Module 2: Conformal mapping and Biliniear
transformation
Dr. P. Vijay Kumar
Assistant Professor, SAS-Mathematics
January 31, 2022
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Geometrical Representation of a complex valued function:
To draw a curve of complex variable (x, y) on z-plane we take two
axes i.e., one real axis and the other imaginary axis. A number of
points (x, y) are plotted on z -plane, by taking different value of z
(different values of x and y ). The curve C is drawn by joining the
plotted points. The diagram obtained is called Argand diagram in
z -plane.
But a complex function w = f (z) i.e., (u + iv) = f (x + iy)
involves four variables x, y and u, v A figure of only three dimensions
(x, y, z) is possible in a plane. A figure of four dimensional region
for x, y, u, v is not possible.
So, we choose two complex planes z -plane and w -plane. In the
w -plane we plot the corresponding points w = u + iv. By joining
these points we have a corresponding curve C 0 in w -plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Transformation
For every point (x, y) in the z -plane, the relation w = f (z)
defines a corresponding point (u, v) in the w -plane. This is called
as ”transformation or mapping of z -plane into w -plane”. If a
point z0 maps into the point w0 , w0 is also known as the image of
z0 .
If the point P (x, y) moves along a curve C in z -plane, the point
will move along a corresponding curve C 0 in w -plane, then
we say that a curve C in the z -plane is mapped into the corresponding curve C 0 in the w -plane by the relation w = f (z).
P 0 (u, v)
For example, the transformation f (z) = z +1 transforms the region
bounded by the lines x = 0, y = 0, x = 1&y = 1 in z−plane into
the region bounded by the lines u = 1, u = 2, c = 0&v = 1 in
w−plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Conformal Transformation
Conformal Mapping:
Let C1 and C2 be two curves in z−plane, intersect each other at a
0
0
point P and their corresponding image curves C1 and C2 in
0
w−plane intersect at P . If the angle of intersection of the curves
C1 and C2 at P in z−plane is equal to the angle of intersection of
0
0
0
the curves C1 and C2 at P in w−plane both in magnitude and
sense, then the mapping is said to be conformal.
1
2
3
The point at which f 0 (z) = 0 is called a critical point of the
transformation. Also the points where dw
dz 6= 0 are called
ordinary points.
The points at which the function maps on to itself are called
as fixed points of the transformation. That is the points for
which f (z) = z are called as fixed points.
An analytic function ceases to be conformal at the points
0
where f (z) = 0 i,e the mapping is conformal at the points
0
where f (z) 6= 0.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Conformal Transformation
1
under conformal mapping, w = f (z), the tangent line to a
smooth curve 0 C 0 in the z−plane is rotated by an angle
0
ψ = arg(f (z)) at z0 in w−plane.
2
The angle ψ is called angle of rotation and |f (z)| is called
coefficient of magnification or scale factor at z0 .
3
A mapping is said to be isogonal, if it preserves only in
magnitude but not direction.
0
Examples:
f (z) = z 2 is conformal everywhere except at z = 0.
f (z) = az + b, a, 6= 0 is always conformal.
f (z) = ez is conformal everywhwere.
f (z) =
1
z
is conformal everywhere except at z = 0.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Contd
Examples:
Determine the angle of rotation at the point z =
the mapping w = z 2 . find its scale factor also.
1+i
2
under
Find the angle of rotation produced by the transformation
w = 2z − 1 + 2i at z0 = 1 + 2i
Standard Transformations:
Translation
Rotation
Mafnification
Inversion
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Standard transformations
Translation:
The transformation w = z + c where c = a + ib is called as
translation.
w = z + c u + iv = x + iy + a + ib
u = x + a and v = y + b
x = u − a and y = v − b
On substituting the values of x and y in the equation of the curve
to be transformed, we get the equation of the image in the w -plane.
The point P (x, y) in the z -plane is mapped onto the point P 0 (x +
a, y + b) in the w -plane. Similarly other points of z -plane are
mapped onto w -plane.
Examples:
Find the image of the region bounded by
x = 0, x = 1, y = 0, y = 1 under the mapping w = z + 1 − i
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Standard transformations
Rotation:
The transformation w = zeiφ is called as rotation.
Under the mapping rotation, the point (r, θ) in z−plane is mapped
to (r, θ + φ) in w−plane.
Example:
Let ABCD be a rectangular region with vertices
A(2, 1), B(3, 1), C(3, 3), D(2, 3) in z-plane. Find the image of
this rectangular region in w− plane under the mapping
iπ
w = e 4 z.
Magnification:
The transformation w = bz is called as Magnification, where b is
a real quantity.
Under the mapping Magnification, the point (r, θ) in z−plane is
mapped to (br, θ) in w−plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Standard transformations
If b > 1 then the image is stretched and if b < 1 then the image is
contracted and for b = 1, the mapping will be identity.
Example:
Determine the region in w-plane on the transformation of
rectangular region enclosed by x = 1, y = 1, x = 2 and y = 2
in the z -plane. The transformation is w = 3z.
Magnification and rotation:
The transformation w = bz is called as Magnification and
rotation, where b is a complex constant.
Under the mapping, the image will be magnified and rotated by an
angle in the w−plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Standard transformations
Inversion:
The transformation w =
1
z
is called as inversion.
Under themapping inversion, the point (r, θ) in z−plane is mapped
to 1r , −θ in w−plane.
Examples:
Find the image of x > 0 under the mapping w = iz + i.
Find the image of x > 0&0 < y < 2 under the mapping
w = iz + 1.
Find the image of the region with vertices (0, 0), (2, 0),
(2, 2)&(0, 2) under the transformation w = (1 + i)z + (2 + i).
Show that under the mapping w = z1 , the circle |z − 3| = 5 is
3
5
mapped onto the circle |w + 16
| = 16
.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Standard transformations
Examples:
Find the image of the strip 2 < x < 3 under the mapping
w = z1 .
Find the image of the triangle with vertices at i, 1 + i, 1 − i in
the z-plane, under the transformation w = 3z + 4 − 2i.
Find the image of the triangle with vertices at i, 1 + i, 1 − i in
5πi
the z-plane, under the transformation w = e 3 (z − 2 + 4i)
Find the image of the circle |z| = 2 under the transformation
w = z + 3 + 2i.
Find and plot the image of the regions under the mapping
w = z1 .
(i) x > 1 (ii) y > 0 (iii) 0 < y < 1/2.
Show that the function w = z4 transforms the straight line
x = c in the z− plane into a circle in the w−plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Transformation w = z 2
w = z2
u + iv = (x + iy)2 = x2 − y 2 + 2ixy
Equating real and imaginary parts, we get u = x2 − y 2 ,
(a) Any line parallel to x -axis, i.e., y = c, maps into
v = 2xy
u = x2 − c2 ,
v = 2cx
Eliminating x, we get v 2 = 4c2 u + c2 . . . (1) which is a parabola.
(b) Any line parallel to y -axis, i.e., x = b, maps into a curve
u = b2 − y 2 ,
Eliminating y, we get v 2 = −4b2
parabola.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
v = 2by
u − b2 ,
. . . (2) which is a
Module 2: Conformal mapping and Biliniear transformation
Transformation w = z 2
(c) In polar co-ordinates: z = reiθ , w = Reiφ
w = z2
= r2 e2iθ
Reiφ
Then
R = r2 , φ = 2θ
In z -plane, a circle r = a maps into R = a2 in w -plane. Thus,
circles with centre at the origin map into circles with centre at the
origin.
(d) If θ = 0, φ = 0, i.e., real axis in z -plane maps into real axis in
w -plane
If θ = π2 , φ = π, i.e., the positive imaginary axis in z -plane maps
into negative real axis in w -plane. Thus, the first quadrant in z
-plane 0 ≤ θ ≤ π2 , maps into upper half of w -plane 0 ≤ φ ≤ π.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Transformation w = ez
u + iv = ex+iy = ex (cos y + i sin y)
Equating real and imaginary parts, we have
u = ex cos y, &v = ex sin y
Now w = ez
⇒ Reiφ = ex+iy = ex · eiy
R = ex or x = loge R and y = φ
Note: Under the mapping w = ez
The straight x = c maps onto the circle R = ec .
The straight x = 0 maps onto the circle R = 1.
Region between y = 0, y = π maps onto upper half plane.
Region between y = 0, y = −π maps onto lower half plane.
Region between y = c, y = c + 2π maps onto whole plane.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Transformation w = ez
Examples:
Find the image of the region bounded by the lines
x = 1, x = 2, y = 1, y = 2 under the mapping w = z 2 .
Find the image of the region bounded by the lines
x = 1, y = 1, x + y = 1 under the mapping w = z 2 .
Find the image of the domain to the left of the straight line
x = −3 under the mapping w = z 2 .
Find the image of a ≤ x ≤ b and c ≤ y ≤ d under the
mapping w = ez .
Find the image of x = 1 and 1 ≤ y ≤ 2 under the mapping
w = ez
Find the image of y = 0 and y = π/2 under the mapping
w = ez
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Bilinear Transformation
Bilinear Transformation (or) Linear fractional transformation (or)
Mobious Transformation:
The transformation w = az+b
cz+d , ad − bc 6= 0 is called as bilinear
transformation, where a, b, c, d are constants.
Note: The inverse mapping of a bilinear transformation is given by
z = −dw+b
cw−a , ad − bc 6= 0 which is also a bilinear transformation.
Fixed points (or) invariant points:
The points at which the transformation remains unchanged are
called as fixed points transformation. The fixed points of a
transformation are given by f (z) = z.
The fixed points of a bilinear transformation are given by
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
az+b
cz+d
=z
Module 2: Conformal mapping and Biliniear transformation
Bilinear Transformation
Cross ratio:
Definition. Let z1 , z2 , z3 , z4 be four distinct.points in the extended
complex plane. The cross ratio of these four points denoted by
(zi, z2 , z3 , z4 ) is defined by
 (z −z )(z −z )
1
3
2
4

if none of z1 , z2 , z3 , z4 is ∞

(z1 −z4 )(z2 −z3 )


z1 −z3

if z2 is ∞
 z1 −z4
z2 −z4
(z1 , z2 , z3 , z4 ) =
if z3 is ∞
z1 −z4


z1 −z3

if z4 is ∞

−z3

 zz22 −z
4
if z1 is ∞
z2 −z3
.
Note:
A bilinear transformation preserves cross ratio of four points,
i, e. (w1 , w2 , w3 , w4 ) = (z1 , z2 , z3 , z4 )
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Bilinear Transformation
The bilinear transformation which maps 3 points z1 , z2 , z3 to
the points 3 points w1 , w2 , w3 is given by
(z, z1 , z2 , z3 ) = (w, w1 , w2 , w3 )
i,e,
(z − z2 ) (z1 − z3 )
(w − w2 ) (w1 − w3 )
=
(z − z3 ) (z1 − z2 )
(w − w3 ) (w1 − w2 )
A bilinear transformation maps circles into circles.
In a bilinear transformation w = az+b
cz+d , ad − bc 6= 0, every
point of z -plane is mapped into unique point in w -plane
except z = − dc
Examples:
Find the fixed points of the transformation
z
(a) w = z−2
(b) w = z−1
z+1 .
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Bilinear Transformation
Examples:
Show that a bilinear transformation w = 5−4z
4z−2 maps a unit
circle |z| = 1 into a circle of radius unity and centre − 21 .
Find the image of the imaginary axis under the mapping
w = z−1
z+1 .
Show that the transformation w = i(1−z)
1+z maps the circle
|z| = 1 into the real axis of the w -plane and the interior of
the circle |z| < 1 into the upper half of the w -plane.
Find the bilinear transformation which transforms the points
z = ∞, i, 0 into the points w = 0, i, ∞ respectively.
Find the bilinear transformation which maps the points
z = 0, −i, −1 into w = i, 1, 0.
Find the bilinear transformation which maps the points
z = −i, 0, i into w = −1, i, 1.
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation
Dr. P. Vijay Kumar Assistant Professor, SAS-Mathematics
Module 2: Conformal mapping and Biliniear transformation