Chapter 8. Conformal Mapping – Engineering Mathematics, Volume III

8

Conformal Mapping

8.1 Introduction

Let y = f(x) be a function in the xy-plane. The pair (x,y) of real numbers x and y can be plotted with the help of two co-ordinate axes Ox and Oy. We thus obtain a graph of the function. It gives a visual representation of the function. This helps in giving a better understanding of the properties of the function.

Now, consider a complex-valued function w = f(z) where z = (x,y) and w = (u, v). The graphical representation as in the case of real-valued functions is not possible in this case. However, if we have two planes, one for each pair of numbers, and the mapping assigns to each point z = (x,y) in the z-plane a corresponding point w = (u, v) in the w-plane, this modified version of the geometrical representation provides a better understanding of the problems. It has several practical applications in potential theory, theory of elasticity, hydrodynamics and other applied fields.

8.1.1 Mapping f: zw

Consider the complex function w = f(z) where z = x + iy or (x,y) and w = u + iv or (u, v).

To represent this function graphically, we need two planes. One plane is the z-plane in which we plot the values of z = (x, y) and the other is the w-plane in which we plot the corresponding values of w = (u, v). Thus a given function f assigns to each z in its domain of definition D a corresponding point w = f(z) in the w-plane (Fig. 8.1).

We say that f defines a mapping of D onto a range of values of w = f(z). If z is any point in D the point w = f(z) is called the image of z under f. For the points of a curve C in D, the image points form the image curve C* of C in the w-plane.

Figure 8.1

Instead of using the phrase “the mapping by a function w = f(z)” we briefly say “the mapping (or transformation) w = f(z)”.

8.1.2 Conformal Mapping

A mapping under which the angles between any two oriented curves are preserved both in magnitude and sense is called a conformal mapping. If a mapping preserves angles in magnitude only and not in sense then it is called an isogonal mapping.

Example 8.1

w = f(z) = . Here |w| = |z| = |z| but arg w = arg() = −arg(z).

Figure 8.2 shows the angle of intersection α of the curves C1 and C2 defined as the angle between the oriented tangents to the curves at the point P(z0) of intersection of C1 and C2.

We know that a curve in the xy-plane has parametric representation.

For example, the equation of the circle x2 + y2 = 1 can be represented by x = cos t, y = sin t (0 ≤ t ≤ 2π) and the equation of the parabola y2 = 4ax can be represented by x = at2, y = 2at (−∞ < t < ∞).

Figure 8.2 Curves C1 and C2 in the z-plane and their images C*1 and C*2 in the w-plane under a conformal mapping

In the complex plane, a curve can be represented by z(t) = x(t) + iy(t) (atb). We assume that C is smooth. That is, at each point z(t) of C, ż(t) exists and is not zero. The positive sense of C is that in which t is increasing, i.e.,

Conformal mapping preserves orthogonality of curves. That is, if ϕ(x,y) = C1, ψ(x,y) = C2 are mutually orthogonal in the z-plane their image curves in the w-plane given by ϕ*(u, v) = C*1 and ψ* (u,v) = C*2 are also mutually orthogonal.

8.2 Conformal Mapping: Conditions for Conformality

The conditions under which the transformation w = f (z) is conformal are stated and proved in the following theorem.

Theorem 8.1    If f(z) is analytic and f′(z) ≠ 0 in a region R of the z-plane, then the mapping w = f(z) is conformal at all points of R.

Proof:    Let P(z) be a point in the region R of the z-plane and P*(w) the corresponding point in the corresponding region G of the w-plane. Suppose that as P moves on a curve C and P′ moves on the corresponding curve Cδ in G. Let Q(z + δz) and Q′(z + δz) be the neighbouring points on C and C′, respectively, so that = δz and = δw.

Then δz is a complex number whose modulus r is the length PQ and amplitude θ is the angle which PQ makes with the x-axis. δz = re and δw = r′ e′, where r′ is the modulus and θ′ the amplitude of δw of the tangent at Pon C, and the tangent at P′ on C makes angles α and α′, respectively, with the real axes x and u. Then as δz → 0, θ α and θ′ → α′. Hence,

Figure 8.3

f ′ (z) ≠ 0 ⇒ f ′ (z) = ρe

so that ρ = |f ′ (z)| and ϕ = amp f ′ (z)

If C1 and C1′ be another pair of curves through P and P′, respectively, in the z- and w-planes and the tangents at these points make angles β and β′ with the real axis then ϕ = β′ − β ⇒ a′ − α = β′ − β or β − α = β′ − α′ = γ.

Figure 8.4

The angle between the curves before and after the mapping is preserved in magnitude and sense. Hence the mapping by the analytic function w = f(z) is conformal at each point where f′(z) ≠ 0.

Note:

  1. A point at which f′(z) = 0 is called a critical point of the transformation.
  2. We have obtained It follows that under the conformal transformation w = f(z), the lengths of arc through P are magnified in the ratio ρ : 1 where ρ = |f′(z)|.
  3. A harmonic function remains harmonic under a conformal transformation.

Example 8.2

For the conformal transformation w = z2 show that

  1. the coefficient of magnification at is and
  2. the angle of rotation at

Solution    w = f(z) = z2f′(z) = 2z and

  1. The coefficient of magnification at
  2. The angle of rotation at

8.3 Conformal Mapping by Elementary Functions

8.3.1 General Linear Transformation

Consider the mapping

    w = f (z) = az + b         (8.3)

    (a ≠ 0, b are arbitrary complex constants.)

It maps conformally the extended complex z-plane onto the extended w-plane since Eq. (8.3) is analytic and f′(z) = a ≠ 0 for any z. If a = 0, then Eq. (8.3) reduces to a constant function.

Special cases of linear transformation
Identity transformation

The mapping

 

w = z          (8.4)

obtained from Eq. (8.3) for a = 1, b = 0 maps a point z onto itself.

Translation transformation

The mapping

 

w = z + b          (8.5)

obtained from Eq. (8.3) for a = 1 translates or shifts z through a distance |b| in the direction of b.

More specifically, if z = x + iy, b = p + iq, w = u + iv, then the transformation becomes

 

   u + iv

=

(x + iy) + (p + iq)

 

=

(x + p) + i(y + q)

so that u = x + p and v = y + q.

Thus the transformation is a mere translation of the axes and it preserves the shape and size of the figure in the z-plane.

For example, the rectangle ABCD in the z-plane is transformed to the rectangle A′ B′C′D′ in the w-plane under the transformation w = z + (2 + i).

Figure 8.5

Rotation transformation

The mapping

 

w = ez          (8.6)

obtained from Eq. (8.3) for a = e and b = 0 rotates a radius vector of a point z through a scalar angle α counter-clockwise if α > 0 and clockwise if α > 0.

For example, the mapping z or w = iz rotates the square ABCD in the z-plane into the square ABCD′, which is rotated through an angle

Figure 8.6

Stretching, scaling or magnification

The mapping

 

w = az          (8.7)

obtained from Eq. (8.3) for b = 0 stretches (magnifies) the radius vector by a factor a if a is real and >1 and contracts if a is real and 0 < a < 1.

8.3.2 Inversion Transformation

Consider the mapping

Put z = re and w = Re then the transformation becomes

Thus under the transformation a point P(r, θ) in the z-plane is mapped into the point

Imagine that the w-plane is superposed on to the z-plane. If P is (r, θ) and P1 is then so that P1 is the r OP inverse of P w.r.t the unit circle |z| = 1.

Note:

The inverse of a point P w.r.t the circle having its centre at 0 and radius r is defined as the point Q on OP such that OP · OQ = r2.

Figure 8.7

The reflection P′ of P in the real axis represents . Thus the transformation is an inversion of z w.r.t the unit circle |z| = 1 followed by reflection of the inverse in the real axis.

Clearly, the transformation maps the interior of the unit circle |z| = 1 into the exterior of the unit circle |w| = 1 and the exterior of |z| = 1 into the interior of |w| = 1.

However, the origin z = 0 is mapped to the point w = ∞, called the point at infinity.

We may discuss the transformation by changing it to cartesian coordinates. For this, we put

 

z = x + iy and w = u + iv

Also,

Now consider the second degree general equation with equal coefficients for x2 and y2 viz.

 

a(x2 + y2) + bx + cy + d = 0          (8.9)

where a, b, c and d are real constants. If a ≠ 0 the above equation represents a circle and if a = 0 it represents a straight line, substituting for x and y in terms of u and v we get,

on multiplication throughout by (u2 + v2).

The following cases arise

  1. a ≠ 0, d ≠ 0. Equations (8.9) and (8.10) show that circles not passing through the origin are mapped into circles not passing through the origin.
  2. a ≠ 0, d = 0. Equations (8.9) and (8.10) show that circles passing through the origin are mapped onto straight lines not passing through the origin.
  3. a = 0, d ≠ 0. Equations (8.9) and (8.10) show that straight lines passing through the origin are mapped onto circles passing through the origin.
  4. a = 0, d = 0. Equations (8.9) and (8.10) show that straight lines passing through the origin are mapped onto straight lines passing through the origin.

Thus, the mapping transforms circles and straight lines into circles and straight lines.

Since a straight line can be considered as a circle with infinite radius, we conclude that under the mapping circles are mapped into circles in the w-plane.

8.4 Some Special Transformations

8.4.1 Transformation w = z2

Polar coordinates

Changing the equation into polar coordinates by putting z = re and w = Re

      we have w = z2

      ⇒ Re = (re)2 = r2e2iθ

      ⇒ R(cosϕ + isinϕ) = r2(cos2θ + isin2θ)

      ⇒ R = r2 and ϕ = 2θ

Hence circles r = r0 (constant) in the z-plane are transformed into circles R = r02 (constant) in the w-plane.

In particular, the region is mapped onto the region Mapping w = z2; lines |z| = constant, arg z = constant and their images are in the w-plane.

Cartesian coordinates

Figure 8.8

Changing the equation w = z2 by putting z = x + iy and w = u + iv where x, y, u, v ∈ ℝ we have w = z2u + iv = (x + iy)2 = x2y + i(2xy). On equating the real and imaginary parts, we get u = x2y2; v = 2xy.

Vertical lines

Consider a vertical line x = c (constant) in the z-plane. It is mapped onto the curve

      u = c2y2, v = 2cy

        ⇒y2 = c2u, 4c2y2 = v2.

Eliminating y between these equations, we get

 

v2 = 4c2 (c2u) = −4c2(uc2)

Figure 8.9

These are parabolas with the negative u-axis as their axis and their openings are to the left.

The region between two vertical lines x = c1 and x = c2 (c2 > c1) is mapped into the region between the parabolas and in the w-plane whose vertices are at and , respectively, and their common axis is the negative u-axis.

Horizontal lines

Consider a horizontal line y = d(constant) in the z-plane. It is mapped onto the curve v2 = 4d2(u+d2), a parabola with its axis along Ou and with its opening to the right.

Figure 8.10

        u = x2d2,   v = 2dx

            ⇒x2 = u + d2,   4d2x2 = v2

Eliminating x between these equations, we get

 

v2 = 4d2(u + d2)

The vertex of the parabola is at (−d2, 0), and the axis is the positive u-axis.

The region between two horizontal lines y = d1 and y = d2(d2 > d1 ≥ 1) is mapped into the region between the parabolas v2 = and v2 = in the w-plane.

8.4.2 Transformation w = zn (n ∈ ℕ)

Differentiating w = zn w.r.t z, we have

So, the mapping is conformal at all points z except at the origin z = 0.

Solving w = zn for z, we have which has n roots. For every non-zero w, this shows that there are n pre-images.

Converting the equation w = zn into polars by putting z = reiθ and w = Reiϕ, we get R = rn and ϕ = .

A circle |z| = r of radius r and centre at the origin in the z-plane is mapped into a circle of radius rn and centre at the origin of the w-plane.

The sectorial region {z = reiθ, α ≤ θ ≤ β} is mapped into the sectorial region {w = Reiϕ, nα ≤ ϕ ≤ } in the w-plane.

In particular, the sector is mapped into the upper half-plane v ≥ 0 in the w-plane given by

{w = Re, 0 ≤ ϕπ} and the region

is mapped into the region

{w = Re, 0 ≤ ϕ ≤ 2π}, which is the entire w-plane.

Figure 8.11 Mapping w = zn

8.4.3 Transformation w = ez

Consider the mapping w = ez. Put z=x+iy, x, y ∈ ℝ

 

w

=

Re = R(cos ϕ + i sin ϕ)

ez

=

exeiy = ex(cosy+i sin y) and

w

=

ez ⇒ |w| = R = ex arg w = ϕ = y.

Figure 8.12 Mapping of rectangle into sectorial region in the w-plane under w = ez

Vertical lines in the z-plane

Straight lines x = c (constant), i.e., vertical lines are mapped onto the circles |w| = R = ec (constant).

If c > 0 then the circles in the w-plane have radius greater than 1, if c < 0 then the circles have radius less than 1 and if c = 0 (y-axis) then the circle is of unit radius.

Figure 8.13 Vertical lines mapped into concentric circles under w = ez

Horizontal lines in the z-plane

Straight lines y = d (constant), i.e., horizontal lines are mapped onto the radial lines ϕ = d (constant) in the w-plane.

Figure 8.14 Horizontal lines mapped into rays under the mapping w = ez

If d > 0 then the ray ϕ = d in the w-plane makes a positively oriented angle and if d < 0 it makes a negatively oriented angle with the real u-axis.

If d = 0 then we have the real axis (x-axis) in the z-plane, and it is mapped onto the real u-axis in the w-plane. The rectangular region ABCD bounded by the straight lines x = c1 and c2, and y = d1 and d2 (c1 < c2, d1 < d2) in the z-plane is mapped onto the sectorial region ABCD′ given by

 

in the w-plane.

8.4.4 Transformation w = sin z

Consider the mapping

 

w = sin z          (8.11)

put z = x + iy, w = u + iv where x,y, u, vR

w = u + iv = sin z = sin (x + iy)

   = sin x coshy + i cos x sinh y       (8.12)

Equating the real and imaginary parts

 

u = sin x coshy v = cos x sinh y       (8.13)

Since these are periodic functions with period 2π therefore (Eq. 8.11) is periodic with period 2π. Hence the mapping is not one-to-one if we consider the whole of the xy-plane.

So, we restrict ourselves to the infinite vertical strip D defined by

Since f′(z) = cos z = 0 at , the mapping is not conformal at these points.

Figure 8.15

Horizontal line y = 0 (the x-axis)

For y = 0, sinh y = 0; cosh y = 1.

Equation (8.13)u = sin x, v = 0.

Since −1 ≤ sin x ≤1, u varies from −1 to 1 ∀ x.

Thus the segment of the x-axis is mapped onto the segment −1 ≤ u ≤ 1 of the u-axis.

Vertical lines

For , we have sin x = ±1, cos x = 0.

Equations (8.13)u = ± cosh y, v = 0.

Since cosh y ≥ 1 and − cosh y ≤ −1 ∀y, the vertical line maps onto u ≥ −1, v = 0 and the vertical line maps onto u ≥ 1, v = 0.

Horizontal line segments y = ±k (k > 0),

Let y = k > 0.

Equation (8.13)u = (cosh k) sin x, v = (sinh k) cos x or = sin x, = cos x, where a = cosh k, b = sinh k.

Squaring and adding, we get

Figure 8.16 Mapping by w = sin z

This equation represents an ellipse. Thus the line segment is mapped onto the upper half portion of the ellipse showing in Fig. 8.16.

Similarly, the line segment y = −k (k > 0), maps onto the lower half of the ellipse shown in Fig. 8.16.

The focii of the ellipse are at w = ±1, which are independent of k.

As k varies, we obtain a family of confocal ellipses.

The rectangular region D defined by k < y < k is thus mapped onto the interior of the ellipse (8.14).

Note that the image of the boundary consists of the ellipse (8.14) and the two segments of the x-axis as shown in Fig. 8.16 with k = 1.

Image of the rectangle k1 < y < k2, −π < x < π

Similarly, the rectangle −π < x < π, k1 < y < k2 maps onto the elliptic ring with a cut along the negative v-axis.

Figure 8.17

Vertical lines x = ±c (c > 0)

Let x = c (constant). Equations (8.13)

 

u = A cosh y, v = B sinh y

where

Squaring and subtracting

The equation represents a hyperbola. Thus the vertical lines x = ±c are mapped onto the confocal hyperbolas (with focii at w = ±1) on the right and left half-planes, respectively, and these curves cut the ellipses orthogonally.

Figure 8.18

8.4.5 Transformation w = cos z

Since cos z = sin the cosine transformation is a composition of the following:

First    : Translation

Second : Mapping w = sin z1, which we have discussed above.

We can also discuss independently as in the case of w sin z.

8.4.6 Transformation w = sinh z

w = sinh z = −i sin(iz)

The transformation is a composition of the following:

  1. Counter-clockwise rotation through    z1 = iz.
  2. Sine mapping         z2 = sin z1.
  3. Clockwise rotation through    w = −iz2.

8.4.7 Transformation w = cosh z

w = cosh z = cos(iz)

The transformation is a composition of the following:

  1. Rotation through in the counter-clockwise direction    z1 = iz
  2. Cosine mapping          w = cos z1

8.4.8 Logarithm

The natural (Naperian) logarithm of z = x + iy is denoted by ln z (sometimes also by log z or log ez). It is defined as the inverse of the exponential function, i.e., w = ln zew = z for z ≠ 0.

(Since ew ≠ 0, z = 0 is impossible)

 

Put   z

=

re, w = u + iv

       ew

=

eu+iv = re

      |ew|

=

eu ∵ |eiv | = 1

arg ew

=

v = 0

      eu

=

ru = ln r

 

 

(real logarithm defined for r > 0).

Hence w = u + iv = ln z is given by

         ln z = ln r + r = |z| > 0,θ = arg z

Since the argument of z is determined only up to integer multiples of 2π, the complex natural logarithm ln z (z ≠ 0) is infinitely many-valued.

The value of ln z corresponding to the principal value Arg z is denoted by Ln z and is called the principal value of ln z.

Thus

 

ln z = ln |z| + iAeg z (z ≠ 0)

For a given z ≠ 0, Arg z is unique. This implies that ln z is single-valued and is therefore a function in the usual sense. Since the other values of arg z differ by integer multiples of 2π the other values of ln z are given by

 

ln z = Ln z ± 2n2π; n = 1, 2, 3….

They all have the same real part and their imaginary parts differ by integer multiples of 2π. If z is positive real then Arg z > 0 and ln z becomes identical with the real natural logarithm of calculus. If z is negative real (natural log of calculus is not defined) then Arg z = π and Ln z = ln |z| + πi.

Conformal mapping by ln z – the principle of inverse mapping

The mapping by the inverse z = f–1(w) of w = f(z) is obtained by interchanging the roles of the z-plane and the w-plane in the mapping by w = f(z).

Example 8.3

Mapping by w = z2 and by its inverse, w = z2 is restricted to the right half-plane x > 0 and maps this half-plane onto the w-plane without the negative half u ≤ 0 of the real u-axis because at the origin, the angles are doubled, so that rays ϕ = Arg z = constant are mapped onto rays ϕ =Arg w = 2ϕ. Hence its inverse w = √z(with √z > 0 for z = x > 0) maps the z-plane without the negative half of the x-axis onto the right half-plane u = Re w > 0.

Natural logarithm

ln z is the inverse relation of the exponential function w = ez. The latter maps a fundamental strip onto the w-plane without z = 0 because ez ≠ 0 for every z. Hence by the principle of inverse mapping, the principal value w = Ln z maps the z-plane with z = 0 omitted and cut along the real axis, when θ = Im(ln z) jumps by 2π onto the horizontal strip −π < v < π of the plane. Since the mapping w =Ln z + 2πi differs from w = Ln z by the translation 2πi (vertically upward), this mapping maps the z-plane (cut as before and 0 omitted) onto the strip −π < v ≤ 3π. Similarly for each of the many mappings

 

w = ln z = ln z ± 2nπi (n = 0, 1, 2,…)

The corresponding horizontal strips of width 2π images of the z-plane under these mappings together cover the whole of the w-plane without overlapping.

8.4.9 Transformation (Joukowski1 Airfoil)

If in the mapping we put z = re = r(cos θ + i sin θ) and w = Re = R(cos ϕ + i sin ϕ) we have w = u + iv = r(cos θ + i sin θ) + – (cos θi sin θ).

Equating the real and imaginary parts, we get

               or u = a cos θ and v = b sin θ

where a = r + and b = r

Consider the circles |z| = r = constant in the z-plane. Since b must not be equal to zero, we have r ≠ 1. These non-unit circles are mapped onto ellipses

The unit circle |z| = r = 1 is mapped onto the line segment −2 ≤ u ≤ 2 of the u-axis. (real axis), since a = = 1 + 1 = 2 and b = 0 yield u = 2 cos θ, v = 0.

Figure 8.19 Mapping w = z + 1/z (Joukowski Airfoil)

Differentiating w.r.t z, we get

Hence = 0 ⇒ z = ±1. These are points at

which the mapping is not conformal.

Solving the equation for z, we get .

So, z is a double-valued function of w. For r = a (constant) as mentioned above, this equation represents an ellipse with focii at

which are independent of r.

The Joukowski’s function maps the family of concentric circles |z| = a of the z-plane onto the family of confocal ellipses in the w-plane with focii at w = ±2.

Case (i) |z| = 1 (Unit circle) As r → 1, v = 0, u = 2 cos θ since | cos θ| ≤ 1, −2 ≤ u ≤ 2.

This shows that the unit circle |z| = 1 in the z-plane is mapped into a degenerated ellipse which flattens to the line segment v = 0, −2 ≤ u ≤ 2 on the real u-axis traversed twice.

Case (ii) As r → 0 the ellipse is transformed into a circle of infinitely large radius. To find the image of the ray arg z = θ0 eliminate the parameter n we get

which represents hyperbola with focii at w = ±2. Thus Joukowski’s function defines the transformation of an orthogonal system of polar coordinates in the z-plane into an orthogonal curvilinear system of coordinates, whose coordinate lines are the confocal families of ellipses and hyperbolas in the w-plane.

Example 8.4

Show that the transformation maps the hyperbola x2y2 = 1 to the lemniscate.

[JNTU 1985, 1998]

Figure 8.20

Solution    The given curve in the z-plane is the hyperbola

 

x2y2 = 1          (1)

The given transformation is

Let z = x + iy, w = u + iv.

In polars z = re, Re w = By Eq. (2) we get

Now let us put Eq. (1) in polar form by substituting

 

x = r cos θ, y = r sin θ

Thus x2y2 = 1 gives

This is the equation in polar coordinates for the lemniscate.

Example 8.5

Show that the transformation maps a circle to a circle or to a straight line in the w-plane if the former goes through the origin.

 

[JNTU 2000]

Solution    Let z = x + iy, w = u + iv where x, y, u, v ∈ ℝ. Then

Equating the real and imaginary parts, we get from Eq. (1)

Squaring and adding

The general equation of a circle in the z-plane is

 

A(x2 + y2) + Bx + Cy + D = 0,          (5)

 

[where A, B,C,D are constants]

Equation (6) represents a circle in the w-plane. Thus a circle (5) in the z-plane is transformed into a circle (6) in the w-plane. If D = 0 then it represents a straight line whose equation is

 

BuCv + A = 0.

Example 8.6

Determine the region in the w-plane corresponding to the triangular region bounded by the lines x = 0, y = 0 and x + y = 1 in the z-plane mapped by the transformation .

Solution    The transformation is

The line x = 0 maps into

The line y = 0 maps into

The line x + y = 1 maps into

Figure 8.21 The region in the z-plane and the w-plane mapped by the transformation

Example 8.7

Find the image of the infinite ship under the transformation and Show the regions graphically.

Solution    The transformation is

Now,

Also,

Figure 8.22

∴ The infinite strip is transformed into the region between the two circles:

u2 + (v + 2)2 = 22; centre c2 = (0, −2) and radius r2 = 2 units.

u2 + (v + 1)2 = 12; centre c1 = (0, −1) and radius r1 = 1 units.

Example 8.8

Plot the image of the region 1 < |z| < 2 under the transformation w = 2iz + 1.

 

[JNTU 1995]

Solution    Let z = x + iy, w = u + iv where x, y, u, v ∈ ℝ.

Now, w = 2iz + 1 ⇒ u + iv = 2i(x + iy) + 1

Figure 8.23

Equating the real and imaginary parts, we get u = − 2y + 1 and v = 2x, i.e., and

The given region 1 < |z| < 2 is the annulus bounded by the circles C1 : |z| = 1 and C2 : |z| = 2.

 

|z| =1 ⇒ x2 + y2 = 1          (1)

 

         ⇒(u − 1)2 + v2 = 12          (2)

The unit circle C1 : x2 + y2 = 1 in the z-plane is mapped to the circle C*1 : (u – 1)2 + v2 = 4 with its centre at u = 1, v = 0 and radius = 2 units.

Again

 

|z| = 2 ⇒ x2 + y2 = 4          (3)

 

                     ⇒ (u − 1)2 + v2 = 4          (4)

The circle C2 : x2 + y2 = 22 with radius = 2 units in the z-plane is mapped to the circle C*2 : (u − 1)2 + v2 = 42 with centre at u = 1, v = 0 and radius = 4 units.

Thus, the annular region 1 < |z| < 2 is mapped to the annular region 2 < |w − 1| < 4.

Example 8.9

Find the image of the triangle with vertices i, 1 + i, 1 − i in the z-plane under the transformation w = 3z + 4 − 2i.

 

[JNTU 2003]

Solution    Let z = x + iy and w = u + iv where x, y u and v ∈ℝ.

 

u + iv = 3(x + iy) + 4 − 2i

Equating the real and imaginary parts, we have

 

u = 3x + 4,          (1)

 

v = 3y − 2          (2)

The coordinates of the vertices of the triangle are

                  z1 = i = (0,1),

                  z2 = 1 + i = (1,1),

                  z3 = 1 − i = (1, −1)

Figure 8.24

The images of the points in the w-plane are

 

(0,1) → (4,1)
(1,1) → (7,1)
(1, −1) → (7, −5)

ΔABC in the z-plane is mapped to the triangle ΔPQR in the w-plane.

Example 8.10

Show that the transformation transforms r = constant in the z-plane into a family of ellipses in the w-plane.

Solution    Let

 

z = re = r(cos θ + i sin θ) and w = u + iv         (1)

This gives

Let r = k(constant) ≠ 1 so that

This equation represents a family of ellipses with semi-major axis , semi-minor axis

If the eccentricity is e, then

The focii of the ellipses are at

The focii of the ellipses are at

Example 8.11

Under the transformation find the image of |z – 2i| = 2.

[JNTU 2001]

Solution    Let

 

z = x + iy,w = u + iv where x, y, u, v ∈ ℝ

Equating the real and imaginary parts, we get

Squaring and adding,

     

 

 

     

by Eq. (3)

(4)

The given curve in the z-plane is the

 

|z − 2i| = 2 ⇒ x2 + (y − 2)2 = 4
     ⇒ x2 + y2 − 4y = 0

which is a circle with centre at(0, 2) and radius = 2.

Substituting for x and y, we have

This is a horizontal straight line in the w-plane.

Example 8.12

If w = , find the image of |z | < 1. [JNTU 2002].

Solution

Now,

Putting w = u + iv, we get

 

(u − 1)2 + v2 < (u + 1)2 + v2
u2 + v2 − 2u + 1 < u2 + v2 + 2u + 1
u > 0

Thus the image of the interior of the circle |z| = 1 is mapped onto the right half of the w-plane.

Example 8.13

If w = cosh z, find the mapping of the region axb onto the w-plane.

Solution    Let z = x + iy and w = u + iv where x, y, u, v ∈ R

 

w

=

cosh zu + iv = cosh(u + iv)

(1)

 

=

cosh x cosh y + i sinh x sinh y

 

u

=

cosh x cosh y, v = sinh x sinh y.

(2)

Vertical line x = c

Figure 8.25

Equation (2) can be written as

where a = cosh c, b = sinh c.

Squaring and adding

Thus the vertical line x = c is mapped onto ellipse (3).

Hence the given vertical strip axb in the z-plane is mapped onto the annular region between two coaxial confocal ellipses (3) with c = a and c = b and focii at ±1.

Example 8.14

Show that the transformation maps the circle x2 + y2 – 4x = 0 into the straight line 4u + 3 = 0.

 

[JNTU 2003]

Solution    Let

The given transformation is

z(w − 2) = 4w + 3, or .

The given equation of the circle is

 

x2 + y2 − 4x = 0 ⇒ z − 2(z + ) = 0

This represents a straight line in the w-plane.

Example 8.15

Find the image of the rectangle R; − π < x < π, under the transformation w = sin z.

 

[JNTU 2003S].

Solution    Let z = x + iy and w = u + iv, then

 

w

=

u + iv = sin(x + iy)

 

=

sin x cosh y + i cos x sinh y

u

=

sin x cosh y, v = cos x sinh y

The horizontal line FED : y = 1 maps onto the outer ellipse

Figure 8.26

and the horizontal line ABC : y = maps onto the inner ellipse

Also, the vertical line segment CD given by maps onto C*D* given by u = 0 and – sinh – sinh 1.

The vertical line segment A*F* given by x = –π, 1 maps onto A*F* given by u = 0 and – sinh

Thus both C *D* and A* f * map onto the same line in the w-plane given by the cut A*F* or C*D* along the imaginary axis Ov.

8.5 Bilinear or Mobius2 or Linear Fractional Transformations

Introduction

Conformal mappings are useful in solving boundary value problems. This is done by mapping complicated regions conformally onto standard regions like circular disks, half-planes, strips. We have already considered many conformal transformations and studied their properties. Here, we discuss an important class of transformations called linear fractional transformations.

Linear fractional transformation (LFT) or Mobius or Bilinear transformation

A transformation of the form

where a, b, c and d are complex or real constants is called a linear fractional transformation (LFT) or Mobius transformation. It is sometimes called simply as linear transformation or linear map.

Cross-multiplying, we can write Eq. 8.16 in the form

 

Awz + Bz + Cw + D = 0          (8.17)

which is linear in z as well as in w. So, it is also called a bilinear transformation.

Differentiating Eq. 8.16 w.r.t z, we have

If (adbc) = 0 then = 0. So, the transformation is not conformal at any z if ad – bc = 0.

Hence we assume that the constants a, b, c, d are such that ad – bc ≠ 0 in which case we have conformal transformation for all z.

The transformation at Eq. 8.16 is very interesting. First of all, for different choices of the constants a, b, c, d we get the following special cases of LFTs:

a =1, c = 0, d = 1

 

w = z + b (translation)          (8.19)

b =0, c = 0, d = 1 ⇒ w = az                   (8.20)

  Simply rotation if |a| = |e| = 1

  Magnification or dimination if

 

a = |a|e(|a| = 1)          (8.21)

c =0, d = 1 w = az + b (linear transformation)          (8.22)

 

a =0, b = 1, c = 1, d = 0

So, the general LFT is composed of all the above types of transformation, which we have already discussed above.

Also, corresponding to each z, Eq. 8.16 gives a unique value of w except when cz + d = 0.

  1. c ≠ 0. In this case, there does not correspond a number w to the value – of z.

    To get over this situation, we attach the number ‘∞’ which we call ‘the point at infinity’. The complex plane then is called the extended complex plane.

  2. c = 0. In this case, a ≠ 0 and d ≠ 0 in view of the fact that ad – bc ≠ 0. In this case, w = ∞ is the image of the point z = ∞.

Inverse mapping of w =

This is the inverse of the transformation (8.16)

  1. c ≠ 0. cw – a = 0 This point corresponds to z = ∞.
  2. c = 0. In view of adbc ≠ 0 we get a ≠ 0 and d ≠ 0. Therefore, w = ∞ corresponds to z = ∞.

Hence the mapping is bijective.

8.6 Fixed Points of the Transformation

A point z is called a fixed point of the transformation if it is such that f (z) = z (not an identity mapping). So, fixed points of Eq. 8.16 are obtained from

This being a quadratic in z has, in general, two values for z. Thus an LFT has two fixed points if it is not an identity.

If an LFT has more than two fixed points, then it must be an identity mapping w = z.

An LFT maps the totality of circles and straight lines in the z-plane into the totality of circles and straight lines in the w-plane.

Though there are four constants in Eq. 8.16, on dividing three of them by the fourth non-zero constant we see that the effective constants are three and the mapping can be determined uniquely through three conditions.

Suppose for example d ≠ 0. Then w = f (z) = can be written as w =

Hence bilinear transformation consists of three independent constants

So, three conditions are required to determine the transformation. That is to say, three distinct points z1, z2, z3 in the z-plane can be mapped into three distinct points w1, w2, w3 in the w-plane.

Cross-ratio of four points

Let z1, z2, z3 and z4 be four distinct points in the extended complex plane.

Then the cross-ratio of these points denoted by(z1, z2, z3, z4) is defined as

if none of z1, z2, z3, z4 are ∞.

Theorem 8.2    Prove that any bilinear transformation is the product of basic transformation namely translation, rotation, magnification or contraction and inversion.

Proof:    Let

Eq. (1) be the given bilinear transformation.

Case (i)    Let c = 0 so that ad – bc = ad ≠ 0 this implies that d ≠ 0.

From Eq. (8.26) we get

Then f1 and f2 are elementary transformations and

Case (ii)    Let c ≠ 0. Equation (8.26) gives

Thus f (z) is the product of f1,f2,f3 and f4.

Hence the theorem.

Corollary    Each elementary transformation as seen earlier, transforms circles and straight lines into circles and straight lines. It may be noted that a bilinear transformation maps circles and straight lines onto circles and straight lines.

If we observe that straight lines are linearising cases of circles with infinite radius we may conclude that a bilinear transformation maps circles into circles.

Theorem 8.3    Any bilinear transformation preserves the cross-ratio.

Proof:    Let

be the given bilinear transformation.

Let z1, z2, z3 and z4 be four distinct points of the z-plane and w1, w2, w3 and w4 be then corresponding images in the w-plane under the transformation (8.27). Suppose all these points are different from ∞. Now we have to prove that (z1, z2, z3, z4) = (w1, w2, w3, w4).

From Eq. (8.27), we have

where k1 =

Similarly, w2w4 = k2(z2z4)

where

 

∴ (w1w3)(w2w4)

=

k1k2(z1z3)(z2z4)

 

=

k (z1z3)(z2z4)

where

Similarly, we can derive the result

Hence the theorem. The proof when one of zi or wi is ∞ is similar.

Note:

Four distinct points z¡ (i = 1, 2, 3, 4) are collinear or concyclic ⇔ (z1, z2, z3, z4) is real. Also any bilinear transformation preserves a cross-ratio. Hence it follows that circles and straight lines are mapped onto circles and straight lines. This gives another proof for the corollary of the above theorem.

Theorem 8.4    The transformation (z1, z2, z3, z4) = (w1, w2, w3, w4) is a bilinear transformation that maps three distinct points z1, z2 and z3 onto three specified distinct points w1, w2 and w3, respectively.

Proof:    Let

 

(z1, z2, z3, z4) = (w1, w2, w3, w4)          (8.28)

        z = z1w = w1   and

        z = z3w = w3   from Eq. (8.29).

Also z = z2

⇒ (ww1)(w2w3)(z2z3)(z1z2)

= (ww3)(w1w2)(z2z1)(z2z3)

⇒ (ww1)(w2w3) = −(ww3)(w1w2).

Cancelling out non-zero factors (z2z3), (z1z2), we get

w(w2w3 + w1w2)

        = w2w1w3w2 + w1w2w1w3

w(w1w3) = w2(w1w3)

w = w2    ∵ w1 6= w3

Hence, the required transformation is given by Eq.(8.28).

Example 8.16

Show that the relation transforms the circle |z| = 1 into a circle of unit radius in the w plane.

 

[JNTU 2003]

Solution    Let z = x + iy and w = uiv so that = xiy, = uiv.

The given transformation is

The given circle is

This is the circle with centre at and radius = 1 in the w-plane.

Example 8.17

Find the bilinear transformation that maps the points ∞, i, 0 into the points 0, i, ∞ respectively.

 

[JNTU 2003S]

Solution    By the invariance of cross-ratios, we can write

Here z1 = ∞, z2 = i, z3 = 0; w1 = 0, w2 = i, w3 = 0

Replacing and by 1, according to the convention, we get as the required bilinear transformation.

Example 8.18

Find the bilinear transformation which maps the points z = 1, i, –1 into the points w = 0, 1, ∞.

Solution    By the invariance of cross ratios, we have

Example 8.19

Show that the transformation w = maps the circle x2 + y2 − 4x = 0 into the straight line 4u+3 = 0.

 

[JNTU 2003]

Solution    Let z = x + iy, w = u + iv so that = xiy, = uiv.

Now x2 + y2 = (x + iy)(xiy) = z and

The given transformation is,

The given equation of the circle is

which is the equation of a straight line in the w-plane.

Example 8.20

Determine the bilinear transformation that maps the points (1-2i, 2+i, 2+3i) into the points (2+i, 1+3i, 4).

 

[JNTU 2003, 2004(Set 4), 2004S (Set 2)]

Solution    Let the required bilinear transformation be

Here

Equation (1) now becomes

(w − 4)(z − 1 + 2i) = (−3w + 6 + 3i)(z − 2 − 3i),

on cross-multiplication

w(z − 1 + 2i + 3z − 6 − 9i) = (6z − 12 − 18i) + (4z − 4 + 8i) + (3iz + 9 − 6i)

Example 8.21

Find the LFT that maps the points –i, 0, i onto the points – 1, i, 1, respectively. [JNTU 1996, 2003S]

Solution    Let the transformation be

Here

z1

=

i, z2 = 0, z3 = i

 

w1

=

−1, w2 = i, w3 = 1

Substituting in Eq. (1), we get

(by componendo–dividendo).

Example 8.22

Find the bilinear transformation that maps the points 2, i, –2 onto 1, i, – 1, respectively. [JNTU 1985S]

Solution    Let the transformation be

Here

z1

=

2, z2 = i, z3 = −2

 

w1

=

1, w2 = i, w3 = −1

Equation (1) becomes

By componendo–dividendo,

Example 8.23

Find the LFT which maps –1, 0, 1 onto 0,”1, ∞,

 

[JNTU 1994S]

Solution    Let the transformation be

Here

z1

=

−1, z2 = 0, z3 = 1

 

w1

=

0,w2 = −1,w3 = ∞

Substituting in Eq. (1), we get

Example 8.24

Find the LFT which maps 0, 1, ∞ onto –1, –i, 1, respectively.

Solution    Let the transformation be

Here

z1

=

0, z2 = 1, z3 = ∞

 

w1

=

−1, w2 = −i, w3 = 1

Substituting in Eq. (1), we get

Example 8.25

Find the bilinear transformation which maps the points z = 1, i, –1, respectively, onto w = i, 0, –x.

 

[JNTU 1999S, 1997S]

Solution    Let

be the required transformation.

When z = 1,w = i

When z = i,w = 0

When z = −1,w = −i

From Eqs. (2) and (4), we get b = ic, a = id

But b = −aI from Eq. (3)

Example 8.26

Show that w = maps the circle x2 + y2 − 4x = 0 onto a straight line 4u + 3 = 0.

 

[JNTU 2003]

Solution    The given transformation is

The given equation of the circle in the z-plane is

 

x2 + y2 − 4x = 0 ⇒ z − 2(z + ) = 0         (2)

(Under the transformation (1), Eq. (2) becomes)

which is a straight line in the w-plane.

Example 8.27

 

If w = find the image of |z| < 1.

[JNTU 2002]

Solution    The given transformation is

Now

                        where w = u + iv

                ⇒ 4u > 0 or u > 0

∴ |z| < 1 is mapped into u > 0, the upper half-plane of the w-plane.

Example 8.28

Find the bilinear transformation which maps the points ∞, i, 0 in the z-plane into –1, – i, 1 in the w-plane.

 

[JNTU 2003S(2)]

Solution    Let the transformation be

Figure 8.27

Here

Substituting these values in Eq. (1)

replacing

by componendo–dividendo.

Example 8.29

Find the bilinear transformation that maps the points (0, 1, ∞) in the z-plane onto the points (–1, –2, –i) in the w-plane.

 

[JNTU 2006(4)]

Solution    Let the transformation be

Here

Substituting these values in Eq. (1)

replacing by 1

which is the required transformation.

Example 8.30

Find the fixed points of the transformation

Solution    The fixed points of the transformation

w = f (z) are such that f (z) = z.

  1. Putting w = f (z) = z, we obtain

    Solving, we get

  2. Putting w = f(z) = z, we obtain
  3. Putting w = f (z) = z, we obtain

Example 8.31

Find the bilinear transformation which maps vertices (1+i, –i, 2–i) of the triangle T of the z-plane into the points (0, 1, i) of the w-plane.

 

[JNTU 2005S(2)]

Solution    Let the transformation be

Here

z1

=

1 + i,    z2 = −i, z3 = 2 − i

 

w1

=

0,    w2 = 1, w3 = i

Substituting these values in Eq. (1), we get

Example 8.32

Determine Mobius transformation having α, β as fixed points and mapping 0 to –1.

Solution    The quadratic equation with α, β as roots is

 

z2 − (α + β)z + αβ = 0

For any γ ∈ ℂ, we have

 

z2 (α + β)z + γzγz + αβ = 0

Adding and subtracting γz

Hence the Mobius transformation with fixed points α,β is given by w =

Example 8.33

Find the linear fractional transformation with fixed points 1 and i and mapping 0 to – 1.

Solution    Take z = 0; w = –1; α = 1, β = i in the transformation w = we have

∴ The required transformation is

EXERCISE 8.1
  1. Find the bilinear transformation that maps zL, z2, z3 onto w1, w2, w3, respectively.
    1. z =–1, 0, 1; w = 0,i,3i.

      Ans: w = –3i(z + 1)/(z – 3)

       

    2. z = 0, –i, –1; w = i, 1, 0.

      Ans: w = –i(z + 1)/(z – 1)

       

    3. z = ∞, i, 0; w = 0, i, ∞.

      Ans:

       

    4. z = 1, 0, –1; w = i, 1, ∞.

      Ans: w = [(–1 + 2i)z + 1]/(z + 1)

  2. Find the fixed points of the transformation.
    1. w = (z – 1)/(z + 1).

      Ans: w = ±i

       

    2. w = (2z – 5)/(z + 4).

      Ans: – 1 ± 2i

  3. Find the bilinear transformation whose fixed
    1. i,–i

      Ans:

    2. 1,–1.

      Ans:

  4. Determine the bilinear transformation which maps the points 0, 1, ∞ in the z-plane into –5, –1, 3, respectively, in the w-plane. Find the invariant points.

    Ans: w = (3z – 5)/(z + 1), z = 1 ± 2i

     

  5. Determine the region into which the rectangular region in the z-plane bounded by x = 0,y = 0, x = 2 and y = 1 is mapped under the transformation w = z + (1 – 2i).

    Ans: Rectangular bounded by u = 1, u = 3, v = –1, v = –2

     

  6. Find the bilinear transformation which maps the points –1, 0, 1 into the points 0, – 1, ∞, respectively.

    Ans:

  7. Find the bilinear transformation which maps the points (–1, 0, 1) into the points (0, i, 3i).

     

    [JNTU 2005]

    Ans: