bilinear transformation engineering

1. MA2002D-Mathematics IV
Bilinear Transformation
National Institute of Technology, Calicut

2. Stereographic Projection:
➢Let 𝒫 be the complex plane and
consider a sphere 𝒮 tangent at 𝒫 at 𝑧 = 0.
The diameter 𝑁𝑆 is perpendicular to 𝒫
and here we call 𝑁 and 𝑆 are the north and
south poles of 𝒮.
➢ Corresponding to any point 𝐴 on 𝒫 we
can construct line 𝑁𝐴 which intersect at the point 𝐴′
to the sphere 𝒮.
➢ Thus to each point in the complex plane 𝒫 there corresponds one and only one point of the
sphere 𝒮 and we can represent only complex no. by a point of the sphere.
➢We say that the point 𝑁 itself corresponds to the “point at infinity” of the plane.
➢ The set of all points of the complex plane including the point at infinity is called entire
complex plane or the extended complex plane.
➢ This method for mapping the plane on the sphere is called Stereographic projection. This
sphere is called the Riemann Sphere.

3. • Bilinear Transformation: A transformation 𝑇 defined by
𝑤 = 𝑇 𝑧 =
𝑎𝑧 + 𝑏
𝑐𝑧 + 𝑑
, 𝑎𝑑 − 𝑏𝑐 ≠ 0
is called Bilinear transformation or linear fractional or Möbius transformation.
It is a mapping from ℂ∞ to ℂ∞, where ℂ∞ = ℂ ∪ {∞}.
Observation: If 𝑤1 =
𝑎𝑧1+𝑏
𝑐𝑧1+𝑑
, 𝑤2 =
𝑎𝑧2+𝑏
𝑐𝑧2+𝑑
𝑤1 − 𝑤2 =
𝑎𝑧1 + 𝑏
𝑐𝑧1 + 𝑑
−
𝑎𝑧2 + 𝑏
𝑐𝑧2 + 𝑑
=
(𝑎𝑑 − 𝑏𝑐)(𝑧1 − 𝑧2)
(𝑐𝑧1 + 𝑑)(𝑐𝑧2 + 𝑑)
When 𝑎𝑑 − 𝑏𝑐 = 0, then 𝑤1 = 𝑤2.

4. • We define 𝑇: ℂ∞ → ℂ∞ by 𝑇 𝑧 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
, 𝑎𝑑 − 𝑏𝑐 ≠ 0.
𝑇 ∞ =
𝑎
𝑐
and 𝑇 −
𝑑
𝑐
= ∞.
• If 𝑇 𝑧 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
, 𝑎𝑑 − 𝑏𝑐 ≠ 0. Then 𝑇(𝑧) is analytic and 𝑇′
𝑧 =
𝑎𝑑−𝑏𝑐
𝑐𝑧+𝑑 2
Therefore, 𝑇′
𝑧 ≠ 0, since 𝑎𝑑 − 𝑏𝑐 ≠ 0.
Remark: Translation, rotation, magnification, Inversion and linear transformation are all
particular cases of bilinear transformation.
H.W. : Prove that every bilinear transformation maps circle or straight line into circle or
straight line respectively.

5. Theorem: Let 𝑇 𝑧 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
be a Möbius transformation.
If 𝑐 = 0, then 𝑇 is a composition of translation and magnification.
If 𝑐 ≠ 0, then 𝑇 is a composition of translation, inversion, magnification followed by another
translation.
Inverse Transformation: Let 𝑇 𝑧 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
, then 𝑇−1
=
−𝑑𝑧+𝑏
𝑐𝑧−𝑎
Fixed Point: A complex number 𝛼 is said to be a fixed point of 𝑇(𝑧) if 𝑇 𝛼 = 𝛼.
Note: Every 𝛼 ∈ ℂ is a fixed point for the identity transformation 𝑇 𝑧 = 𝑧.
Remark: Every Möbius transformation has at most two fixed points.

6. Problem: Find the image of 𝑢 = 𝑅𝑒 𝑤 < 0 under the transformation 𝑤 =
𝑧
𝑧−2
.
Solution: Put 𝑤 = 𝑢 + 𝑖𝑣 and 𝑧 = 𝑥 + 𝑖𝑦 so that
𝑤 = 𝑢 + 𝑖𝑣 =
𝑧
𝑧 − 2
=
𝑥 + 𝑖𝑦
𝑥 + 𝑖𝑦 − 2
=
𝑥 𝑥 − 2 + 𝑦2
+ 𝑖(−2𝑦)
𝑥 − 2 2 + 𝑦2
Comparing the real and imaginary parts from both sides we get,
𝑢 =
𝑥 𝑥 − 2 + 𝑦2
𝑥 − 2 2 + 𝑦2
, 𝑢 < 0 ⇒ 𝑥 𝑥 − 2 + 𝑦2
< 0
⇒ 𝑥2
+ 𝑦2
− 2𝑥 < 0
⇒ 𝑥 − 1 2
+ 𝑦2
< 1

7. Hence the image of 𝑢 = 𝑅𝑒 𝑤 < 0 is mapped into the open disc with centre (1,0) and
radius 1.

8. Problem : Show that if 𝑎 is real and 𝛼 < 1. Then 𝑤 = 𝑒𝑖𝑎 𝑧−𝛼
ഥ
𝛼𝑧−1
maps 𝑧 < 1 onto 𝑤 <
1 with 𝑧 = 𝛼 mapping onto 𝑤 = 0.
Problem: If 𝛼 = 𝑎 + 𝑖𝑏 with 𝑏 > 0, then show that 𝑤 =
𝑧−𝛼
𝑧−ഥ
𝛼
maps 𝑦 ≥ 0 is a 1-1 and onto,
𝑤 ≤ 1 with 𝑧 = 𝛼 going to 𝑤 = 0.
Cross Ratio: There exists a linear fractional transformation which maps three given distinct
points 𝑧1, 𝑧2, 𝑧3 onto three specified distinct points 𝑤1, 𝑤2, 𝑤3 respectively.
In fact,
(𝑤−𝑤1)(𝑤2−𝑤3)
(𝑤1−𝑤2)(𝑤3−𝑤)
=
(𝑧−𝑧1)(𝑧2−𝑧3)
(𝑧1−𝑧2)(𝑧3−𝑧)
defines such a transformation when none of these points is ∞.

9. Note: In the cross ratio, point at ∞, can be introduced as one the prescribed points. If 𝑤2 =
∞, then replace 𝑤2 by
1
𝑤2
then the cross ratio becomes
(𝑤−𝑤1)
(𝑤−𝑤3)
=
(𝑧−𝑧1)(𝑧2−𝑧3)
(𝑧1−𝑧2)(𝑧3−𝑧)
Problem: Find the linear fractional transformation which maps 𝑧1 = −1, 𝑧2 = 0, 𝑧3 = 1 onto
𝑤1 = −𝑖, 𝑤2 = 1, 𝑤3 = 𝑖 respectively.
Solution: We have
(𝑤−𝑤1)(𝑤2−𝑤3)
(𝑤1−𝑤2)(𝑤3−𝑤)
=
(𝑧−𝑧1)(𝑧2−𝑧3)
(𝑧1−𝑧2)(𝑧3−𝑧)
, which becomes
(𝑤+𝑖)(1−𝑖)
(−𝑖−1)(𝑖−𝑤)
=
(𝑧+1)(0−1)
(−1−0)(1−𝑧)
⇒
(𝑤+𝑖)(1−𝑖)
(1+𝑖)(𝑤−𝑖)
= −
𝑧+1
𝑧−1

10. ⇒
𝑤+𝑖 1−𝑖 2
2(𝑤−𝑖)
= −
𝑧+1
𝑧−1
⇒
𝑖(𝑤 + 𝑖)
(𝑤 − 𝑖)
=
𝑧 + 1
𝑧 − 1
⇒ 𝑤 − 𝑖 𝑧 + 1 = 𝑖 𝑤 + 𝑖 𝑧 − 1
⇒ 𝑤 𝑧 + 1 − 𝑖 𝑧 + 1 = 𝑖𝑤 𝑧 − 1 − 𝑧 − 1
⇒ 𝑤 𝑧 + 1 − 𝑖 𝑧 − 1 = 𝑖 𝑧 + 1 − 𝑧 − 1
⇒ 𝑤 =
𝑖 𝑧 + 1 − (𝑧 − 1)
𝑧 + 1 − 𝑖(𝑧 − 1)
⇒ 𝑤 =
𝑖−𝑧
𝑖+𝑧
, this is the required linear fractional transformation.

11. Problem: Find the linear fractional transformation which maps 𝑧1 = ∞, 𝑧2 = 𝑖, 𝑧3 = 0 onto
𝑤1 = 0, 𝑤2 = 𝑖, 𝑤3 = ∞.
Solution:
(𝑤−𝑤1)(𝑤2−𝑤3)
(𝑤1−𝑤2)(𝑤3−𝑤)
=
(𝑧−𝑧1)(𝑧2−𝑧3)
(𝑧1−𝑧2)(𝑧3−𝑧)
Or,
(𝑤−𝑤1)(
𝑤2
𝑤3
−1)
(1−
𝑤
𝑤3
)(𝑤1−𝑤2)
=
(
𝑧
𝑧1
−1)(𝑧2−𝑧3)
(1−
𝑧2
𝑧1
)(𝑧3−𝑧)
Or,
(𝑤−𝑤1)(0−1)
(0−𝑖)
=
(0−1)(𝑖−0)
(0−𝑧)
Or,
𝑤
𝑖
=
𝑖
𝑧
⇒ 𝑤𝑧 = −1 ⇒ 𝑤 = −
1
𝑧
This is the required linear fractional transformation.

12. Problem 1: Find the linear fractional transformation in which −1, ∞, 𝑖 maps into 𝑖, 1, 1 + 𝑖
respectively.
Problem 2: Find the linear fractional transformation for which 1, 𝑖 are fixed and 0 goes to
− 1.
Problem 3: Find the bilinear transformation which maps 𝑧1, 𝑧2, 𝑧3 onto 𝑤1 = 0, 𝑤2 =
1, 𝑤3 = ∞.
Problem 4: Show that 𝑤 =
2𝑧+3
𝑧−4
maps the circle 𝑥2
+ 𝑦2
− 4𝑥 = 0 onto a straight line 4𝑢 +
3 = 0.
Problem 5: Determine all linear fractional transformation that maps 𝜋+
(upper half plane)
onto the open disc 𝑤 < 1 and boundary 𝐼𝑚 𝑧 = 0 onto the boundary of 𝑤 = 1.

13. Theorem: Let 𝑤1, 𝑤2, 𝑤3, 𝑤4 are the images of four distinct points say 𝑧1, 𝑧2, 𝑧3, 𝑧4 in the 𝑧-
plane under the bilinear transformations 𝑤 =
𝑎𝑧+𝑏
𝑐𝑧+𝑑
, 𝑎𝑑 − 𝑏𝑐 ≠ 0. Then their cross ratios are
equal.
Solution: We have 𝑤𝑖 =
𝑎𝑧𝑖+𝑏
𝑐𝑧𝑖+𝑑
, 𝑖 = 1,2,3,4, 𝑎𝑑 − 𝑏𝑐 ≠ 0
Now 𝑤1 − 𝑤2 =
𝑎𝑧1+𝑏
𝑐𝑧1+𝑑
−
𝑎𝑧2+𝑏
𝑐𝑧2+𝑑
=
𝑎𝑑−𝑏𝑐 𝑧1−𝑧2
𝑐𝑧1+𝑑 𝑐𝑧2+𝑑
And 𝑤3 − 𝑤4 =
𝑎𝑑−𝑏𝑐 𝑧3−𝑧1
𝑐𝑧3+𝑑 𝑐𝑧4+𝑑
Therefore, 𝑤1 − 𝑤2 𝑤3 − 𝑤4 =
𝑎𝑑−𝑏𝑐 2(𝑧1−𝑧2)(𝑧3−𝑧4)
ς𝑖=1
4 (𝑐𝑧𝑖+𝑑)

14. Similarly, 𝑤2 − 𝑤3 𝑤4 − 𝑤1 =
𝑎𝑑−𝑏𝑐 2(𝑧2−𝑧3)(𝑧4−𝑧1)
ς𝑖=1
4 (𝑐𝑧𝑖+𝑑)
Dividing we get,
(𝑤1 − 𝑤2)(𝑤2 − 𝑤4)
(𝑤2 − 𝑤3)(𝑤4 − 𝑤1)
=
(𝑧1 − 𝑧2)(𝑧3 − 𝑧4)
(𝑧2 − 𝑧3)(𝑧4 − 𝑧1)
Hence the theorem.