Conformal Mappings

Mustafa Akman

January 21, 2010

1

Contents

1 Introduction 1

2 Basic Properties of Conformal Mappings 2

3 Bilinear Transformations 5

3.1 The Implicit Formula . . . . . . . . . . . . . . . . . . . . . . . 6

4 Mappings Involving Elementary Functions 12

5 Mapping by Trigonometric Functions 15

6 Conclusion 19

1 Introduction

In mathematics, a conformal map is a function which preserves angels. In the

most common case the function is between domains in the complex plane.

More formally, a map f : U → V is called conformal (or angel preserving)

at u0 if it preserves oriented angles between curves through u0 with respect

to their orientation. In 1569, the Flemish cartographer Gerardus Merca-

tor devised a cylindrical map projection that preserves angles. The Mercator

projection is still used today for world maps. Another map projection known

to the ancient Greeks is the stereographic projection and both examples are

conformal. In Complex analysis a function preserves angles if and only if

it is analytic or anti analytic. A significant result, known as the Riemann

mapping theorem, states that any simply connected domain can be mapped

conformally onto the disk. Conformal mappings is very important in Com-

plex Analysis, as well as in many areas of physics and engineering. In this

project, I study about the some definitions and information about conformal

mapping and i am going to explain with some examples.

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2 Basic Properties of Conformal Mappings

Definition 2.1. The mapping f(z) = f(x+ iy) = u(x, y) + iv(x, y) is called

conformal if

• It is one to one. (Bijection)

• It and its inverse continuously differentiable.

• Image of every smooth curve of G1 is a smooth curve of G2.

• Angle between any two curves equals angle between their images.

Theorem 2.1. Let f : A → B be analytic and let f ′(zo) 6= 0 for each z0 ∈A. Then f is conformal.

Proof. We let C1 and C2 be two smooth curves passing through z0 with

tangents given by T1 and T2, respectively. We let β1 and β2 denote the

angles of inclination of T1 and T2, respectively. The image curves K1 and K2

that pass through the point w0 = f(z0) have tangents denoted T ∗1 and T ∗2 ,

respectively.

Then the angles of inclination γ1 and γ2 of T ∗1 and T ∗2 are related to β1 and

β2 by equations γ1 = α + β1 and γ2 = α + β2, where α = Arg f ′(z0). Hence

γ2 − γ1 = β2 − β1. That is, the angle γ2 − γ1 from K1 to K2 is the same in

magnitude and orientation as the angle β2 − β1 from C1 to C2. Therefore,

the mapping w = f(z) is conformal at z0.

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Example 2.1. Show that the mapping w = f(z) = cos z is conformal at the

points z1 = i, z2 = 1, and z3 = π + i, and determine the angle of rotation

given by α = Arg f ′(z) at the given points.

Solution. Because f ′(z) = − sin z, we conclude that the mapping w = cos z

is conformal at all points except z = nπ, where n is an integer. Calculation

reveals that f ′(i) = − sin(i) = −i sinh 1, f ′(1) = − sin 1 sinh 1, and

f ′(π+ i) = − sin(π+ i) = i sinh 1. Therefore, the angle of rotations are given

by α1 = Arg f ′(i) = −π2

, α2 = Arg f ′(1) = π, α3 = Arg f ′(π + 1) =π

2.

Proposition 2.1. Let f : A → B and g : B → C be two conformal maps.

Then f−1 and g ◦ f are conformal, too.

Proof. Since f is bijective, the mapping f−1 exists. Then, f−1 is analytic

with df−1(w)/dw = 1/[df(z)/dz] where w = f(z). Thus df−1(w)/dw 6= 0, so

f−1 is conformal.

Certainly g◦f is bijective and analytic, since f and g are. Then derivative

of g ◦ f at z is g′(f(z))f ′(z) 6= 0. Therefore g ◦ f is conformal.

Theorem 2.2 (Riemann Mapping Theorem). Let A be a connected and sim-

ply connected region other than the whole complex plane. Then there exists

a bijective conformal map f : A → D where D = {z such that |z| < 1}.Furthermore, for any fixed z0 ∈ A, we can find and f such that f(z0) = 0

and f ′(z0) > 0. With such a specification, f is unique.

Proof. (In this proof, we only show the uniqueness part) Suppose f and g

are bijective conformal maps of A onto D with f(z0) = g(z0) = 0, f ′(z0) > 0,

and g′(z0) > 0. We want to show that f(z) = g(z) for all z in A. To do

this, define h on D by h(w) = g(f−1(w)) for w ∈ D. Then h : D → D and

h(0) = g(f−1(0)) = g(z0) = 0. By the Schwarz Lemma, |h(w)| ≤ |w| for all

w ∈ D. Exactly the same argument to h−1 = f ◦ g−1, so that |h−1(ζ)| ≤ |ζ|for all ζ ∈ D. With ζ = h(w), this gives |w| ≤ |h(w)|. Combining these

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inequalities, we get |h(w)| = |w| for all w ∈ D. The Schwarz Lemma now

tells us that h(w) = cw for a constant c with |c| = 1. Thus cw = g(f−1(w)).

With z = f−1(w) we obtain cf(z) = g(z) for all z ∈ A. In particular,

cf ′(z0) = g′(z0). Since both f ′(z0) and g′(z0) are positive real numbers, so is

c. Thus, c = 1 and so f(z) = g(z), as desired.

Example 2.2. Show that the mapping w = f(z) = z2 maps the unit square

S = {x + iy : 0 < x < 1, 0 < y < 1} onto the region in the upper half-plane

Im(w) > 0, which lies under the parabolas u = 1− 1

4v2 and u = −1 +

1

4v2.

Solution. z2 = (reiθ)2 = r2e2iθ doubles the angles and squares the norm.

Let `1 : z = x + iy, 0 ≤ x ≤ 1, y = 0, `2 : z = x + iy, x = 1, 0 ≤ y ≤ 1,

`3 : z = x + iy, 0 ≤ x ≤ 1, y = 1 and `4 : z = x + iy, x = 0, 0 ≤ y ≤ 1.

Thus, it is easy to find the images of `1 and `4. If z ∈ `2, then z = 1 + iy,

0 ≤ y ≤ 1 and u+ iv = z2 = 1− y2 + 2iy. Then,

u = 1− y2 = 1−(

2y

2

)2

= 1−(v

2

)2

= 1− v2

4, 0 ≤ v ≤ 2

. If z ∈ `2, then z = x+ i, 0 ≤ x ≤ 1 and u+ iv = z2 = x2 − 1 + 2ix. Then,

u = x2 − 1 =

(2x

2

)2

− 1 =(v

2

)− 1 =

v2

4− 1 = −1 +

v2

4, 0 ≤ v ≤ 1

Note that, the angles at 1 and -1 are preserved but the angle at 0 is doubled.

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Theorem 2.3. Let A be a bounded region with f : A→ C a bijective confor-

mal map onto its image f(A). Suppose that f extends to be continuous on

A and that f maps the boundary of A onto a circle of radius R. Then f(A)

equals the inside of that circle. More generally, if B is a bounded region that,

together with its boundary, can be mapped conformally onto the unit disk and

its boundary and if f maps ∂(A) onto ∂(B), then f(A) = B.

Proof. By composing f with the conformal map h that takes B to the unit

disk it is sufficient to consider the special case in which B equals D = {z :

|z| < 1}. On ∂(A), |f(z)| = 1, so by the Maximum Modulus Theorem,

|f(z)| ≤ 1 on A. Since f cannot be constant at no z ∈ A is the maximum

|f(z)| = 1 reached. We assumed that f(∂(A)) = ∂(D), but this is also

equal to ∂(f(A)). To see this, use compactness of A, continuity of f , and

D∩∂(D) = f(A)∩∂(f(A)) = �. Thus our earlier argument applies to show

that f(A) = D.

3 Bilinear Transformations

A bilinear transformation is a mapping of the form

T (z) =az + b

cz + d

where a, b, c, d are fixed complex numbers and ad− bc 6= 0 because otherwise

T would be a constant. T id also called a mobius transformation, or fractional

linear transformation.

Proposition 3.1. The map T defined in the preceding display is bijective

and conformal from the set

A = {z ∈ C : z 6= −dc} onto B = {w|w 6= a

c}

The inverse of T is also a fractional linear transformation given by

T−1(w) =−dw + b

cw − a

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Proof. Certainly T is analytic on A and S(w) = (−dw + b)/(cw − a) is

analytic on B. The map T will be bijective if we can show that T ◦ S and

S ◦ T are the identities since this means that T has S as its inverse. Indeed,

this is seen in this computation.

T (S(w)) =a(−dw+b

cw−a ) + b

c(−dw+bcw−a ) + d

=−adw + ab+ bcw − ab−cdw + bc+ dcw − da

=(bc− ad)w

bc− ad= w

We can cancel because cw−a 6= 0 and bc−ad 6= 0. Similarly, S(T (z)) = z.

Finally, T ′(z) 6= 0 because

1 =d

dz[z] =

d

dz[S(T (z))] = S ′(T (z)).T ′(z),

so T ′(z) 6= 0.

3.1 The Implicit Formula

Theorem 3.1 (The Implicit Formula). There exist a unique bilinear trans-

formation that maps three distinct points, z1, z2 and z3, onto three distinct

points, w1, w2 and w3, respectively. An implicit formula for the mapping is

given by

z − z1

z − z3

z2 − z3

z2 − z1

=w − w1

w − w3

w2 − w3

w2 − w1

Proof. Now we will solve w in terms of z. The result is an expression for w

that has the form of the first formula of bilinear transformation, where the

coefficients a, b, c and d involve various combination of the values z1, z2, z3,

w1, w2 and w3. If we set z = z1 and w = w1 in the theorem, then both sides

of the equation are zero, showing w1 is the image of z1. If we set z = z2

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and w = w2 in theorem, then both sides of the equation take on the value 1.

Hence w2 is the image of z2. Taking reciprocals we write the theorem in the

form of,

z − z3

z − z1

z2 − z1

z2 − z3

=w − w3

w − w1

w2 − w1

w2 − w3

If we set z = z3 and w = w3 in the equation above, then both sides of the

equation are zero. Therefore w3 is the image of z3, and we have shown that

the transformation has the required properties.

Corollary 3.1. (The implicit formula with a point at infinity) In Implicit

Theorem, the point at infinity can be introduced as one of the prescribed points

in either the z plane or the w plane.

Proof.

Case 1 If z3 =∞, then we can writez2 − z3

z − z3

=z2 −∞z −∞

= 1 and substitute

this expression in to implicit theorem to obtain

z − z1

z2 − z1

=w−1

w − w3

w2 − w3

w2 − w1

Case 2 If w3 =∞, then we can writew2 − w3

w − w3

=w2 −∞w −∞

= 1 and substitute

this expression in to implicit theorem to obtain

z−1

z − z3

z2 − z3

z2 − z1

=w − w1

w2 − w1

Example 3.1. a) Find the image of the unit circle |z| = 1 by the following

transformation w = uz − vvz − 1

, u, v are complex numbers and |z| 6= 1

b) Which bilinear transformation maps the unit circle |z| = 1 on the unit

circle |w| = 1? What is the image of the unit disc |z| ≤ 1 by this transfor-

mation?

Solution. a) Since

ww = uuzz − vz − vz + vv

vvzz − vz − vz + 1

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for zz = 1 we have that ww = uu. Therefore the unit circle |z| = 1 is mapped

on the circle |w| = |u|.b) Using a) we obtain that the bilinear transformation

w = uz − vvz − 1

v 6= 1,

The unit disc |z| ≤ 1, for |v| < 1, is mapped on the unit disc |w| ≤ 1, and

for |v| > 1 on the region |w| ≥ 1.

Example 3.2. Show that w = s(z) =i(1− z)

1 + zmaps the unit disk D : |z| < 1

one to one and onto the upper half-plane Im(w) > 0.

Solution. We first consider the unit circle C : |z| = 1, which forms the

boundary of the disk and find its image in the w plane. If we write S(z) =iz + i

z + 1, then wee see that a = −i, b = i, c = 1 and d = 1. Using the inverse

transformation equation we find that the inverse is given by

z = s−1(w) =−dw + b

cw − a=−(1)w + i

(1)w − (−i)=−w + i

w + i

If |z| = 1, then the equation on above implies that the images of points on

the unit circle satisfy |−w + i

w + i| = 1 which yields the equation

|w + i| = | − w + i|

Squaring both sides, we obtain u2 + (1 + v)2 = u2 + (1 − v)2, which can be

simplified to yield v = 0, which is the equation of the u-axis in the w plane.

The circle C divides the z plane in to two portions, and its image is the u

axis, which divides the w plane into two portions. The image of the point

z = 0 is w = S(0) = i, so we expect that the interior of the circle C is

mapped onto the portion of the w plane that lies above the u axis. To show

that this outcome is true, we let |z| < 1. Then the equation of inverse S is

implies that the image values must satisfy the inequality | −w+ i| < |w+ i|,

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which we write as d1 = |w − i| < |w − (−i)| = d2.

If we interpret d1 as the distance from w to i and d2 as the distance from

w to −i, then a geometric argument shows that the image point w must lie

in the upper half-plane Im(w) > 0. As S(z) is one to one and onto in the

extended complex plane, it follows that S(z) maps the disk onto the half

plane.

Example 3.3. Find the bilinear transformation w = S(z) that maps the

points z1 = −2, z2 = −1, z3 = 0 onto the points w1 = −1, w2 = 0, w3 = 1,

respectively.

Solution. We use the implicit formula, and write

(z − (−2))((−1− i)− 0)

(z − 0)((−1− i)− (−2))=

(w − (−1))(0− 1)

(w − 1)(0− (−1))

(z + 2(−1− i)(z)(−1− i+ 2)

=(w + 1)(−1)

(w − 1)(1)z + 2

z

−1− i1− i

=1 + w

1− w

Using the fact that−1− i1− i

=1

i, we rewrite this equation as

z + 2

iz=

1 + w

1− w.

We now expand the equation and obtain (1-i)z+2=w((1+i)z+2) which can

be solved for w in terms of z, giving the desired solution

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w = S(z) =(1− i)z + 2

(1 + i)z + 2

Example 3.4. Show that the mapping

w = S(z) =(1− i)z + 2

(1 + i)z + 2

maps the disk D : |z + 1| < 1 onto the upper half plane Im(w) > 0

Solution. For simplicity, we choose the ordered triple z1 = −2, z2 = −1− i,z3 = 0, which gives the circle C : |z + 1| = 1 a positive orientation and the

disk D a left orientation. From previous example image points are

w1 = S(z1) = S(−2) = −1

w2 = S(z2) = S(−1− i) = 0

w3 = S(z3) = S(0) = 1

Because the ordered triple of points w1 = −1, w2 = 0, w3 = 1, lie on

the u axis, it follows that the image of circle C is the u axis. The points

w1 = −1, w2 = 0, w3 = 1 give the upper half plane G : Im(w) > 0 a left

orientation. Therefore w = S(z) =(1− i)z + 2

(1 + i)z + 2maps the disk D onto the

upper half plane G. To check our work, we choose a point z0 that lies in D

and find the half plane in which its image, w0 lies. The choice z0 = −1 yields

w0 = S(z0) = i. Hence the upper half plane is correct image.

Example 3.5. Find the bilinear transformation that maps the crescent-

shaped region that lies inside the disk |z − 2| < 2 and outside the circle

|z − 1| = 1 onto a horizontal strip.

Solution. For simplicity, we choose z1 = 4, z2 = 2 + 2i and z3 = 0 and the

image values w1 = 0, w2 = 1 and w3 = ∞, respectively. The ordered triple

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z1, z2 and z3 gives the circle |z − 2| = 2 a positive orientation and the disk

|z − 2| < 2 has a left orientation. The image points w1, w2 and w3 all lie on

the extended u axis, and they determine a left orientation for the upper half

plane Im(w) > 0. Therefore, we can use the case 2 from corollary to write

z − 4

z − 0

2 + 2i− 0

2 + 2i− 4=w − 0

1− 0

which determines a mapping of the disk |z − 2| < 2 onto the upper half

plane Im(w) > 0. We simplify the preceding equation to obtain the desired

solution

w = S(z) =−iz + 4i

z

A straightforward calculation shows that the points z4 = 1 − i, z5 = 2, and

z6 = 1 + i are mapped onto the points

w4 = S(1− i) = −2 + i

w5 = S(2) = i

w6 = S(1 + i) = 2 + i

The points w4, w5, w6 lie on the horizontal line Im(w) = 1 in the upper half

plane. Therefore, the crescent shaped region is mapped onto the horizontal

strip 0 < Im(w) < 1.

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4 Mappings Involving Elementary Functions

The function w = f(z) = exp z is a one to one mapping of the fundamental

period strip −π < y ≤ π in the z plane onto the w plane with the point w = 0

deleted. Because f ′(z) 6= 0, the mapping w =exp z is a conformal mapping

at each point z in the complex plane. The family of horizontal lines y = c

for −π < c ≤ π and the segments x = a for −π < y ≤ π from an orthogonal

grid in the fundamental period strip. Their images under the mapping w =

exp z are the rays ρ > 0 and φ = c and the circles |w| = ea, respectively.

The images from an orthogonal curvilinear grid in the w plane, as shown in

the shape.

If −π < c < d ≤ π, then the rectangle R = {x+iy : a < x < b, c < y < d}is mapped one to one and onto the regionG = {ρeiφ : ea < ρ < eb, c < φ < d}.The inverse mapping is the principal branch of the logarithm z = Logw.

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Example 4.1. Show that the transformation

w = f(z) =ez − iez + i

is a one to one conformal mapping of the horizontal strip 0 < y < π onto the

disk |w| < 1. Furthermore, the x axis is mapped onto the lower semicircle

bounding the disk, and the line y = π is mapped onto the upper semicircle.

Solution. The function f is the composition of Z =exp z followed by w =Z − iZ + i

. The transformation Z =exp z maps the horizontal strip 0 < y < π

onto the upper half plane Im(Z) > 0, the x axis is mapped on to the positive

X axis and the line y = π is mapped onto the negative X axis. Then the bi-

linear transformation w =Z − iZ + i

maps the upper half plane Im(Z) > 0 onto

the disk |w| < 1, the positive X axis is mapped onto the lower semicircle,

and the negative X axis onto the upper semicircle. We can see composite

mapping in the shape.

Example 4.2. Show that the transformation w = f(z) = Log(1 + z

1− z) is a

one to one conformal mapping of the unit disk |z| < 1 onto the horizontal

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strip |v| < π

2. Furthermore, the upper semicircle of the disk is mapped onto

the line v =π

2and the lower semicircle onto v =

−π2

.

Solution. The function w = f(z) is the composition of the bilinear trans-

formation Z =1 + z

1− zfollowed by the logarithmic mapping w = Log z. The

image of the disk |z| < 1 under the bilinear mapping Z =1 + z

1− zis the right

half plane Re(Z) > 0, the upper semicircle is mapped onto the positive Y axis.

The logarithmic function w = Log z then maps the right half plane onto the

horizontal strip, the image of the positive Y axis is the line v =π

2and the

image of the negative Y axis is the line v =−π2

. We can see composite

mapping in the shape.

Example 4.3. Show that the transformation w = f(z) = (1 + z

1− z)2 is a one to

one conformal mapping of the portion of the disk |z| < 1 that lies in the upper

half plane Im(z) > 0 onto the upper half plane Im(w) > 0. Furthermore,

show that the image of the semicircular portion of the boundary is mapped

onto the negative uaxis, and the segment −1 < x < 1, y = 0 is mapped onto

the positive uaxis.

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Solution. The function w = f(z) is the composition of the bilinear trans-

formation Z =1 + z

1− zfollowed by the mapping w = Z2. The image of the

half disk under the bilinear mapping Z =1 + z

1− zis the first quadrant X > 0,

Y > 0, the image of the segment y = 0, −1 < x < 1, is the positive

Xaxis, and the image of the semicircle is the positive Y axis. The mapping

w = Z2 then maps the first quadrant in the Z plane onto the upper half

plane Im(w) > 0, as we can see in the figure.

Example 4.4. Consider the function w = f(z) = (z2 − 1)

1

2 , which is the

composition of the functions Z = z2 − 1 and w = Z

1

2 , where the branch of

the square root is Z

1

2 = R

1

2 (cosϕ

2+ i sin

ϕ

2), where 0 ≤ ϕ < 2π. Show that

the transformation w = f(z) maps the upper half plane Im(z) > 0 one to

one and onto the upper half plane Im(w) > 0 slit along the segment u = 0,

0 < v ≤ 1.

Solution. The function Z = z2 − 1 maps the upper half plane Im(z) > 0

one to one and onto the Z plane slit along the ray Y = 0, X ≥ −1. Then

the function w = Z

1

2 maps the slit plane onto the slit half plane, as shown

in the figure.

5 Mapping by Trigonometric Functions

The trigonometric functions can be expressed with compositions that involve

the exponential function followed by a bilinear function. We can find images

of certain regions by following the shapes of successive images in the com-

posite mapping.

Example 5.1. Show that the transformation w = tan z is a one to one

conformal mapping of the vertical strip |x| < π

4onto the unit disk |w| < 1.

15

Figure 1: Example 4.3

Figure 2: Example 4.4

Solution. Using equations sin z =1

2i(eiz − e−iz) and cosz =

1

2(eiz + e−iz),

we write

w = tan z =1

i

eiz − e−iz

eiz + e−iz=−iei2z + i

ei2z + 1

Then, mapping w = tan z can be considered to be the composition

w =−iZ + i

Z + 1andZ = ei2z.

The function Z = exp (i2z) maps the vertical strip |x| < π

4one to one

and onto the right half plane Re(Z) > 0. Then the bilinear transformation

16

w =−iZ + i

Z + 1maps the half plane one to one and onto the disk, as show in

the picture.

Example 5.2. Show that the transformation w = f(z) = sin z is a one to

one conformal mapping of the vertical strip |x| < π

2onto the w plane slit

along the rays u ≤ −1, v = 0, and u ≥ 1, v = 0.

Solution. Because f ′(z) =cosz 6= 0 for values of z satisfying the inequality−π2

< Re(z) <π

2, it follows that e = sin z is a conformal mapping.

u+ iv = sin z = sinx cosh y + i cosx sinh y

If |a| < π

2, then the image of the vertical line x = a is the curve in the w

plane given by the parametric equations

u = sin a cosh y and v = cos a sinh y

for −∞ < y <∞. Next, we rewrite these equations as

cosh y =u

sin aand sinh y =

v

cos a

17

We now eliminate y from these equations by squaring and using the hyper-

bolic identity cosh2 y − sinh2 y = 1. The result is the single equation

u2

sin2 a− v2

cos2 a= 1

The curve given byu2

sin2 a− v2

cos2 a= 1 is identified as a hyperbola in the uw

plane that has foci at the points (±1, 0). Therefore, the vertical line x = a

is mapped one to one onto the branch of the hyperbolau2

sin2 a− v2

cos2 a= 1

that passes through the points (sin a, 0). If 0 < a <π

2, then it is the right

branch, if−π2

< a < 0, it is the left branch. The image of the y axis, which

is the line x = 0, is the v axis. The images of several vertical lines are shown

in the first part of shape.

The image of the horizontal segment−π2

< x <π

2, y = b is the curve in the

w plane given by the parametric equations

u = sinx cosh b and v = cosx sinh b

for−π2

< x <π

2. We rewrite them as

sinx =u

cosh band cosx =

v

sinh b

We now eliminate x from the equations by squaring and using the trigono-

metric identity sin2 x+ cos2 x = 1. The result is the single equation

u2

cosh2 b+

v2

sinh2 b= 1

The curve given byu2

cosh2 b+

v2

sinh2 b= 1 is identified as an ellipse in the

uw plane that passes through the points (± cosh b, 0) and (0,± sinh b) and

has foci at the points (±1, 0). Therefore, if b > 0, then v = cosx sinh b > 0,

and the image of the horizontal segment is the portion of the ellpse given byu2

cosh2 b+

v2

sinh2 b= 1 that lies in the upper half plane im(w) > 0. If b < 0,

then it is the portion that lies in the lower half plane. The images of several

segments are show in the second part of the figure.

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6 Conclusion

In this project, we have studied the bilinear transformations, mappings in-

volving elementary functions, mapping by trigonometric functions. Before

giving details of bilinear transformation, we’ve shown basic properties and

defined conformal mapping theorem and riemann mapping theorem. Then,

we have given the definiton of implicit formula. Then we have shown map-

pings involving elemantary functions and mappings of trigonometric func-

tions with some examples. Moreover, we have shown on figures how functions

are mapped.

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References

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H. Freeman, 1999

[3] Theory and Problems of Complex Variables, Murray R.Spiegel,

McGraww-Hill, 1981

[4] Complex Analysis through Examples and Exercises, Endre Pap, Kluwer

Academic Publisher, 1999

[5] Complex Analysis, Theodore W.Gomelin, Springer, 2001

[6] math.fullerton.edu/mathews/c2003/ComplexUndergradMod.html

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