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Conformal Mapping
UNIT 9 CONFORMAL MAPPING
Structure
9.1 Introduction Objectives 9.2 Geometrical Representation of Function of z 9.3 The Conformal Mapping 9.4 The Bilinear Transformation 9.5 The Schwarz-Christoffel Transformation 9.6 Summary 9.7 Answers to SAQs
9.1 INTRODUCTION
Earlier we define that the quantity x + i y , in which x and y are independent real variables, is known as a complex variable and is, generally denoted by z. Also we defined that if there is a second complex variable w = u + i v , which has one or more values for each value of z , then w is called a function of z and is written as w = f ( z ). In this unit, we shall study functions mappings or transformations which preserve certain properties. We first of all give geometrical representation of a function of z. We then discuss some Simple Transformations, followed by Bilinear Transformations. The transformations or mappings which preserve angles between any two oriented curves interesting at a point, known as Conformal Mappings, form an important part of this unit. The analytic property of f ( z ), at z = z 0 with f ′ ( z 0 ) ≠ 0 maps the mapping f ( z ) as conformal at z 0. In addition, a harmonic function remains harmonic under as one-to-one conformal mapping. This property of conformal mapping has applications in solving two-dimensional Laplace’s equation with given boundary values. We shall discuss all these in detail in this unit. Finally, we shall study some special mappings which map real line in the z -plane into a polygon into the w -plane.
Objectives
After studying this unit, you should be able to
- determine the images of a given set under the mapping by a function of a complex variable,
- find some special mappings using bilinear transformations,
- define conformal mapping and discuss its properties and applications, and
- map certain given regions to certain other given regions using some special transformations.
9.2 GEOMETRICAL REPRESENTATION OF
FUNCTION OF z
In the conventional method of Cartesian plotting, a complex number z = x + i y is geometrically represented in an Argand diagram by a point ( x , y ) in the xy -plane; a curve is also associated with a real function of one variable y = g ( x ) in xy -plane and we associate a surface in xyz -rectangular system with a real function of two variables z = h ( x , y ). In the complex domain, a relation w = f ( z ), defined for all z , that is
w = u + i v = f ( x + i y )
involves four real variables, namely, two independent variables x and y and the two dependent variables u and v. Hence, a space of four dimensions is required if we are to pole w = f ( z ) in the cartesian fashion. To avoid difficulties inhert in such a device, we make a decided departure from the conventional method of cartesian plotting and choose to proceed as follows :
Complex Variables-II
We plot the independent variables x , y in one-plane, called the z -plane. The dependent variables u , v is plotted in another plane and call this second plane as w -plane (Figure 9.1). y
O (^) z-Plane x
z (x, y)
Figure 9.1 : Argand Diagrams Representing z -plane and w -plane A function w = f ( z ) is now represented not by a focus of point in a space of four dimensions, but by a correspondence between the points of these two cartesian planes. Wherever a point z = x + i y is given in the z -plane, the function w = u + i v = f ( z ) determines one or more values of u + i v and, hence, one ore more point in w -plane. As z ranges over any configuration in the z -plane (called the domain of function w = f ( z )), the corresponding point u + i v describes some configuration in the w -plane. The function w = f ( z ) this defines a Mapping or a transformation of the z -plane into the w -plane and, in turn, is represented geometrically by this mapping. If the point z 0 maps into the point w 0 = f ( z 0 ) in w -plane, then z 0 is known as the image of z 0. Consider for example the relation w = z^2 , i.e.
w = u + i v = z^2 = ( x + i y ) 2 = x^2^ − y^2 + i 2 xy
This given u = x^2 − y^2 , v = 2 xy... (9.1)
These are the equation of transformation from the z -plane to the w -plane. From them, many features of the mapping can easily be infessed. Here the origin in the z -plane maps into the origin in the w -plane; the points A (2, 0), B (2, 2) and C (0, 2) in the z -plane respectively map into A ′ (4, 0), (0, 8) and (− 4, 0). The curves
x^2 − y^2 = c , 2 xy = c ′
in the z -plane map into straight lines u = c , v = c ′ parallel to the coordinate axes in the w -plane. Figure 9.2 shows the lines u = 1, u = 4, u = 2, u = 4 and the corresponding curves in the z -plane corresponding to there. The shaded region in the two figures of Figure 9.2 map into each other. Again, lines parallel to x -axis, i.e. lines with equation x = c , map into curves in w -plane whose parametric equation, from (9.1), are 2 2 u = c 1 (^) − y , v = 2 c y 1
w (u, v)
-Plane u
y
O′ (^) w
y v
v = 4
u = 1 u = 4 C
B
Complex Variables-II (^) v = 2 x (2 x + 1) = 4 x (^2) + 2 x
If we regard these as two simultaneous equations in quantities x and x^2 , we can solve getting (^4 3 4) , 2 2 10 3
u v u v 1 x x
Hence,
2 2 1 4 3 4 3 10
u + v + ⎛ u + v + ⎞ = ⎜ ⎟ ⎝ − ⎠
or 16 u^2 + 24 u v + 9 v^2 + 12 u − 16 v = 4 which is the equation of a parabola. For the converse relation w = z^2 , although w is a single-valued function of z , the converse is not true. In fact, when w is given, z may be either of two square roots of w. Because of this, the mapping from the z -plane to the w -plane covers the latter twice, as Figure 9. suggests. This, of course, is nothing but the graphic representation of the familiar fact
Figure 9.4 : The Two-valued Character of the Mapping z = w 1/ that the angle of complex numbers are doubled when the number are squared. We now take up some example of simple transformations.
y
x
v
u
y
x
v
u
y
x
v
u
y
x
v
u
Z-plane W-plane
Conformal Mapping
Example 9.
Discuss the mapping w = z + z 0 , when z 0 is a complex constant.
Solution
Let z 0 (^) = x 0 (^) + i y 0 (^) , z = x + i y
Then w = u + i v = z + z 0 (^) = ( x + i x ) + ( x 0 (^) + i y 0 (^) ) = ( x + x 0 ) + i ( y + y
Thus the point z = ( x , y ) is translated to the point w = ( , u v ) = ( x + x 0 (^) , y + y 0 ).
The image of any figure in the z -plane is exactly the same figure in the w -plane, except that the location is changed with respect to the origin. The transformation w = z + z 0 preserves both distances and angles and is mere translation of axes. If z 0 = 0, the mapping becomes w = z and is called the Identity mapping. It maps every point to itself.
z = x + iy w = z + z^ o
y
o (^) z
v
o′ (^) x + xo u
y + yo
Figure 9.5 : Transition of Axes
Example 9.
Discuss the mapping w = b z , where (a) b is a real constant. (b) b is a complex constant.
Solution
(a) Let w = b z , where b is a real constant. Then | w | = | b z | = b | z | and arg w = arg ( b z ) = arg ( b z ) = arg b + arg z = arg z since arg b = 0 because b is real. Hence the image in the w -plane of a figure in z -plane is similar and similarly located about the origins, but the figure in w -plane is magnified b times (Figure 9.6). This mapping is called a magnification.
y
o x
θ (^) r
z
v
o′
u
θ br
w
Figure 9.6 : Magnification ( w = b z , b is a Real Constant) (b) Let w = b z , where b is a complex constant. Taking the polar form, let x = r ei^ θ, b = r , ei^ α, and w = p ei^ φ.
Conformal Mapping 2 2
1 u i v x i y u i v (^) u v
so that 2 2 , 2
u v x y u v u v
Hence the equation
a ( x^2 + y^2 )+ bx + cy + d =... (9.2)
transforms of 2 2
( 2 2 )^2 2 2 2 2
u v bu cv a d u v u v u v
⎝ +^ ⎠ +^ +
i.e. (^) ( u^2 + v^2 )+ bu − cv + a = 0
This shows that a circle in the z -plane, in general, transforms to another circle in the w-plane. The exceptional case occurs when d = 0. A circle through the origin transforms to a straight line. Again, when a = 0, Eq. (9.2) represents a straight line and we see that straight lines in the z -plane, not passing through the origin, transform into circles in w -plane, while straight lines through the origin transform into straight lines.
Example 9.
Discuss the mapping (^) w = ez.
Solution
Let z = x + i y , w = u + i v so that u + i v = ex^ +^ i y^ = ex^ cos y + i ex^ sin y
i.e. u = ex cos y and v = e x sin y
Also if w = ρ ei φ, then f = ex^ and φ = y
Therefore lines x = c transform to circles f = ec^ and the lines y = c ′ transform to radial lines φ = c ′. In particular, the y -axis ( x = 0) transforms into the unit circle f = 1. The x -axis ( y = 0) transforms into positive u -axis (φ = 0) and the line y = π transforms into the negative u -axis (φ = π). Thus the region between the lines y = 0 and y = π maps into the upper half of w -plane. Similarly, the region between the lines y = 0 and y = − π maps into the lower half of w -plane. In fact ez^ is a periodic function with period 2π i. Therefore any horizontal strip between the lines y = c and y = c + 2π will transform into the whole w -plane.
Let us use the knowledge gained to attempt the following SAQs.
SAQ 1
(a) What is the image of y = x – 2 under the mapping w = z^2? (b) What is the image of circle x^2 + y^2 = a^2 under the mapping w = z^2?
SAQ 2
What is the image of | z | < | and | z | > | under the mapping
w z
Complex Variables-II SAQ 3
Discuss the mapping w = z^3. Plot the image of the line u – 1. What is the equation of the image of the line x = 1?
In the section, we saw that various functions of a complex variable map the xy -plane onto the uv -plane. In the next section we purpose to investigate in more general terms the character of transmission for which the mapping function is analytic. This is done through the study of conformal mapping.
w = u ( , x y ) + i v ( , x y )
9.3 THE CONFORMAL MAPPING
It is important to know when the transformation equations can be solved (at least theoretically) for x and y as single-valued functions of u and v , that is, when the transformation has a single-valued inverse. This is achieved through the following theorem. Theorem 1 Statement If f ( z ) is analytic, the function w = f ( z ) will have a single valued inverse in the neighbourhood of any point where the derivative of the mapping function is different from zero. Proof If z = x + i y and w = u + i v = f ( z ) = f ( x + i y ), then the condition for function f ( z ) to have a single-valued inverse, as established in most text on advanced calculus (Refer to pp. 363-366, Advanced Calculus , by R. C. Buck, 3rd^ Edition, McGraw-Hill, New Your, 1978) is simply that the Jacobian determinant of the transformation is different from zero, that is,
u v u v x^ y J x y v v x y
⎛ ⎞ ∂^ ∂
⎝ ⎠ ∂^ ∂
Since w = u + i v = f ( z ) is assumed to be analytic, therefore, u and v must satisfy the Cauchy-Reimann equations, i.e.
and
u v u v x y y x
Hence substituting into the Jacobian determinant, we have
,^2
u v u v x^ y u v J x y v^ u x x y
⎛ ⎞ ∂^ ∂ ⎛ ∂ ⎞ ⎛ ∂ ⎞
⎝ ⎠ ∂^ ∂^ ⎝^ ∂^ ⎠^ ⎝^ ∂ ⎠
x
u v^2 i x x
=| f ′( ) | z^2
Complex Variables-II (^) Since infinitesimal lengths are magnified by the factor | f ′ ( z )|, it follows that
infinitesimal areas are magnified by the factor | f ′ ( ) | z^2 , that is, by
u v J x y
Similarly, we conclude from (9.4), that, in general, the difference between the angles of an infinitesimal segment and its image is independent of the direction of the segment and depends only on the point from which the segment is drawn. In particular, two infinitesimal segments forming an angle will both be rotated in the same direction by the same amount. Hence, the measure of the angle between them will, in general, be left invariant by the transformation.
However, when (^) f ′ (^) ( ) z = 0 , then arg (^) f ′ (^) ( ) z is undefined and we cannot assert that angles are preserved. To investigate this case, suppose that f ′ ( ) z has n -fold zero at z = z 0. Then f ′( )^ z must contain the factor ( z – z 0 ) n , and hence we cannot write
1 ( ) ( 1) ( 0 ) ( 2) ( 0 )... f ′^ z = n + a z − z n^ + n + b z − z n + +
where a , b are complex coefficients, of no common to us, and the factors ( n +1) ( n + 2)... have been inserted for convenience in integrating f ′ ( ) z to obtain f ( z ).
Integrating, we get
1 2 ( ) ( 0 ) ( 0 ) ( 0 ). f z = f z + a z − z n^ +^ + b z − z n + +..
Let ( z − z 0 = Δ z , f ( ) z − f ( z 0 )= Δ w and dividing by , we get a ( z Δ) n +^1
( ) n^1 .
w b z a w + a
As Δ z → 0, the right hand side approaches 1. Therefore,
1 0 0
Lt ( ) Lt [ ( ) n ] 1 0 z z
arg w arg a z + arg Δ → Δ →
or, to an arbitrary degree of approximation,
arg ( Δ w ) = arg + ( n + 1) arg (Δ z... (9.5)
Now let Δ z 1 and Δ z 2 be two infinitesimal segments which make an angle θ with each other, and let Δ w 1 and Δ w 2 be their images. From Eq. (9.5), we have
arg Δ w 1 (^) = arg a + ( n + 1) arg Δ z 1
arg Δ w 2 (^) = arg a + ( n + 1) arg Δ z 2
Hence subtracting, we get
arg w 2 (^) arg w 1 (^) ( n 1) ( arg z 2 (^) arg z 1 ) ( n 1) Δ − Δ = + Δ − Δ = + θ
Thus we have established that.
Theorem 3
In the mapping defined by an analytic function w = f ( z ), angle are, in general, preserved in magnitude and in sence. The only exception to this occurs when the vertex of the angle is an n -fold zero of f ′^ ( ) z , in which case the measure of the angle is altered by the factor n + 1.
Proof^ Conformal Mapping
Example considered in Section 9.2, namely, the transformation w = z^2 is an excellent illustration of the behaviour described in Theorem 3 above. The mapping function is everywhere analytic, and as Figures 9.2 and 9. suggest, angle measures are in general preserved. However, the derivative
w = f ( ) z = z^2
f ′ ( ) z = 2 z has a simple zero at z = 0 and as Figure 9.4 indicates, the angle with vertex at origin are not preserved but are doubled. We now give some definitions.
Definition 1
A transformation which preserves the magnitude of the angles is said to be ISOGONAL.
Definition 2
A transformation which preserves the sense as well as the magnitude of the angles is said to be CONFORMAL.
Alternatively, A transformation w = f (z) is said to be CONFORMAL to z 0 if a small triangle in the neighbourhood of z 0 transforms into a similar triangle in the w-plane.
This is not true, however, for large configurations, which may bear little or no resemblance to their images.
We now prove some theorems on such transformations.
Theorem 4
A transformation w = f ( z ) is conformal if f ( z ) is analytic and f ′^ ( ) z is not zero at z = z 0.
y
o (^) x
(z (^) o)
(z 1 )
(z 2 ) P (^2) P (^1)
P (^) o
z-plane
v
o′ (^) u
(wo)
(w 1 )
(w 2 ) P′ (^2) P′ 1
P′o
w-plane
Figure 9.
Proof
Let the triangle p 0 p 1 p 2 in the z -plane have its vertices at z 0 z 1 z 2 where z 1 and z 2 are points in the neighbourhood of z 0. Let the corresponding points P 0 ′ P P 1 ′ 2 ′ in w -plane be w 0 , w 1 , w 2. Then upto a first approximation.
1 0 2 0 1 0 2 0
w w w w w 0 z z z z z
′ Δ^ −^ −
f z... (9.6)
provided f ′^ ( ) z is analytic at z 0.
It f ′^ ( ) z ≠ 0 , this gives
2 0 2 0 1 0 1 0
w w
... (9.7)
z z w w z z
then when is φ ( x , y ) transformed into a function of u and v by a conformal^ Conformal Mapping transformation, it will satisfy the equation 2 2
u^2^ v^2
∂ φ ∂ φ
everywhere except possibly at the images of the points where the derivatives of the mapping function is equal to zero.
Proof
Let define a conformal transformation by means of which φ ( x , y ) is transformed into a function of u and v. Then
w = u ( , x y ) + i v ( , x y )
.. and
u v u v x u x v x y u y v y
∂φ ∂φ ∂ ∂φ ∂ ∂φ ∂φ ∂ ∂φ ∂ = + = + ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂ ∂
A second differentiation of each of and x y
∂φ ∂φ ∂ ∂
yields
2 2 2 2 2.^2 .^.
u u v u x v u x v u x x
2
x x u v^2
∂ φ ∂φ ∂ ⎛^ ∂ φ ∂ ∂ φ ∂ ⎞∂ ∂ ∂ = + (^) ⎜⎜ + (^) ⎟⎟ + ∂ ∂^ ∂ (^) ⎝ ∂ ∂^ ∂^ ∂^ ∂^ ⎠ ∂ ∂^ ∂ x
2 2 2
u v u v x (^) v x x
⎛ (^) ∂ φ ∂ ∂ φ ∂ ⎞∂
- (^) ⎜⎜ + ⎟⎟ ⎝ ∂^ ∂^ ∂^ ∂ ∂^ ⎠∂
v
2 2 2 2 2 2 2
u u v u y y u^ y u y^ v^ u^ y^ y
∂ φ ∂φ ∂ ⎛^ ∂ φ ∂ ∂ φ ∂ ⎞∂ ∂ ∂ = + (^) ⎜⎜ + (^) ⎟⎟ + ∂ ∂^ ∂ (^) ⎝ ∂ ∂^ ∂^ ∂^ ∂^ ⎠∂^ ∂
2 2
v v (^) ∂ y
2 2 2
u v u v y (^) v y y
⎛ (^) ∂ φ ∂ ∂ φ ∂ ⎞∂
- (^) ⎜⎜ + ⎟⎟ ⎝ ∂^ ∂^ ∂^ ∂ ∂^ ⎠∂
v
Adding, we get
2 2 2 2 2 2 2 2 2 2 2 2
u u u u x y u^ x y u x y
∂ φ ∂ φ ∂φ ⎛^ ∂ ∂ ⎞^ ∂ φ ⎡^ ⎛ ∂ (^) ⎞ ⎛ ∂ ⎞ ⎤
- = (^) ⎜⎜ + (^) ⎟⎟ + ⎢ (^) ⎜ ⎟ +⎜ ⎟ ⎥ ∂ ∂ ∂^ ⎝ ∂ ∂ (^) ⎠ ∂ ⎢⎝⎣ ∂ ⎠ (^) ⎝ ∂^ ⎠ ⎥⎦
2 2 2 2 2 2 (^2 2 2 )
u v u v v v v v u v x x y y v (^) x y v x y
∂ φ ⎛ ∂ ∂ ∂ ∂ ⎞ ∂φ ⎡^ ∂ ∂ ⎤ ∂ φ ⎡^ ⎛ ∂ (^) ⎞ ⎛∂ ⎞ ⎤
- (^) ⎜ + (^) ⎟ + (^) ⎢ + (^) ⎥+ ⎢ (^) ⎜ ⎟ +⎜ ⎥ ∂ ∂ (^) ⎝ ∂ ∂ ∂ ∂ (^) ⎠ ∂ (^) ⎢⎣ ∂ ∂ (^) ⎥⎦ ∂ ⎢⎝⎣ ∂ (^) ⎠ ⎝∂ ⎟⎠ ⎥⎦
Since w = u + i v is analytic, by hypothesis, therefore u and v themselves satisfy Laplace’s equation. Hence, the first and fourth group of terms on the right vanish identically. Moreover, u and v satisfy the Cauchy-Riemann equations, hence the 3rd group of term also vanishes identically. By using the Cauchy-Riemann equations again in the remaining terms, we get
2 2 2 2 2 2 2 2 2 2 2
u v v u x y u x^ x^ v x^ x
∂ φ ∂ φ ∂ φ ⎡^ ⎛ ∂ (^) ⎞ ⎛ ∂ (^) ⎞ ⎤^ ∂ φ ⎡⎛ ∂ (^) ⎞ ⎛ ∂ ⎞
- = ⎢ (^) ⎜ ⎟ + (^) ⎜ − (^) ⎟ ⎥ + ⎢⎜ ⎟ +⎜ ⎟ ∂ ∂ ∂ (^) ⎢⎣⎝ ∂^ ⎠ ⎝ ∂^ ⎠ (^) ⎥⎦ ∂ ⎢⎣⎝ ∂^ ⎠ ⎝ ∂ ⎠
2 2 2 2 2 2 2 2 2 |^ ( ) | 2 2
u v f z u v x^ x u v
⎡ (^) ∂ φ ∂ φ ⎤ ⎡^ ⎛ ∂ ⎞ ⎛ ∂ ⎞ ⎤ ⎡∂ φ ∂ φ = (^) ⎢ + (^) ⎥ ⎢ (^) ⎜ ⎟ + (^) ⎜ ⎟ ⎥= ′ ⎢ + ⎢⎣ ∂^ ∂^ ⎥⎦ ⎢⎣⎝ ∂^ ⎠^ ⎝^ ∂ ⎠ ⎥⎦ ⎢⎣∂^ ∂
Thus, at any point where the transformation is conformal, that is, where f ′ ( ) z ≠ 0 ,
Complex Variables-II 2 2 2 2
x^2 y^2^ 0 implies^ u^2 v^20
∂ φ (^) + ∂ φ (^) = ∂ φ (^) + ∂ φ= ∂ ∂ ∂ ∂
as asserted.
Suppose now that it is required to solve Laplace’s equation subject to certain boundary conditions, within a region R. Unless R is of very simple shape, the solution of the problem, directly, will usually be exceedingly difficult. However, it may be possible to find a conformal transformation which will convert region R into some simple region R^1 , such as a circle or a half-plane, in which Laplace’s equation can be solved, subject, of course, to transformed boundary conditions. If this is the case, the resulting solution, when carried back to R by the inverse transformation, will be the required solution of the original problem.
Let us take up some examples on conformal mapping.
Example 9.
For the conformal transformation w = z^2 , show that
(a) coefficient of magnification at z = 2 + i is 2 5.
(b) the angle of relation at z = 2 + i as tan– 1^ (0.5). (c) the circle | z – 1 | = 1 transforms into the cardiod , where
ρ = 2 (1 + cos φ) w = ρ e^ i φ in the w -plane.
Solution (a) For conformal transformation w = f ( z ), the coefficient of a magnification at z = z 0 is given by | f ′^ ( z 0 ) |.
Here f ( ) z = z^2 , ∴ f ′( ) z = 2 z
∴ Coefficient of magnification at z = 2 + i
= | 2 (2 + i )| = | 4 =2 | i
= 4 2 + 22 = 2 5. Ans. I
(b) For conformal transformation w = f ( z ), the angle of solution at z = z 0 is arg { f ′( z 0 )}.
Here f ( ) z = z^2 , ∴ f ′( ) z = 2 z
∴ Angle of relation at z = 2 + i
(2 ) (4 2 ) tan 1 2 tan 1 (0.5) 4
arg f ′^ + i = arg + i = −^ ⎛⎜^ ⎞⎟= − ⎝ ⎠
Ans. II
(c) Let z = r ei θ^ , w = ρ ei φ
Then for given transformation w = z^2 , we get
ρ ei φ^ = r^2 e^2 i φ= ρ = r^2 and φ = 2 θ
Now the circle | z – 1 | = 1 is transformed to
| z − 1 | = 1 ⇒ | r ei θ− 1 | ⇒ | r cos θ − 1 | + i ( r sin θ) | = 1
⇒ r^2 cos 2 θ − 2 r cos θ + r^2 sin^2 θ + 1 = 1
Complex Variables-II The line y = x and x = 2 intersect at 4
π and their point of intersection
is 2 + 2 i.
Also f ′( ) |^ z (^) 2 + 2 i = 2 z | 2 + 2 i = 4 + 4 i ≠ 0
∴ By conformality of w = z^2 , we conclude that the images u = 0 and
v^2 = − 16 ( u − 4)of the lines y = x and x = 2 also intersect at an angle 4
π in
w -plane.
You may now try the following SAQs.
SAQ 4
Find the images of lines x = 0 and y = 0 under the transformation w = z^2 and find the angle of their intersection.
SAQ 5
If w = f ( z ) is analytic at z = z 0 and f ′^ ( z 0 ) ≠ 0 , then f has an inverse in some neighbourhood of z 0. Show that this inverse is an analytic function. Further show that the inverse mapping is also conformal.
SAQ 6
Determiner the critical points of the transformation w = f ( z ), where
(a) w = z^2 + 2 z
(b) w =cos z
SAQ 7
Find the magnification coefficient of the conformal transformation w = z^3 at the point 1 + i.
The simplest class of conformal transformations, yet one of the most important, is the class of bilinear transformations, which we take up in the next section.
9.4 THE BILINEAR TRANSFORMATION
Let us first give the definition of bilinear transformation.
Definition^ Conformal Mapping
The transformation
a z b w ad b c z d
c ≠... (9.9)
where a , b , c , d are real or complex constants, is called a “Bilinar or Linear Fractional or Mobins Transformations.”
Note 1
This transformation was first studied by German geometer A. F. Mobins (1790-1868) and is also named for him.
Note 2
The condition ad – bc ≠ 0 is essential, because if ad – bc = 0, then
a c b d
= and the
transformation becomes
a z cz b b b b w cz d d c d z d
+ ⎜^ ⎟
= = ⎝^ ⎠ =
which is a constant.
Note 3
If a = b = c = d = 0, the expression
cz b w cz d
becomes meaningless.
The bilinear transformation is frequently denoted by w = T ( z ). The inverse
transformation T – 1^ of T is obtained by solving
az b w cz d
for z. it is easily seen that
z wd^ b T^1 ( ) cw a
= −^ + =^ −^ w −
Thus T – 1^ ( w ) is also a bilinear mapping.
We may obverse that transformations defined in Examples 9.1, 9.2 and 9.3 of this unit represent bilinear transformations. For example, if we take a = 1, b = z 0 , c = 0, d = 0, we get the bilinear transformation given by (9.9) becomes w = z + z 0 of Example 9.1.
Except of
d z c
to every value of z , these corresponds first one finite value of w , i.e.
the transformation (9.9) gives a one-to-one mapping of the z -plane into the w -plane. The
point
d z c
maps into the ‘point at infinity.’
The fixed points of bilinear transformation (i.e. points which do not change their position) are given by
( ) ( )
az b z cz d
i.e. cz^2^ − ( a − d ) z − b = 0
This is quadratic equation and has at the most two roots. Hence a bilinear transformation has at most two fixed points.
Conformal Mapping Thus
z p l z q r
, where p and q are inverse points, is an alternatively form of the
equation of the circle | z – a | = r.
This circle is called the Circle of Appllonius.
Conversely any equation (^1
z p k k z q
) (^) represent a circle with p and q as inverse
points.
To prove this, the equation gives
2 − (^) = 2 or | − | (^2) = 2 | − | −
z p k z p k z p z q
2
or ( z − p ) ( z − q ) = k^2 ( z − q ) ( z − q )
or ( z − p ) ( z − p ) = k^2 ( z − q ) ( z − q )
or (^) | z |^2 − ( z p + z p ) + | p |^2 = k^2 [| z |^2 − ( z q + z q ) +| q | ]^2
or | z |^2 − 2 Re ( p z ) + | p |^2 = k^2 [| z |^2 − 2 Re ( z q ) +| q | ]^2
or
2 2 2 2 2 2
2 Re [( ) ] [| | | | | | 0 1 1
p k q z p k q z k k
2
or
2 2 2 2 2 2 2 2 2
2 2
p k q p k q p k p | z k k k
2 2 2 2 2 2 2 2
| | (1 ) [| | |
p k q k p k q | k
2 2 2 2 2 2
p k q p k q k pp k qq ) k
since zz =| z |^2
2 2 2 2
k p q k
Hence
2 2 2
p k q k p q z k k
This equation is of the form | z – a | = r and hence it represents a circle with centre 2
1 2
p k q a k
and radius (^2)
k q p k
Also
2 2
k q p p a k
and (^2) 1
q p q a k
Since 2
p a k q a
, which is real and positive; it follows that p and q are on the same side
of centre a.
Complex Variables-II Also
2 2 2 2 2
| | | | (^ )
p a q a k^ q^ p r k
Hence p and q are inverse points of the circle.
If k = 1, then | z – p | = | z – q |. This means that z is equidistant from p and q , and therefore lies on the perpendicular bisector giving p and q.
Remark
The inverse point of the centre of the circle is the point at infinity.
We now prove an interesting theorem.
Theorem 9. Under a general bilinear transformation circles and straight lines are mapped into circles and straight lines.
Remark
By this theorem, a circle is transformed only to a circle or a straight line by a bilinear transformation. Similarly a straight line is transformed only to a circle or a straight line.
We given two proofs of this theorem.
Proof 1
It is convenient to investigate the general bilinear transformation by considering first the three special cases :
(a) w = z + z 0
(b) w = b z
(c)
w = z
In case (a), w is found by adding a constant z 0 to each z. This transformation is just a translation in the direction defined by arg z 0 through a distance equal to | z 0 | and this rigid motion necessarily transforms circles and straight lines into straight lines.
In case (b), it is found by rotating each z through a fixed angle equal to arg b and then multiplying its length by the factor | b |. The equation straight circle in z -plane is
a 1 (^) ( x^2^ + y^2 ) + b x 1 + c y 1 + d 1 (^) =0, a b c d 1 1 1 1 real and b 1^2 + c 12 (^) ≥ 4 a 1 d 1... (9.10)
We know that , ,^2 2 2
z z z z x y x y i
z z.
Then Eq. (9.10), becomes
1 1 1 1 1 2 2 1 0
b i c b i c a z z z z d i
or, renaming the coefficients,
( A + A ) z z + BZ + BZ + ( D + D ) = 0... (9.11)
where now A , B and D are arbitrary complex number and the condition b 1^2 + c 12 (^) ≥ 4 a 1 d 1 becomes BB ≥ ( A + A ) ( D + D )− this condition ensure that radius of the circle is real.
