑑(𝑗𝑧+𝑘) =(𝑎ℎ+𝑏𝑗)𝑧+(𝑎𝑖+𝑏𝑘)(𝑐ℎ+𝑑𝑗)𝑧+(𝑐𝑖+𝑑𝑘)```

and so is the inverse.

`𝑤(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑⇒solve for 𝑧(𝑤)`
• Given three distinct points 𝑧1, 𝑧2, 𝑧3 and three distinct points 𝑤1, 𝑤2, 𝑤3, there exists a unique Mobius transformation 𝑓:to that maps 𝑧𝑗 to 𝑤𝑗, 𝑗=1, 2, 3,.

Proof:

Let 𝑓1 be the Mobius transformation that maps 𝑧1, 𝑧2, 𝑧3, to 0, 1, ∞.

Let 𝑓2 be the Mobius transformation that maps 𝑤1, 𝑤2, 𝑤3, to 0, 1, ∞.

Then 𝑓−12∘𝑓1 maps 𝑧1, 𝑧2, 𝑧3 to 𝑤𝑗, 𝑗=1, 2, 3,.

## Examples

Find the Mobius transformation 𝑓 that maps 0 to −1, 𝑖 to 0, and ∞ to 1.

• For 𝑓1(𝑧)

`𝑓1(𝑧)=𝑧−0"𝑧−∞"⋅"𝑖−∞"𝑖−0=𝑧𝑖` maps 0, 𝑖, ∞ to 0, 1, ∞.
• For 𝑓2(𝑧)

`𝑓2(𝑧)=𝑧+1"𝑧−1"⋅0−1"0+1=𝑧+1−𝑧+1` maps −1, 0, 1 to 0, 1, ∞.
• For 𝑓−12

```𝑤=𝑧+1−𝑧+1⇔𝑤(−𝑧+1)=𝑧+1  ⇔−𝑤𝑧+𝑤=𝑧+1  ⇔𝑤−1=𝑧(1+𝑤)  ⇔𝑧=𝑤−1𝑤+1 ``` maps 0, 1, ∞ to −1, 0, 1.
• For 𝑓=𝑓−12∘𝑓1

```𝑓(𝑧)=(𝑓−12∘𝑓1)(𝑧)=𝑧+1−𝑧+1= 𝑧𝑖−1𝑧𝑖+1=𝑧−𝑖−𝑧+𝑖```

Another approach to find the Mobius transformation 𝑓 that maps 0 to −1, 𝑖 to 0, and ∞ to 1.

• 𝑓 is of the form 𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑
• Since z(𝑖)=0, 𝑎≠0 and thus assume that 𝑎=1
• Thus 𝑓(𝑧)=𝑧+𝑏𝑐𝑧+𝑑. Since 𝑓(𝑖)=0, and thus b=−𝑖
• Thus 𝑓(𝑧)=𝑧−𝑖𝑐𝑧+𝑑. Since 𝑓(∞)=1, and thus 𝑐=1
• Thus 𝑓(𝑧)=𝑧−𝑖𝑧+𝑑. Since 𝑓(0)=-1, and thus d=𝑖
• Thus 𝑓(𝑧)=𝑧−𝑖𝑧+𝑖.

## Fact

By fact. Every Mobius transformation is the composition of maps of the type

• 𝑧⟼𝑎𝑧 (rotation and dilation
• 𝑧⟼𝑧+𝑏 (translation)
• 𝑧⟼1𝑧 (invesion)

Proof: Let 𝑓 be a Mobius transformation.

suppose first that 𝑓(∞)=∞. Then 𝑓(𝑧)=𝑎𝑧+𝑏. this corresponds to the following composition

`𝑧rot. & dil.⟼𝑎𝑧translation⟼𝑎𝑧+𝑏`

Suppose next that 𝑓(∞)≠∞. Then 𝑓(𝑧)=𝑎𝑧+𝑏𝑐𝑧+𝑑 with 𝑐≠0

`𝑓(𝑧)=𝑎𝑐𝑧+𝑏𝑐𝑧+𝑑𝑐`

Thus assume that 𝑐=1, So

`𝑓(𝑧)=𝑎𝑧+𝑏𝑧+𝑑=𝑎(𝑧+𝑑)+(𝑏−𝑎𝑑)𝑧+𝑑=𝑎+(𝑏−𝑎𝑑)𝑧+𝑑`

This corresponds to the following composition:

```𝑧trans.⟼𝑧+𝑑inv.⟼ 1𝑧+𝑑dil. & rot.⟼ 𝑏−𝑎𝑑𝑧+𝑑trans.⟼𝑎+ 𝑏−𝑎𝑑𝑧+𝑑```

## Images of Circles and Lines

As mentioned earlier the fact that Mobius transformations map circles and lines to circles and lines. How to prove this? By the previous composition result it suffices to prove this fact for the three standard types (translation, rotation and dilation, inversion).

Clearly, dilations, rotations and translations preserve circles and lines as circles and lines, and so all that is left to show is that 𝑓(𝑧)=1𝑧 maps circles and lines to circles and lines. The main ideas on how to do so is same as previous types.

ID: 190400016 Last Updated: 16/4/2019 Revision: 0

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