8
Conformal Mapping
8.1 Introduction
Let y = f(x) be a function in the xyplane. The pair (x,y) of real numbers x and y can be plotted with the help of two coordinate 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 complexvalued function w = f(z) where z = (x,y) and w = (u, v). The graphical representation as in the case of realvalued 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 zplane a corresponding point w = (u, v) in the wplane, 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: z → w
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 zplane in which we plot the values of z = (x, y) and the other is the wplane 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 wplane (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 wplane.
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 C_{1} and C_{2} defined as the angle between the oriented tangents to the curves at the point P(z_{0}) of intersection of C_{1} and C_{2}.
We know that a curve in the xyplane has parametric representation.
For example, the equation of the circle x^{2} + y^{2} = 1 can be represented by x = cos t, y = sin t (0 ≤ t ≤ 2π) and the equation of the parabola y^{2} = 4ax can be represented by x = at^{2}, y = 2at (−∞ < t < ∞).
Figure 8.2 Curves C_{1} and C_{2} in the zplane and their images C^{*}_{1} and C^{*}_{2} in the wplane under a conformal mapping
In the complex plane, a curve can be represented by z(t) = x(t) + iy(t) (a ≤ t ≤ b). 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) = C_{1}, ψ(x,y) = C_{2} are mutually orthogonal in the zplane their image curves in the wplane 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 zplane, 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 zplane and P^{*}(w) the corresponding point in the corresponding region G of the wplane. 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 xaxis. δz = re^{iθ} and δw = r′ e^{iδ}′, 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^{iϕ}
so that ρ = f ′ (z) and ϕ = amp f ′ (z)
If C_{1} and C_{1}′ be another pair of curves through P and P′, respectively, in the z and wplanes 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:
 A point at which f′(z) = 0 is called a critical point of the transformation.
 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).
 A harmonic function remains harmonic under a conformal transformation.
Example 8.2
For the conformal transformation w = z^{2} show that
 the coefficient of magnification at is and
 the angle of rotation at
Solution w = f(z) = z^{2} ⇒ f′(z) = 2z and
 The coefficient of magnification at
 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 zplane onto the extended wplane 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
obtained from Eq. (8.3) for a = 1, b = 0 maps a point z onto itself.
Translation transformation
The mapping
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 zplane.
For example, the rectangle ABCD in the zplane is transformed to the rectangle A′ B′C′D′ in the wplane under the transformation w = z + (2 + i).
Figure 8.5
Rotation transformation
The mapping
obtained from Eq. (8.3) for a = e^{iα} and b = 0 rotates a radius vector of a point z through a scalar angle α counterclockwise if α > 0 and clockwise if α > 0.
For example, the mapping z or w = iz rotates the square ABCD in the zplane into the square A′B′C′D′, which is rotated through an angle
Figure 8.6
Stretching, scaling or magnification
The mapping
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^{iθ} and w = Re^{iϕ} then the transformation becomes
Thus under the transformation a point P(r, θ) in the zplane is mapped into the point
Imagine that the wplane is superposed on to the zplane. If P is (r, θ) and P_{1} is then so that P_{1} 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 = r^{2}.
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
Also,
Now consider the second degree general equation with equal coefficients for x^{2} and y^{2} viz.
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 (u^{2} + v^{2}).
The following cases arise
 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.
 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.
 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.
 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 wplane.
8.4 Some Special Transformations
8.4.1 Transformation w = z^{2}
Polar coordinates
Changing the equation into polar coordinates by putting z = re^{iθ} and w = Re^{iϕ}
we have w = z^{2}
⇒ Re^{iϕ} = (re^{iθ})^{2} = r^{2}e^{2iθ}
⇒ R(cosϕ + isinϕ) = r^{2}(cos2θ + isin2θ)
⇒ R = r^{2} and ϕ = 2θ
Hence circles r = r_{0} (constant) in the zplane are transformed into circles R = r_{0}^{2} (constant) in the wplane.
In particular, the region is mapped onto the region Mapping w = z^{2}; lines z = constant, arg z = constant and their images are in the wplane.
Cartesian coordinates
Figure 8.8
Changing the equation w = z^{2} by putting z = x + iy and w = u + iv where x, y, u, v ∈ ℝ we have w = z^{2} ⇒ u + iv = (x + iy)^{2} = x^{2} − y + i(2xy). On equating the real and imaginary parts, we get u = x^{2} − y^{2}; v = 2xy.
Vertical lines
Consider a vertical line x = c (constant) in the zplane. It is mapped onto the curve
u = c^{2} − y^{2}, v = 2cy
⇒y^{2} = c^{2} − u, 4c^{2}y^{2} = v^{2}.
Eliminating y between these equations, we get
Figure 8.9
These are parabolas with the negative uaxis as their axis and their openings are to the left.
The region between two vertical lines x = c_{1} and x = c_{2} (c_{2} > c_{1}) is mapped into the region between the parabolas and in the wplane whose vertices are at and , respectively, and their common axis is the negative uaxis.
Horizontal lines
Consider a horizontal line y = d(constant) in the zplane. It is mapped onto the curve v^{2} = 4d^{2}(u+d^{2}), a parabola with its axis along Ou and with its opening to the right.
Figure 8.10
u = x^{2} − d^{2}, v = 2dx
⇒x^{2} = u + d^{2}, 4d^{2}x^{2} = v^{2}
Eliminating x between these equations, we get
The vertex of the parabola is at (−d^{2}, 0), and the axis is the positive uaxis.
The region between two horizontal lines y = d_{1} and y = d_{2}(d_{2} > d_{1} ≥ 1) is mapped into the region between the parabolas v^{2} = and v^{2} = in the wplane.
8.4.2 Transformation w = z^{n} (n ∈ ℕ)
Differentiating w = z^{n} w.r.t z, we have
So, the mapping is conformal at all points z except at the origin z = 0.
Solving w = z^{n} for z, we have which has n roots. For every nonzero w, this shows that there are n preimages.
Converting the equation w = z^{n} into polars by putting z = re^{iθ} and w = Re^{iϕ}, we get R = r^{n} and ϕ = nθ.
A circle z = r of radius r and centre at the origin in the zplane is mapped into a circle of radius r^{n} and centre at the origin of the wplane.
The sectorial region {z = re^{iθ}, α ≤ θ ≤ β} is mapped into the sectorial region {w = Re^{iϕ}, nα ≤ ϕ ≤ nβ} in the wplane.
In particular, the sector is mapped into the upper halfplane v ≥ 0 in the wplane given by
{w = Re^{iϕ}, 0 ≤ ϕ ≤ π} and the region
is mapped into the region
{w = Re^{iϕ}, 0 ≤ ϕ ≤ 2π}, which is the entire wplane.
Figure 8.11 Mapping w = z^{n}
8.4.3 Transformation w = e^{z}
Consider the mapping w = e^{z}. Put z=x+iy, x, y ∈ ℝ
w 
= 
Re^{iϕ} = R(cos ϕ + i sin ϕ) 
⇒ e^{z} 
= 
e^{x}e^{i}y = e^{x(cosy+i sin y)} and 
w 
= 
e^{z} ⇒ w = R = e^{x} arg w = ϕ = y. 
Figure 8.12 Mapping of rectangle into sectorial region in the wplane under w = e^{z}
Vertical lines in the zplane
Straight lines x = c (constant), i.e., vertical lines are mapped onto the circles w = R = e^{c} (constant).
If c > 0 then the circles in the wplane have radius greater than 1, if c < 0 then the circles have radius less than 1 and if c = 0 (yaxis) then the circle is of unit radius.
Figure 8.13 Vertical lines mapped into concentric circles under w = e^{z}
Horizontal lines in the zplane
Straight lines y = d (constant), i.e., horizontal lines are mapped onto the radial lines ϕ = d (constant) in the wplane.
Figure 8.14 Horizontal lines mapped into rays under the mapping w = e^{z}
If d > 0 then the ray ϕ = d in the wplane makes a positively oriented angle and if d < 0 it makes a negatively oriented angle with the real uaxis.
If d = 0 then we have the real axis (xaxis) in the zplane, and it is mapped onto the real uaxis in the wplane. The rectangular region ABCD bounded by the straight lines x = c_{1} and c_{2}, and y = d_{1} and d_{2} (c_{1} < c_{2}, d_{1} < d_{2}) in the zplane is mapped onto the sectorial region A′B′C′D′ given by
8.4.4 Transformation w = sin z
Consider the mapping
put z = x + iy, w = u + iv where x,y, u, v ∈ R
w = u + iv = sin z = sin (x + iy)
= sin x coshy + i cos x sinh y (8.12)
Equating the real and imaginary parts
Since these are periodic functions with period 2π therefore (Eq. 8.11) is periodic with period 2π. Hence the mapping is not onetoone if we consider the whole of the xyplane.
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 xaxis)
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 xaxis is mapped onto the segment −1 ≤ u ≤ 1 of the uaxis.
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 xaxis as shown in Fig. 8.16 with k = 1.
Image of the rectangle k_{1} < y < k_{2}, −π < x < π
Similarly, the rectangle −π < x < π, k_{1} < y < k_{2} maps onto the elliptic ring with a cut along the negative vaxis.
Figure 8.17
Vertical lines x = ±c (c > 0)
Let x = c (constant). Equations (8.13)
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 halfplanes, 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 z_{1}, 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
The transformation is a composition of the following:
 Counterclockwise rotation through z_{1} = iz.
 Sine mapping z_{2} = sin z_{1}.
 Clockwise rotation through w = −iz_{2}.
8.4.7 Transformation w = cosh z
The transformation is a composition of the following:
 Rotation through in the counterclockwise direction z_{1} = iz
 Cosine mapping w = cos z_{1}
8.4.8 Logarithm
The natural (Naperian) logarithm of z = x + iy is denoted by ln z (sometimes also by log z or log e^{z}). It is defined as the inverse of the exponential function, i.e., w = ln z ⇔ e^{w} = z for z ≠ 0.
(Since e^{w} ≠ 0, z = 0 is impossible)
Put z 
= 
re^{iθ}, w = u + iv 
e^{w} 
= 
e^{u+iv} = re^{iθ} 
e^{w} 
= 
e^{u} ∵ e^{iv}  = 1 
arg e^{w} 
= 
v = 0 
e^{u} 
= 
r ⇒ u = ln r 


(real logarithm defined for r > 0). 
Hence w = u + iv = ln z is given by
ln z = ln r + iθ 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 manyvalued.
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
For a given z ≠ 0, Arg z is unique. This implies that ln z is singlevalued 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
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 zplane and the wplane in the mapping by w = f(z).
Example 8.3
Mapping by w = z^{2} and by its inverse, w = z^{2} is restricted to the right halfplane x > 0 and maps this halfplane onto the wplane without the negative half u ≤ 0 of the real uaxis 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 zplane without the negative half of the xaxis onto the right halfplane u = Re w > 0.
Natural logarithm
ln z is the inverse relation of the exponential function w = e^{z}. The latter maps a fundamental strip onto the wplane without z = 0 because e^{z} ≠ 0 for every z. Hence by the principle of inverse mapping, the principal value w = Ln z maps the zplane 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 zplane (cut as before and 0 omitted) onto the strip −π < v ≤ 3π. Similarly for each of the many mappings
The corresponding horizontal strips of width 2π images of the zplane under these mappings together cover the whole of the wplane without overlapping.
8.4.9 Transformation (Joukowski^{1} Airfoil)
If in the mapping we put z = re^{iθ} = r(cos θ + i sin θ) and w = Re^{iθ} = R(cos ϕ + i sin ϕ) we have w = u + iv = r(cos θ + i sin θ) + – (cos θ − i 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 zplane. Since b must not be equal to zero, we have r ≠ 1. These nonunit circles are mapped onto ellipses
The unit circle z = r = 1 is mapped onto the line segment −2 ≤ u ≤ 2 of the uaxis. (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 doublevalued 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 zplane onto the family of confocal ellipses in the wplane 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 zplane is mapped into a degenerated ellipse which flattens to the line segment v = 0, −2 ≤ u ≤ 2 on the real uaxis 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 = θ_{0} 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 zplane into an orthogonal curvilinear system of coordinates, whose coordinate lines are the confocal families of ellipses and hyperbolas in the wplane.
Example 8.4
Show that the transformation maps the hyperbola x^{2} − y^{2} = 1 to the lemniscate.
[JNTU 1985, 1998]
Figure 8.20
Solution The given curve in the zplane is the hyperbola
The given transformation is
Let z = x + iy, w = u + iv.
In polars z = re^{iθ}, Re^{iϕ} w = By Eq. (2) we get
Now let us put Eq. (1) in polar form by substituting
Thus x^{2} − y^{2} = 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 wplane 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 zplane is
[where A, B,C,D are constants]
Equation (6) represents a circle in the wplane. Thus a circle (5) in the zplane is transformed into a circle (6) in the wplane. If D = 0 then it represents a straight line whose equation is
Example 8.6
Determine the region in the wplane corresponding to the triangular region bounded by the lines x = 0, y = 0 and x + y = 1 in the zplane 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 zplane and the wplane 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:
u^{2} + (v + 2)^{2} = 2^{2}; centre c_{2} = (0, −2) and radius r_{2} = 2 units.
u^{2} + (v + 1)^{2} = 1^{2}; centre c_{1} = (0, −1) and radius r_{1} = 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 C_{1} : z = 1 and C_{2} : z = 2.
The unit circle C_{1} : x^{2} + y^{2} = 1 in the zplane is mapped to the circle C^{*}_{1} : (u – 1)^{2} + v^{2} = 4 with its centre at u = 1, v = 0 and radius = 2 units.
Again
The circle C_{2} : x^{2} + y^{2} = 2^{2} with radius = 2 units in the zplane is mapped to the circle C^{*}_{2} : (u − 1)^{2} + v^{2} = 4^{2} 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 zplane 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.
Equating the real and imaginary parts, we have
The coordinates of the vertices of the triangle are
z_{1} = i = (0,1),
z_{2} = 1 + i = (1,1),
z_{3} = 1 − i = (1, −1)
Figure 8.24
The images of the points in the wplane are
ΔABC in the zplane is mapped to the triangle ΔPQR in the wplane.
Example 8.10
Show that the transformation transforms r = constant in the zplane into a family of ellipses in the wplane.
Solution Let
This gives
Let r = k(constant) ≠ 1 so that
This equation represents a family of ellipses with semimajor axis , semiminor 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
Equating the real and imaginary parts, we get
Squaring and adding,




by Eq. (3) 
(4) 
The given curve in the zplane is the
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 wplane.
Example 8.12
If w = , find the image of z  < 1. [JNTU 2002].
Solution
Now,
Putting w = u + iv, we get
Thus the image of the interior of the circle z = 1 is mapped onto the right half of the wplane.
Example 8.13
If w = cosh z, find the mapping of the region a ≤ x ≤ b onto the wplane.
Solution Let z = x + iy and w = u + iv where x, y, u, v ∈ R
∴ w 
= 
cosh z ⇒ u + 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 a ≤ x ≤ b in the zplane 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 x^{2} + y^{2} – 4x = 0 into the straight line 4u + 3 = 0.
[JNTU 2003]
Solution Let
The given transformation is
The given equation of the circle is
This represents a straight line in the wplane.
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 wplane given by the cut A*F* or C*D* along the imaginary axis Ov.
8.5 Bilinear or Mobius^{2} 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, halfplanes, 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.
Crossmultiplying, we can write Eq. 8.16 in the form
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 (ad – bc) = 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
b =0, c = 0, d = 1 ⇒ w = az (8.20)
Simply rotation if a = e^{iθ} = 1
Magnification or dimination if
c =0, d = 1 w = az + b (linear transformation) (8.22)
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.
 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.
 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)
 c ≠ 0. cw – a = 0 This point corresponds to z = ∞.
 c = 0. In view of ad – bc ≠ 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 zplane into the totality of circles and straight lines in the wplane.
Though there are four constants in Eq. 8.16, on dividing three of them by the fourth nonzero 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 z_{1}, z_{2}, z_{3} in the zplane can be mapped into three distinct points w_{1}, w_{2}, w_{3} in the wplane.
Crossratio of four points
Let z_{1}, z_{2}, z_{3} and z_{4} be four distinct points in the extended complex plane.
Then the crossratio of these points denoted by(z_{1}, z_{2}, z_{3}, z_{4}) is defined as
if none of z_{1}, z_{2}, z_{3}, z_{4} 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 f_{1} and f_{2} are elementary transformations and
Case (ii) Let c ≠ 0. Equation (8.26) gives
Thus f (z) is the product of f_{1},f_{2},f_{3} and f_{4}.
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 crossratio.
Proof: Let
be the given bilinear transformation.
Let z_{1}, z_{2}, z_{3} and z_{4} be four distinct points of the zplane and w_{1}, w_{2}, w_{3} and w_{4} be then corresponding images in the wplane under the transformation (8.27). Suppose all these points are different from ∞. Now we have to prove that (z_{1}, z_{2}, z_{3}, z_{4}) = (w_{1}, w_{2}, w_{3}, w_{4}).
From Eq. (8.27), we have
where k_{1} =
Similarly, w_{2} − w_{4} = k_{2}(z_{2} − z_{4})
where
∴ (w_{1} − w_{3})(w_{2} − w_{4}) 
= 
k_{1}k_{2}(z_{1} − z_{3})(z_{2} − z_{4}) 

= 
k (z_{1} − z_{3})(z_{2} − z_{4}) 
where
Similarly, we can derive the result
Hence the theorem. The proof when one of z_{i} or wi is ∞ is similar.
Note:
Four distinct points z¡ (i = 1, 2, 3, 4) are collinear or concyclic ⇔ (z_{1}, z_{2}, z_{3}, z_{4}) is real. Also any bilinear transformation preserves a crossratio. 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 (z_{1}, z_{2}, z_{3}, z_{4}) = (w_{1}, w_{2}, w_{3}, w_{4}) is a bilinear transformation that maps three distinct points z_{1}, z_{2} and z_{3} onto three specified distinct points w_{1}, w_{2} and w_{3}, respectively.
Proof: Let
z = z_{1} ⇒ w = w_{1} and
z = z_{3} ⇒ w = w_{3} from Eq. (8.29).
Also z = z_{2}
⇒ (w − w_{1})(w_{2} − w_{3})(z_{2} − z_{3})(z_{1} − z_{2})
= (w − w_{3})(w_{1} − w_{2})(z_{2} − z_{1})(z_{2} − z_{3})
⇒ (w − w_{1})(w_{2} − w_{3}) = −(w − w_{3})(w_{1} − w_{2}).
Cancelling out nonzero factors (z_{2} – z_{3}), (z_{1} – z_{2}), we get
w(w_{2} − w_{3} + w_{1} − w_{2})
= w_{2}w_{1} − w_{3}w_{2} + w_{1}w_{2} − w_{1}w_{3}
⇒w(w_{1} − w_{3}) = w_{2}(w_{1} − w_{3})
⇒w = w_{2} ∵ w_{1} 6= w_{3}
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 = u ≠ iv so that = x − iy, = u − iv.
The given transformation is
The given circle is
This is the circle with centre at and radius = 1 in the wplane.
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 crossratios, we can write
Here z_{1} = ∞, z_{2} = i, z_{3} = 0; w_{1} = 0, w_{2} = i, w_{3} = 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 x^{2} + y^{2} − 4x = 0 into the straight line 4u+3 = 0.
[JNTU 2003]
Solution Let z = x + iy, w = u + iv so that = x − iy, = u − iv.
Now x^{2} + y^{2} = (x + iy)(x − iy) = z and
The given equation of the circle is
which is the equation of a straight line in the wplane.
Example 8.20
Determine the bilinear transformation that maps the points (12i, 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 crossmultiplication
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 
z_{1} 
= 
−i, z_{2} = 0, z_{3} = i 

w_{1} 
= 
−1, w_{2} = i, w_{3} = 1 
Substituting in Eq. (1), we get
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 
z_{1} 
= 
2, z_{2} = i, z_{3} = −2 

w_{1} 
= 
1, w_{2} = i, w_{3} = −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 
z_{1} 
= 
−1, z_{2} = 0, z_{3} = 1 

w_{1} 
= 
0,w_{2} = −1,w_{3} = ∞ 
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 
z_{1} 
= 
0, z_{2} = 1, z_{3} = ∞ 

w_{1} 
= 
−1, w_{2} = −i, w_{3} = 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 x^{2} + y^{2} − 4x = 0 onto a straight line 4u + 3 = 0.
[JNTU 2003]
Solution The given transformation is
The given equation of the circle in the zplane is
(Under the transformation (1), Eq. (2) becomes)
which is a straight line in the wplane.
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 halfplane of the wplane.
Example 8.28
Find the bilinear transformation which maps the points ∞, i, 0 in the zplane into –1, – i, 1 in the wplane.
[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 zplane onto the points (–1, –2, –i) in the wplane.
[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.
 Putting w = f (z) = z, we obtain
Solving, we get
 Putting w = f(z) = z, we obtain
 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 zplane into the points (0, 1, i) of the wplane.
[JNTU 2005S(2)]
Solution Let the transformation be
Here 
z_{1} 
= 
1 + i, z_{2} = −i, z_{3} = 2 − i 

w_{1} 
= 
0, w_{2} = 1, w_{3} = 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
For any γ ∈ ℂ, we have
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
 Find the bilinear transformation that maps z_{L}, z_{2}, z_{3} onto w_{1}, w_{2}, w_{3}, respectively.
 z =–1, 0, 1; w = 0,i,3i.
Ans: w = –3i(z + 1)/(z – 3)
 z = 0, –i, –1; w = i, 1, 0.
Ans: w = –i(z + 1)/(z – 1)
 z = ∞, i, 0; w = 0, i, ∞.
Ans:
 z = 1, 0, –1; w = i, 1, ∞.
Ans: w = [(–1 + 2i)z + 1]/(z + 1)
 z =–1, 0, 1; w = 0,i,3i.
 Find the fixed points of the transformation.
 w = (z – 1)/(z + 1).
Ans: w = ±i
 w = (2z – 5)/(z + 4).
Ans: – 1 ± 2i
 w = (z – 1)/(z + 1).
 Find the bilinear transformation whose fixed
 i,–i
Ans:
 1,–1.
Ans:
 i,–i
 Determine the bilinear transformation which maps the points 0, 1, ∞ in the zplane into –5, –1, 3, respectively, in the wplane. Find the invariant points.
Ans: w = (3z – 5)/(z + 1), z = 1 ± 2i
 Determine the region into which the rectangular region in the zplane 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
 Find the bilinear transformation which maps the points –1, 0, 1 into the points 0, – 1, ∞, respectively.
Ans:
 Find the bilinear transformation which maps the points (–1, 0, 1) into the points (0, i, 3i).
[JNTU 2005]
Ans: