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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:
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 nonconstant.
 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 onetoone 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 =1𝑟 ℯ^{−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 +9𝑤^{2}=𝑤^{2}
⇔8𝑤^{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:𝑧3=1}
Let now 𝐿 be the line {𝑧:𝑧=𝑡+𝑖𝑡, ∞<𝑡<∞}. If 𝑧=𝑡+𝑖𝑡, then
𝑓(𝑧)=1𝑧 =𝑧𝑧^{2} =
𝑡𝑖𝑡2𝑡^{2} =12𝑡 𝑖12𝑡 =
𝑠𝑖𝑠
Thus the image of the line {𝑡+𝑖𝑡:𝑡∈ℝ} is the line {𝑠𝑖𝑠: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 𝑧_{1}, 𝑧_{2}, 𝑧_{3}∈ℂ ,
there exists a unique Mobius transformation 𝑓 such that
𝑓(𝑧_{1})=0, 𝑓(𝑧_{2})=1, and 𝑓(𝑧_{3})=∞.
This Mobius transformation is
𝑓(𝑧)=𝑧𝑧_{1}𝑧𝑧_{3} ⋅
𝑧_{2}𝑧_{3}𝑧_{2}𝑧_{1}
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 𝑧_{1},𝑧_{2},𝑧_{3}, the Mobius transformation
𝑓(𝑧)=𝑧−𝑧_{1}𝑧−𝑧_{3} ⋅
𝑧_{2}−𝑧_{3}𝑧_{2}−𝑧_{1}
maps 𝑧_{1},𝑧_{2},𝑧_{3} 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 𝑧_{1}, 𝑧_{2}, 𝑧_{3} and three distinct points
𝑤_{1}, 𝑤_{2}, 𝑤_{3}, there exists a unique Mobius transformation
𝑓:ℂto ℂ that maps 𝑧_{𝑗} to 𝑤_{𝑗}, 𝑗=1, 2, 3,.
Proof:
Let 𝑓_{1} be the Mobius transformation that maps 𝑧_{1}, 𝑧_{2}, 𝑧_{3}, to 0, 1, ∞.
Let 𝑓_{2} be the Mobius transformation that maps 𝑤_{1}, 𝑤_{2}, 𝑤_{3}, to 0, 1, ∞.
Then 𝑓^{−1}_{2}∘𝑓_{1} maps 𝑧_{1}, 𝑧_{2}, 𝑧_{3} 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 𝑓_{1}(𝑧)
𝑓_{1}(𝑧)=𝑧−0"𝑧−∞" ⋅"𝑖−∞"𝑖−0 =𝑧𝑖
maps 0, 𝑖, ∞ to 0, 1, ∞.
For 𝑓_{2}(𝑧)
𝑓_{2}(𝑧)=𝑧+1"𝑧−1" ⋅0−1"0+1 =𝑧+1−𝑧+1
maps −1, 0, 1 to 0, 1, ∞.
For 𝑓^{−1}_{2}
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^{−1}_{2} ∘𝑓_{1}
𝑓(𝑧)=(𝑓^{−1}_{2} ∘𝑓_{1})(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+𝑑 dil. & rot.⟼
𝑏−𝑎𝑑𝑧+𝑑 trans.⟼ 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.
©sideway
ID: 190400016 Last Updated: 16/4/2019 Revision: 0

