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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=ℂ∪{∞} to the extended plane ℂ

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 ℂ to ℂ. To prove this, pick 𝑤∈ℂ and observe

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

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

Examples

• 𝑐=0, 𝑑=1: Then 𝑓(𝑧)=𝑎𝑧+𝑏 (𝑎≠0). These are also called affine transformations. They map ∞ to ∞ and therefore map ℂ to ℂ. They are hence conformal mappings from ℂ to ℂ (and in fact, these are the only conformal mappings from ℂ to ℂ).
  • 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=𝑟ℯ^𝑖𝜃 then 1𝑧=1𝑟ℯ^−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.

Facts About Mobius Transformations

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∈ℂ, 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=ℂ∪{∞} to ℂ.
• 𝑓 is a conformal map from ℂto ℂ
• If 𝑐=0, 𝑑=1:𝑓(𝑧)=𝑎𝑧+𝑏 is a conformal map from ℂ to ℂ
• 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:ℂto ℂ that maps 𝑧_𝑗 to 𝑤_𝑗, 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 𝑤_𝑗, 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.