conformal mapping

736

C H A P T E R 1 7

Conformal Mapping

Conformal mappings are invaluable to the engineer and physicist as an aid in solvingproblems in potential theory. They are a standard method for solving boundary valueproblems in two-dimensional potential theory and yield rich applications in electrostatics,heat flow, and fluid flow, as we shall see in Chapter 18.

The main feature of conformal mappings is that they are angle-preserving (except atsome critical points) and allow a geometric approach to complex analysis. More detailsare as follows. Consider a complex function defined in a domain D of the z–plane;then to each point in D there corresponds a point in the w-plane. In this way we obtain amapping of D onto the range of values of in the w-plane. In Sec. 17.1 we show thatif is an analytic function, then the mapping given by is a conformal mapping,that is, it preserves angles, except at points where the derivative is zero. (Such pointsare called critical points.)

Conformality appeared early in the history of construction of maps of the globe.Such maps can be either “conformal,” that is, give directions correctly, or “equiareal,”that is, give areas correctly except for a scale factor. However, the maps will alwaysbe distorted because they cannot have both properties, as can be proven, see [GenRef8]in App. 1. The designer of accurate maps then has to select which distortion to takeinto account.

Our study of conformality is similar to the approach used in calculus where we studyproperties of real functions and graph them. Here we study the properties of conformalmappings (Secs. 17.1–17.4) to get a deeper understanding of the properties of functions, mostnotably the ones discussed in Chap. 13. Chapter 17 ends with an introduction to Riemannsurfaces, an ingenious geometric way of dealing with multivalued complex functions such as

and So far we have covered two main approaches to solving problems in complex analysis.

The first one was solving complex integrals by Cauchy’s integral formula and was broadlycovered by material in Chaps. 13 and 14. The second approach was to use Laurent seriesand solve complex integrals by residue integration in Chaps. 15 and 16. Now, in Chaps. 17and 18, we develop a third approach, that is, the geometric approach of conformal mappingto solve boundary value problems in complex analysis.

Prerequisite: Chap. 13.Sections that may be omitted in a shorter course: 17.3 and 17.5.References and Answers to Problems: App. 1 Part D, App. 2.

w � ln z.w � sqrt (z)

y � f (x)

f r(z)w � f (z)f (z)

f (z)

w � f (z)

c17.qxd 11/1/10 7:35 PM Page 736

y

x

2

1

0210 –4 –3 –2 –1 0 1 2 3 4

v

u

(z-plane) (w-plane)

17.1 Geometry of Analytic Functions: Conformal Mapping

We shall see that conformal mappings are those mappings that preserve angles, except atcritical points, and that these mappings are defined by analytic functions. A critical pointoccurs wherever the derivative of such a function is zero. To arrive at these results, wehave to define terms more precisely.

A complex function

(1)

of a complex variable z gives a mapping of its domain of definition D in the complex z-plane into the complex w-plane or onto its range of values in that plane.1 For any point in D the point is called the image of with respect to f. More generally, forthe points of a curve C in D the image points form the image of C; similarly for otherpoint sets in D. Also, instead of the mapping by a function we shall say morebriefly the mapping .

E X A M P L E 1 Mapping

Using polar forms and we have Comparing moduli and arguments givesand Hence circles are mapped onto circles and rays onto rays

Figure 378 shows this for the region which is mapped onto the region

In Cartesian coordinates we have and

Hence vertical lines are mapped onto From this we can eliminate y. Weobtain and Together,

(Fig. 379).

These parabolas open to the left. Similarly, horizontal lines are mapped onto parabolas openingto the right,

(Fig. 379). �v2 � 4k2(k2 � u)

y � k � const

v2 � 4c2(c2 � u)

v2 � 4c2y2.y2 � c2 � uu � c2 � y2, v � 2cy.x � c � const

u � Re (z2) � x2 � y2, v � Im (z2) � 2xy.

z � x � iy1 � ƒ w ƒ � 9

4 , p>3 � u � 2p>3.1 � ƒ z ƒ � 3

2 , p>6 � u � p>3,� � 2u0 .u � u0R � r 0

2r � r0� � 2u.R � r 2w � z2 � r 2e2iu.w � Rei�,z � reiu

w � f (x) � z2

w � f (z)w � f (z)

z0w0 � f (z0)z0

(z � x � iy)w � f (z) � u(x, y) � iv(x, y)

SEC. 17.1 Geometry of Analytic Functions: Conformal Mapping 737

Fig. 378. Mapping w � z2. Lines �z� � const, arg z � const and their images in the w-plane

1The general terminology is as follows. A mapping of a set A into a set B is called surjective or a mapping ofA onto B if every element of B is the image of at least one element of A. It is called injective or one-to-one ifdifferent elements of A have different images in B. Finally, it is called bijective if it is both surjective and injective.

c17.qxd 11/1/10 7:35 PM Page 737

(z-plane) (w-plane)

z0

f (z0)

C1

C1*

C2

C2*

α

α

Conformal MappingA mapping is called conformal if it preserves angles between oriented curves inmagnitude as well as in sense. Figure 380 shows what this means. The anglebetween two intersecting curves and is defined to be the angle between their orientedtangents at the intersection point And conformality means that the images and of and make the same angle as the curves themselves in both magnitude and direction.

T H E O R E M 1 Conformality of Mapping by Analytic Functions

The mapping by an analytic function f is conformal, except at criticalpoints, that is, points at which the derivative is zero.

P R O O F has a critical point at where and the angles are doubled (seeFig. 378), so that conformality fails.

The idea of proof is to consider a curve

(2)

in the domain of and to show that rotates all tangents at a point (wherethrough the same angle. Now is tangent to C in

(2) because this is the limit of (which has the direction of the secant z1 � z0(z1 � z0)>¢tz�(t) � dz>dt � x�(t) � iy�(t)f r(z0) � 0)

z0w � f (z)f (z)

C: z(t) � x(t) � iy(t)

f r(z) � 2z � 0z � 0,w � z2

f rw � f (z)

C2C1

C*2C*1z0.

C2C1

a (0 � a � p)w � f (z)

738 CHAP. 17 Conformal Mapping

v

u

2

4

–4

–2

–5 5

y = 2 y = 1

x = 1

y = 0

x = 2

y =

x =

x =

12

32

12

Fig. 379. Images of x � const, y � const under w � z2

Fig. 380. Curves and and their respective images and under a conformal mapping w ƒ(z)�C*2C*1

C2C1

c17.qxd 11/1/10 7:35 PM Page 738

y

x

v

u

/nπ

Fig. 382. Mapping by w z n�

Tangent

Curve C

z1 = z(t

0 + Δt)

z0 = z(t

0)

z(t0)

Fig. 381. Secant and tangent of the curve C

in Fig. 381) as approaches along C. The image of C is By the chainrule, Hence the tangent direction of is given by the argument (use (9)in Sec. 13.2)

(3)

where gives the tangent direction of C. This shows that the mapping rotates alldirections at a point in the domain of analyticity of f through the same angle arg which exists as long as But this means conformality, as Fig. 381 illustratesfor an angle between two curves, whose images and make the same angle (becauseof the rotation). �

C*2C*1a

f r(z0) � 0.f r(z0),z0

arg z�

arg w� � arg f r � arg z�

C*w� � f r(z(t))z�(t).w � f (z(t)).C*z0z1


In the remainder of this section and in the next ones we shall consider various conformalmappings that are of practical interest, for instance, in modeling potential problems.

E X A M P L E 2 Conformality of

The mapping is conformal, except at where For this isshown in Fig. 378; we see that at 0 the angles are doubled. For general n the angles at 0 are multiplied by afactor n under the mapping. Hence the sector is mapped by onto the upper half-plane (Fig. 382). �

v � 0zn0 � u � p>n

n � 2wr � nzn�1 � 0.z � 0,w � zn, n � 2, 3, Á ,

w � zn

E X A M P L E 3 Mapping /z. Joukowski Airfoil

In terms of polar coordinates this mapping is

By separating the real and imaginary parts we thus obtain

where

Hence circles are mapped onto ellipses The circle is mappedonto the segment of the u-axis. See Fig. 383.�2 � u � 2

r � 1x2>a2 � y2>b2 � 1.ƒ z ƒ � r � const � 1

a � r �1r

, b � r �1r

.u � a cos u, v � b sin u

w � u � iv � r (cos u � i sin u) �1r

(cos u � i sin u).

w � z � 1

c17.qxd 11/1/10 7:35 PM Page 739

y

x

1

00 1

0.5

v

u

2

3

0 1 2 3–3 –2 –1

D

A

C

B D*A*

C*

= 1

.0φ

= 0.5φB*1

Fig. 385. Mapping by w ez�

y

x

v

u–1 –21 2

C

Now the derivative of w is

which is 0 at These are the points at which the mapping is not conformal. The two circles in Fig. 384pass through The larger is mapped onto a Joukowski airfoil. The dashed circle passes through both and 1 and is mapped onto a curved segment.

Another interesting application of (the flow around a cylinder) will be considered in Sec. 18.4. �w � z � 1>z

�1z � �1.z � �1.

wr � 1 �1

z2�

(z � 1)(z � 1)

z2


1 2 –2 2

v

u

y

x

Fig. 383. Example 3

Fig. 384. Joukowski airfoil

x

y

π

0 0 1–1

v

u

(z-plane) (w-plane)

Fig. 386. Mapping by w ez�

E X A M P L E 4 Conformality of

From (10) in Sec. 13.5 we have and Arg Hence maps a vertical straight line onto the circle and a horizontal straight line onto the ray arg The rectanglein Fig. 385 is mapped onto a region bounded by circles and rays as shown.

The fundamental region of in the z-plane is mapped bijectively and conformally ontothe entire w-plane without the origin (because for no z). Figure 386 shows that the upper half

of the fundamental region is mapped onto the upper half-plane the left half being mapped inside the unit disk and the right half outside (why?). �ƒ w ƒ � 1

0 arg w � p,0 y � pez � 0w � 0

ez�p Arg z � p

w � y0.y � y0 � constƒ w ƒ � ex0

x � x0 � constezz � y.ƒ ezƒ � ex

w � ez

c17.qxd 11/1/10 7:35 PM Page 740

E X A M P L E 5 Principle of Inverse Mapping. Mapping

Principle. The mapping by the inverse of is obtained by interchanging the roles of the z-plane and the w-plane in the mapping by

Now the principal value of the natural logarithm has the inverse FromExample 4 (with the notations z and w interchanged!) we know that maps the fundamental regionof the exponential function onto the z-plane without (because for every w). Hence maps the z-plane without the origin and cut along the negative real axis (where jumps by conformally onto the horizontal strip of the w-plane, where

Since the mapping differs from by the translation (vertically upward), thisfunction maps the z-plane (cut as before and 0 omitted) onto the strip Similarly for each of theinfinitely many mappings The corresponding horizontal strips of width

(images of the z-plane under these mappings) together cover the whole w-plane without overlapping.

Magnification Ratio. By the definition of the derivative we have

(4)

Therefore, the mapping magnifies (or shortens) the lengths of short lines byapproximately the factor The image of a small figure conforms to the originalfigure in the sense that it has approximately the same shape. However, since variesfrom point to point, a large figure may have an image whose shape is quite different fromthat of the original figure.

More on the Condition From (4) in Sec. 13.4 and the Cauchy–Riemannequations we obtain

that is,

(5)

This determinant is the so-called Jacobian (Sec. 10.3) of the transformation written in real form Hence implies that the Jacobianis not 0 at This condition is sufficient that the mapping in a sufficiently smallneighborhood of is one-to-one or injective (different points have different images). SeeRef. [GenRef4] in App. 1.

z0

w � f (z)z0.f r(z0) � 0u � u(x, y), v � v(x, y).

w � f (z)

ƒ f r(z) ƒ2 � 4 0u

0x 0u

0y

0v0x

0v0y

4 � 0(u, v)

0(x, y).

ƒ f r(z) ƒ2 � ` 0u

0x � i

0v0x

` 2 � a 0u0xb2 � a 0v

0xb2 �

0u0x

0v0y

�0u0y

0v0x

(5r)

f �(z) � 0.

f r(z)ƒ f r(z0) ƒ .

w � f (z)

limz:z0

` f (z) � f (z0)z � z0

` � ƒ f r(z0) ƒ .

�2pw � ln z � Ln z � 2npi (n � 0, 1, 2, Á ).

p v � 3p.2piw � Ln zw � Ln z � 2pi

w � u � iv.�p v � p2p)u � Im Ln z

w � f (z) � Ln zew � 0z � 0f �1

(w) � ewz � f �1

(w) � ew.w � f (z) � Ln zw � f (z).

w � f (z)z � f �1 (w)

w � Ln z


1. On Fig. 378. One “rectangle” and its image are colored.Identify the images for the other “rectangles.”

2. On Example 1. Verify all calculations.

3. Mapping Draw an analog of Fig. 378 forw � z3.

w � z3.

4. Conformality. Why do the images of the straight linesand under a mapping by an

analytic function intersect at right angles? Samequestion for the curves and Are there exceptional points?

arg z � const.ƒ z ƒ � const

y � constx � const

P R O B L E M S E T 1 7 . 1

c17.qxd 11/1/10 7:35 PM Page 741

5. Experiment on Find out whether pre-serves angles in size as well as in sense. Try to proveyour result.

6–9 MAPPING OF CURVESFind and sketch or graph the images of the given curvesunder the given mapping.

6.

7. Rotation. Curves as in Prob. 6,

8. Reflection in the unit circle.

9. Translation. Curves as in Prob. 6,

10. CAS EXPERIMENT. Orthogonal Nets. Graph theorthogonal net of the two families of level curves Re and Im where (a)(b) (c) (d)

Why do these curves generally intersect atright angles? In your work, experiment to get the bestpossible graphs. Also do the same for other functionsof your own choice. Observe and record shortcomingsof your CAS and means to overcome such deficiencies.

11–20 MAPPING OF REGIONSSketch or graph the given region and its image under thegiven mapping.

11.

12.

13.

14.

15.

16.

17.

18. �1 � x � 2, �p y p, w � ez

�Ln 2 � x � Ln 4, w � ez

ƒ z ƒ 12 , Im z 0, w � 1>z

ƒ z � 12 ƒ � 1

2 , w � 1>z

x � 1, w � 1>z

2 � Im z � 5, w � iz

1 ƒ z ƒ 3, 0 Arg z p>2, w � z3

ƒ z ƒ � 12 , �p>8 Arg z p>8, w � z2

(1 � iz).f (z) � (z � i)>f (z) � 1>z2,f (z) � 1>z,

f (z) � z4,f (z) � const,f (z) � const

w � z � 2 � i

Arg z � 0, �p>4, �p>2, �3p>2ƒ z ƒ � 1

3 , 12 , 1, 2, 3,

w � iz

x � 1, 2, 3, 4, y � 1, 2, 3, 4, w � z2

w � zw � z .


19.

20.

21–26 FAILURE OF CONFORMALITYFind all points at which the mapping is not conformal. Givereason.

21. A cubic polynomial

22.

23.

24.

25.

26.

27. Magnification of Angles. Let be analytic at Suppose that Then themapping magnifies angles with vertex at bya factor k. Illustrate this with examples for

28. Prove the statement in Prob. 27 for general Hint. Use the Taylor series.

29–35 MAGNIFICATION RATIO, JACOBIANFind the magnification ratio M. Describe what it tellsyou about the mapping. Where is ? Find theJacobian J.

29.

30.

31.

32.

33.

34.

35. w � Ln z

w �z � 1

2z � 2

w � ez

w � 1>z2

w � 1>z

w � z3

w � 12 z2

M � 1

2, Á .k � 1,

k � 2, 3, 4.z0w � f (z)

f r(z0) � 0, Á , f (k�1) (z0) � 0.

z0.f (z)

sin pz

cosh z

exp (z5 � 80z)

z � 12

4z2 � 2

z2 � 1>z2

12 � ƒ z ƒ � 1, 0 � u p>2, w � Ln z

1 ƒ z ƒ 4, p>4 u � 3p>4, w � Ln z

17.2 Linear Fractional Transformations (Möbius Transformations)

Conformal mappings can help in modeling and solving boundary value problems by firstmapping regions conformally onto another. We shall explain this for standard regions(disks, half-planes, strips) in the next section. For this it is useful to know properties ofspecial basic mappings. Accordingly, let us begin with the following very important class.

The next two sections discuss linear fractional transformations. The reason for ourthorough study is that such transformations are useful in modeling and solving boundaryvalue problems, as we shall see in Chapter 18. The task is to get a good grasp of which

c17.qxd 11/1/10 7:35 PM 42

conformal mappings map certain regions conformally onto each other, such as, saymapping a disk onto a half-plane (Sec. 17.3) and so forth. Indeed, the first step in themodeling process of solving boundary value problems is to identify the correct conformalmapping that is related to the “geometry” of the boundary value problem.

The following class of conformal mappings is very important. Linear fractionaltransformations (or Möbius transformations) are mappings

(1)

where a, b, c, d are complex or real numbers. Differentiation gives

(2)

This motivates our requirement It implies conformality for all z and excludesthe totally uninteresting case once and for all. Special cases of (1) are

(3)

(Translations)

(Rotations)

(Linear transformations)

(Inversion in the unit circle).

E X A M P L E 1 Properties of the Inversion /z (Fig. 387)

In polar forms and the inversion is

and gives

Hence the unit circle is mapped onto the unit circle For a generalz the image can be found geometrically by marking on the segment from 0 to z andthen reflecting the mark in the real axis. (Make a sketch.)

Figure 387 shows that maps horizontal and vertical straight lines onto circles or straight lines. Eventhe following is true.

maps every straight line or circle onto a circle or straight line.w � 1>z

w � 1>z

ƒ w ƒ � R � 1>rw � 1>zƒ w ƒ � R � 1; w � ei� � e�iu.ƒ z ƒ � r � 1

R �1

r , � � �u.Rei� �

1

reiu �1

r e�iu

w � 1>zw � Rei�z � reiu

w � 1

w � 1>z

w � az � b

w � az with ƒ a ƒ � 1

w � z � b

wr � 0ad � bc � 0.

wr �a(cz � d) � c(az � b)

(cz � d)2 �

ad � bc

(cz � d)2.

(ad � bc � 0)w �az � b

cz � d

SEC. 17.2 Linear Fractional Transformations (Möbius Transformations) 743

x u

yy = –

v12

–1 –11 1–2 –22 2

1 1

–1

2

–2

–1

x = 12

y = 12

y = 0

x = 0

x = – 12

Fig. 387. Mapping (Inversion) w � 1>z

c17.qxd 11/1/10 7:35 PM Page 743

Proof. Every straight line or circle in the z-plane can be written

(A, B C, D real).

gives a straight line and a circle. In terms of z and this equation becomes

Now Substitution of and multiplication by gives the equation

or, in terms of u and v,

This represents a circle (if or a straight line (if in the w-plane.

The proof in this example suggests the use of z and instead of x and y, a general principlethat is often quite useful in practice.

Surprisingly, every linear fractional transformation has the property just proved:

T H E O R E M 1 Circles and Straight Lines

Every linear fractional transformation (1) maps the totality of circles and straightlines in the z-plane onto the totality of circles and straight lines in the w-plane.

P R O O F This is trivial for a translation or rotation, fairly obvious for a uniform expansion orcontraction, and true for as just proved. Hence it also holds for composites ofthese special mappings. Now comes the key idea of the proof: represent (1) in terms ofthese special mappings. When this is easy. When the representation is

where

This can be verified by substituting K, taking the common denominator and simplifying;this yields (1). We can now set

and see from the previous formula that then This tells us that (1) is indeeda composite of those special mappings and completes the proof.

Extended Complex PlaneThe extended complex plane (the complex plane together with the point in Sec. 16.2)can now be motivated even more naturally by linear fractional transformations as follows.

To each z for which there corresponds a unique w in (1). Now let Then for we have so that no w corresponds to this z. This suggeststhat we let be the image of z � �d>c.w � �

cz � d � 0,z � �d>cc � 0.cz � d � 0

�

�

w � w4 � a>c.

w1 � cz, w2 � w1 � d, w3 �1

w2, w4 � Kw3,

K � � ad � bc

c .w � K 1

cz � d �ac

c � 0,c � 0,

w � 1>z,

z

�D � 0)D � 0)

A � Bu � Cv � D(u2 � v2) � 0.

A � B

w � w

2 � C

w � w

2i � Dww � 0

wwz � 1>ww � 1>z.

Azz � B

z � z

2 � C

z � z

2i � D � 0.

zA � 0A � 0

A (x2 � y2) � Bx � Cy � D � 0


c17.qxd 11/1/10 7:35 PM Page 744

Also, the inverse mapping of (1) is obtained by solving (1) for z; this gives again alinear fractional transformation

(4)

When then for and we let be the image of Withthese settings, the linear fractional transformation (1) is now a one-to-one mapping of theextended z-plane onto the extended w-plane. We also say that every linear fractionaltransformation maps “the extended complex plane in a one-to-one manner onto itself.”

Our discussion suggests the following.

General Remark. If then the right side of (1) becomes the meaningless expressionWe assign to it the value if and if

Fixed PointsFixed points of a mapping are points that are mapped onto themselves, are “keptfixed” under the mapping. Thus they are obtained from

The identity mapping has every point as a fixed point. The mapping hasinfinitely many fixed points, has two, a rotation has one, and a translation nonein the finite plane. (Find them in each case.) For (1), the fixed-point condition is

(5) thus

For this is a quadratic equation in z whose coefficients all vanish if and only if themapping is the identity mapping (in this case, Hence we have

T H E O R E M 2 Fixed Points

A linear fractional transformation, not the identity, has at most two fixed points. Ifa linear fractional transformation is known to have three or more fixed points, it mustbe the identity mapping

To make our present general discussion of linear fractional transformations even moreuseful from a practical point of view, we extend it by further facts and typical examples,in the problem set as well as in the next section.

w � z.

a � d � 0, b � c � 0).w � zc � 0

cz2 � (a � d)z � b � 0.z �az � b

cz � d ,

w � zw � 1>z

w � zw � z

w � f (z) � z.

w � f (z)

c � 0.w � �c � 0w � a>c(a # � � b)>(c # � � d).z � �,

z � �.a>cw � a>c,cw � a � 0c � 0,

z �dw � b

�cw � a.

SEC. 17.2 Linear Fractional Transformations (Möbius Transformations) 745

1. Verify the calculations in the proof of Theorem 1,including those for the case

2. Composition of LFTs. Show that substituting a linearfractional transformation (LFT) into an LFT givesan LFT.

c � 0.3. Matrices. If you are familiar with matrices,

prove that the coefficient matrices of (1) and (4) areinverses of each other, provided that andthat the composition of LFTs corresponds to themultiplication of the coefficient matrices.

ad � bc � 1,

2 � 2

P R O B L E M S E T 1 7 . 2

c17.qxd 11/1/10 7:35 PM Page 745

4. Fig. 387. Find the image of underHint. Use formulas similar to those in

Example 1.

5. Inverse. Derive (4) from (1) and conversely.

6. Fixed points. Find the fixed points mentioned in thetext before formula (5).

7–10 INVERSEFind the inverse Check by solving for w.

7.

8.

9.

10. w �z � 1

2 i

�12 iz � 1

w �z � i

3iz � 4

w �z � iz � i

w �i

2z � 1

z(w)z � z(w).

w � 1>z.x � k � const


11–16 FIXED POINTSFind the fixed points.

11.

12.

13.

14.

15.

16.

17–20 FIXED POINTSFind all LFTs with fixed point(s).

17. 18.

19.

20. Without any fixed points

z � �i

z � �1z � 0

w �aiz � 1

z � ai , a � 1

w �iz � 4

2z � 5i

w � az � b

w � 16z5

w � z � 3i

w � (a � ib)z2

17.3 Special Linear Fractional TransformationsWe continue our study of linear fractional transformations. We shall identify linear fractionaltransformations

(1)

that map certain standard domains onto others. Theorem 1 (below) will give us a tool forconstructing desired linear fractional transformations.

A mapping (1) is determined by a, b, c, d, actually by the ratios of three of these constantsto the fourth because we can drop or introduce a common factor. This makes it plausiblethat three conditions determine a unique mapping (1):

T H E O R E M 1 Three Points and Their Images Given

Three given distinct points can always be mapped onto three prescribeddistinct points by one, and only one, linear fractional transformation

This mapping is given implicitly by the equation

(2)

(If one of these points is the point , the quotient of the two differences containingthis point must be replaced by 1.)

P R O O F Equation (2) is of the form with linear fractional F and G. Hencewhere is the inverse of F and is linear fractional (see (4) inF�1w � F�1(G(z)) � f (z),

F(w) � G(z)

�

w � w1

w � w3 #

w2 � w3

w2 � w1 �

z � z1

z � z3 #

z2 � z3

z2 � z1 .

w � f (z).w1, w2, w3

z1, z2, z3

(ad � bc � 0)w �az � b

cz � d

c17.qxd 11/1/10 7:35 PM Page 746

Sec. 17.2) and so is the composite (by Prob. 2 in Sec. 17.2), that is, is linear fractional. Now if in (2) we set on the left and onthe right, we see that

From the first column, thus Similarly, This proves the existence of the desired linear fractional transformation.

To prove uniqueness, let be a linear fractional transformation, which alsomaps onto Thus Hence where Together, a mapping with the three fixed points By Theorem 2in Sec. 17.2, this is the identity mapping, for all z. Thus for allz, the uniqueness.

The last statement of Theorem 1 follows from the General Remark in Sec. 17.2.

Mapping of Standard Domains by Theorem 1Using Theorem 1, we can now find linear fractional transformations of some practicallyuseful domains (here called “standard domains”) according to the following principle.

Principle. Prescribe three boundary points of the domain D in the z-plane.Choose their images on the boundary of the image of D in the w-plane.Obtain the mapping from (2). Make sure that D is mapped onto not onto itscomplement. In the latter case, interchange two w-points. (Why does this help?)

D*,D*w1, w2, w3

z1, z2, z3

�

f (z) � g(z)g�1( f (z)) � zz1, z2, z3.g�1( f (z j)) � z j,

wj � f (z j).g�1(wj) � z j,wj � g(z j).wj, j � 1, 2, 3.z j

w � g(z)w3 � f (z3).

w2 � f (z2),w1 � F�1(G(z1)) � f (z1).F(w1) � G(z1),

F(w1) � 0, F(w2) � 1, F(w3) � �

G(z1) � 0, G(z2) � 1, G(z3) � �.

z � z1, z2, z3w � w1, w2, w3

w � f (z)F�1(G (z))

SEC. 17.3 Special Linear Fractional Transformations 747

v

u

x = –2

x = –1 x = 1

x = 2

y = 5

y = 0

x = 0

y = 1

y =

x = – x =

1

12

12

12

Fig. 388. Linear fractional transformation in Example 1

E X A M P L E 1 Mapping of a Half-Plane onto a Disk (Fig. 388)

Find the linear fractional transformation (1) that maps onto respectively.

Solution. From (2) we obtain

w � (�1)

w � 1 #

�i � 1

�i � (�1) �

z � (�1)

z � 1 #

0 � 1

0 � (�1) ,

w3 � 1,w1 � �1, w2 � �i,z1 � �1, z2 � 0, z3 � 1

c17.qxd 11/1/10 7:35 PM Page 747

thus

Let us show that we can determine the specific properties of such a mapping without much calculation. Forwe have thus so that the x-axis maps onto the unit circle. Since

gives the upper half-plane maps onto the interior of that circle and the lower half-plane onto the exterior.go onto so that the positive imaginary axis maps onto the segment S:

The vertical lines map onto circles (by Theorem 1, Sec. 17.2) through (the image of ) andperpendicular to (by conformality; see Fig. 388). Similarly, the horizontal lines map ontocircles through and perpendicular to S (by conformality). Figure 388 gives these circles for and for

they lie outside the unit disk shown.

E X A M P L E 2 Occurrence of

Determine the linear fractional transformation that maps onto respectively.

Solution. From (2) we obtain the desired mapping

This is sometimes called the Cayley transformation.2 In this case, (2) gave at first the quotient which we had to replace by 1.

E X A M P L E 3 Mapping of a Disk onto a Half-Plane

Find the linear fractional transformation that maps onto respectively, such that the unit disk is mapped onto the right half-plane. (Sketch disk and half-plane.)

Solution. From (2) we obtain, after replacing by 1,

Mapping half-planes onto half-planes is another task of practical interest. For instance,we may wish to map the upper half-plane onto the upper half-plane Thenthe x-axis must be mapped onto the u-axis.

E X A M P L E 4 Mapping of a Half-Plane onto a Half-Plane

Find the linear fractional transformation that maps onto respectively.

Solution. You may verify that (2) gives the mapping function

What is the image of the x-axis? Of the y-axis?

Mappings of disks onto disks is a third class of practical problems. We may readilyverify that the unit disk in the z-plane is mapped onto the unit disk in the w-plane by thefollowing function, which maps onto the center w � 0.z0

�

w �z � 1

2z � 4

w1 � �, w2 � 14 , w3 � 3

8 ,z1 � �2, z2 � 0, z3 � 2

v � 0.y � 0

�w � �

z � 1

z � 1 .

(i � �)>(w � �)

w1 � 0, w2 � i, w3 � �,z1 � �1, z2 � i, z3 � 1

�(1 � �)>(z � �),

w �z � i

z � i .

w3 � 1,w1 � �1, w2 � �i,z1 � 0, z2 � 1, z3 � �

�

�y 0y � 0,w � i

y � constƒ w ƒ � 1z � �w � ix � const

u � 0, �1 � v � 1.w � �i, 0, i,z � 0, i, �w � 0,

z � iƒ w ƒ � 1,w � (x � i)>(�ix � 1),z � x

w �z � i

�iz � 1 .


2ARTHUR CAYLEY (1821–1895), English mathematician and professor at Cambridge, is known for hisimportant work in algebra, matrix theory, and differential equations.

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(3)

To see this, take obtaining, with as in (3),

Hence

from (3), so that maps onto as claimed, with going onto 0, as thenumerator in (3) shows.

Formula (3) is illustrated by the following example. Another interesting case will begiven in Prob. 17 of Sec. 18.2.

E X A M P L E 5 Mapping of the Unit Disk onto the Unit Disk

Taking in (3), we obtain (verify!)

(Fig. 389). �w �2z � 1

z � 2

z0 � 12

z0ƒ w ƒ � 1,ƒ z ƒ � 1

ƒ w ƒ � ƒ z � z0 ƒ > ƒ cz � 1 ƒ � 1

� ƒ zz � cz ƒ � ƒ 1 � cz ƒ � ƒ cz � 1 ƒ .

� ƒ z ƒ ƒ z � c ƒ

ƒ z � z0 ƒ � ƒ z � c ƒ

c � z0ƒ z ƒ � 1,

ƒ z0 ƒ 1.c � z0,w �z � z0

cz � 1 ,

SEC. 17.3 Special Linear Fractional Transformations 749

y

x

y = –x =

x = –

y =

B A1 1 2

A* B*

x = 0

y = 0

v

u

12

12

12

12

Fig. 389. Mapping in Example 5

E X A M P L E 6 Mapping of an Angular Region onto the Unit Disk

Certain mapping problems can be solved by combining linear fractional transformations with others. For instance,to map the angular region D: (Fig. 390) onto the unit disk we may map D by

onto the right Z-half-plane and then the latter onto the disk by

combined �w � i

z3 � 1

z3 � 1 .w � i

Z � 1

Z � 1 ,

ƒ w ƒ � 1Z � z3ƒ w ƒ � 1,�p>6 � arg z � p>6

c17.qxd 11/1/10 7:35 PM Page 749

This is the end of our discussion of linear fractional transformations. In the next sectionwe turn to conformal mappings by other analytic functions (sine, cosine, etc.).


/6π

(z-plane) (Z -plane) (w-plane)

Fig. 390. Mapping in Example 6

1. CAS EXPERIMENT. Linear Fractional Transfor-mations (LFTs). (a) Graph typical regions (squares,disks, etc.) and their images under the LFTs inExamples 1–5 of the text.

(b) Make an experimental study of the continuousdependence of LFTs on their coefficients. For instance,change the LFT in Example 4 continuously and graphthe changing image of a fixed region (applying animationif available).

2. Inverse. Find the inverse of the mapping in Example 1.Show that under that inverse the lines arethe images of circles in the w-plane with centers on theline

3. Inverse. If is any transformation that has aninverse, prove the (trivial!) fact that f and its inversehave the same fixed points.

4. Obtain the mapping in Example 1 of this section fromProb. 18 in Problem Set 17.2.

5. Derive the mapping in Example 2 from (2).

6. Derive the mapping in Example 4 from (2). Find itsinverse and the fixed points.

7. Verify the formula for disks.

w � f (z)

v � 1.

x � const

8–16 LFTs FROM THREE POINTS AND IMAGES Find the LFT that maps the given three points onto the threegiven points in the respective order.

8. 0, 1, 2 onto

9. onto

10. onto

11. onto

12. onto

13. onto

14. onto

15. onto

16. onto

17. Find an LFT that maps onto so thatis mapped onto Sketch the images of

the lines and

18. Find all LFTs that map the x-axis onto the u-axis.

19. Find an analytic function that maps the regiononto the unit disk

20. Find an analytic function that maps the second quadrantof the z-plane onto the interior of the unit circle in thew-plane.

ƒ w ƒ � 1.0 � arg z � p>4w � f (z)

w(z)

y � const.x � constw � 0.z � i>2

ƒ w ƒ � 1ƒ z ƒ � 1

0, 32 , 1�32 , 0, 1

0, �i � 1, �12 1, i, 2

1, 1 � i, 1 � 2i�1, 0, 1

�, 1, 00, 1, �

�1, 0, �0, 2i, �2i

�i, �1, i�1, 0, 1

�1, 0, �0, �i, i

i, �1, �i1, i, �1

1, 12 , 13

P R O B L E M S E T 1 7 . 3

17.4 Conformal Mapping by Other FunctionsWe shall now cover mappings by trigonometric and hyperbolic analytic functions. So farwe have covered the mappings by and (Sec. 17.1) as well as linear fractionaltransformations (Secs. 17.2 and 17.3).

Sine Function. Figure 391 shows the mapping by

(1) (Sec. 13.6).w � u � iv � sin z � sin x cosh y � i cos x sinh y

ezzn

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Hence

(2)

Since is periodic with period the mapping is certainly not one-to-one if weconsider it in the full z-plane. We restrict z to the vertical strip inFig. 391. Since at the mapping is not conformal at these twocritical points. We claim that the rectangular net of straight lines and in Fig. 391 is mapped onto a net in the w-plane consisting of hyperbolas (the images ofthe vertical lines and ellipses (the images of the horizontal lines intersecting the hyperbolas at right angles (conformality!). Corresponding calculations aresimple. From (2) and the relations and weobtain

(Hyperbolas)

(Ellipses).

Exceptions are the vertical lines which are “folded” onto andrespectively.

Figure 392 illustrates this further. The upper and lower sides of the rectangle are mappedonto semi-ellipses and the vertical sides onto and

respectively. An application to a potential problem will be given in Prob. 3 ofSec. 18.2.(v � 0),

1 � u � cosh 1�cosh 1 � u � �1

u � 1 (v � 0),u � �1x � �

12 px � 1

2 p,

u2

cosh2 y �

v2

sinh2 y � sin2 x � cos2 x � 1

u2

sin2 x �

v2

cos2 x � cosh2 y � sinh2 y � 1

cosh2 y � sinh2 y � 1sin2 x � cos2 x � 1

y � const)x � const)

y � constx � constz � �1

2 p,f r(z) � cos z � 0S: �1

2 p � x � 12 p

2p,sin z

v � cos x sinh y.u � sin x cosh y,

SEC. 17.4 Conformal Mapping by Other Functions 751

v

u–

1

–1

–2

–1

1

2

(z-plane) (w-plane)

y

xπ2

π2

1–1

Fig. 391. Mapping w � u � iv � sin z

y

x

v

uA

BC

D

E F

1

–1

C*E*

B*F*

D* A*1–1π

2π2

–

Fig. 392. Mapping by w � sin z

c17.qxd 11/1/10 7:35 PM Page 751

Cosine Function. The mapping could be discussed independently, but since

(3)

we see at once that this is the same mapping as preceded by a translation to the rightthrough units.

Hyperbolic Sine. Since

(4)

the mapping is a counterclockwise rotation through (i.e., followed by thesine mapping followed by a clockwise -rotation

Hyperbolic Cosine. This function

(5)

defines a mapping that is a rotation followed by the mapping Figure 393 shows the mapping of a semi-infinite strip onto a half-plane by

Since the point is mapped onto For real isreal and increases with increasing x in a monotone fashion, starting from 1. Hence thepositive x-axis is mapped onto the portion of the u-axis.

For pure imaginary we have Hence the left boundary of the stripis mapped onto the segment of the u-axis, the point corresponding to

On the upper boundary of the strip, and since and itfollows that this part of the boundary is mapped onto the portion of the u-axis.Hence the boundary of the strip is mapped onto the u-axis. It is not difficult to see thatthe interior of the strip is mapped onto the upper half of the w-plane, and the mapping isone-to-one.

This mapping in Fig. 393 has applications in potential theory, as we shall see in Prob. 12of Sec. 18.3.

u � �1cos p � �1,sin p � 0y � p,

w � cosh ip � cos p � �1.

z � pi1 � u � �1cosh iy � cos y.z � iy

u � 1

z � x � 0, cosh zw � 1.z � 0cosh 0 � 1,w � cosh z.

w � cos Z.Z � iz

w � cosh z � cos (iz)

w � �iZ*.90°Z* � sin Z,90°),1

2 pZ � iz

w � sinh z � �i sin (iz),

12p

sin z

w � cos z � sin (z � 12 p),

w � cos z


0

v

ux

y

B* A*

1–1A

Bπ

Fig. 393. Mapping by w � cosh z

Tangent Function. Figure 394 shows the mapping of a vertical infinite strip onto theunit circle by accomplished in three steps as suggested by the representation(Sec. 13.6)

w � tan z �sin z

cos z �

(eiz � e�iz)>i

eiz � e�iz �(e2iz � 1)>i

e2iz � 1 .

w � tan z,

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Hence if we set and use we have

(6)

We now see that is a linear fractional transformation preceded by an exponentialmapping (see Sec. 17.1) and followed by a clockwise rotation through an angle

The strip is and we show that it is mapped onto the unit disk inthe w-plane. Since we see from (10) in Sec. 13.5 that

Hence the vertical lines are mapped onto the raysrespectively. Hence S is mapped onto the right Z-half-plane. Also

if and if Hence the upper half of S is mapped insidethe unit circle and the lower half of S outside as shown in Fig. 394.

Now comes the linear fractional transformation in (6), which we denote by

(7)

For real Z this is real. Hence the real Z-axis is mapped onto the real W-axis. Furthermore,the imaginary Z-axis is mapped onto the unit circle because for pure imaginary

we get from (7)

The right Z-half-plane is mapped inside this unit circle not outside, becausehas its image inside that circle. Finally, the unit circle is mapped

onto the imaginary W-axis, because this circle is so that (7) gives a pure imaginaryexpression, namely,

From the W-plane we get to the w-plane simply by a clockwise rotation through see (6).Together we have shown that maps onto the unit

disk with the four quarters of S mapped as indicated in Fig. 394. This mappingis conformal and one-to-one.

ƒ w ƒ 1,S: �p>4 Re z p>4w � tan z

p>2;

g(ei�) �ei� � 1

ei� � 1 �

ei�>2 � e�i�>2

ei�>2 � e�i�>2 �i sin (�>2)

cos (�>2) .

Z � ei�,ƒ Z ƒ � 1g(1) � 0Z � 1

ƒ W ƒ � 1,

ƒ W ƒ � ƒ g(iY) ƒ � ` iY � 1iY � 1

` � 1.

Z � iYƒ W ƒ � 1

W � g(Z) �Z � 1Z � 1

.

g(Z):ƒ Z ƒ � 1,ƒ Z ƒ � 1

y 0.ƒ Z ƒ 1y 0ƒ Z ƒ � e�2y 1Arg Z � �p>2, 0, p>2,

x � �p>4, 0, p>4Arg Z � 2x.ƒ Z ƒ � e�2y,Z � e2iz � e�2y�2ix,

S : � 14 p x 1

4 p,

12p(90°).

w � tan z

Z � e2iz.W �Z � 1Z � 1

,w � tan z � �iW,

1>i � �i,Z � e2iz

SEC. 17.4 Conformal Mapping by Other Functions 753

y

x

v

u

(z-plane) (Z-plane) (W-plane) (w-plane)

Fig. 394. Mapping by w � tan z

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17.5 Riemann Surfaces. OptionalOne of the simplest but most ingeneous ideas in complex analysis is that of Riemannsurfaces. They allow multivalued relations, such as or to becomesingle-valued and therefore functions in the usual sense. This works because the Riemannsurfaces consist of several sheets that are connected at certain points (called branch points).Thus will need two sheets, being single-valued on each sheet. How many sheetsdo you think needs? Can you guess, by recalling Sec. 13.7? (The answer willbe given at the end of this section). Let us start our systematic discussion.

The mapping given by

(1) (Sec. 17.1)w � u � iv � z2

w � ln zw � 1z

w � ln z,w � 1z


CONFORMAL MAPPING

1. Find the image of under

2. Find the image of under

3–7 Find and sketch the image of the given regionunder

3.

4.

5.

6.

7.

8. CAS EXPERIMENT. Conformal Mapping. If yourCAS can do conformal mapping, use it to solve Prob. 7.Then increase y beyond say, to or Statewhat you expected. See what you get as the image.Explain.

CONFORMAL MAPPING

9. Find the points at which is not conformal.

10. Sketch or graph the images of the lines under the mapping

11–14 Find and sketch or graph the image of the givenregion under

11.

12.

13.

14.

15. Describe the mapping in terms of the map-ping and rotations and translations.

16. Find all points at which the mapping isnot conformal.

w � cosh 2pz

w � sin zw � cosh z

0 x p>6, �� y �

0 x 2p, 1 y 3

�p>4 x p>4, 0 y 1

0 x p>2, 0 y 2

w � sin z.

w � sin z.�p>2�p>3,�p>6,x � 0,

w � sin z

w � sin z

100p.50pp,

0 x 1, 0 y p

0 x �, 0 y p>2

�� x �, 0 � y � 2p

0 x 1, 12 y 1

�12 � x � 1

2 , �p � y � p

w � ez.

w � ez.y � k � const, �� x � �,

w � ez.x � c � const, �p y � p,

w � ez 17. Find an analytic function that maps the region Rbounded by the positive x- and y-semi-axes and thehyperbola in the first quadrant onto the upperhalf-plane. Hint. First map R onto a horizontal strip.

CONFORMAL MAPPING

18. Find the images of the lines under themapping

19. Find the images of the lines under themapping

20–23 Find and sketch or graph the image of the givenregion under the mapping

20.

21. directly and from Prob. 11

22.

23.

24. Find and sketch the image of the region under the mapping

25. Show that maps the upper half-plane

onto the horizontal strip as shown inthe figure.

0 � Im w � p

w � Ln z � 1

z � 1

w � Ln z.p>4 � u � p>22 � ƒ z ƒ � 3,

p x 2p, y 0

�1 x 1, 0 � y � 1

0 x p>2, 0 y 2

0 x 2p, 12 y 1

w � cos z.

w � cos z.x � c � const

w � cos z.y � k � const

w � cos z

xy � p

P R O B L E M S E T 1 7 . 4

BA C D E

0 1–1(z-plane)

(∞) (∞)

C*

0(w-plane)

E* = A*D*(∞) B*(∞)

πi

Problem 25

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is conformal, except at where At angles are doubled underthe mapping. Thus the right z-half-plane (including the positive y-axis) is mapped ontothe full w-plane, cut along the negative half of the u-axis; this mapping is one-to-one.Similarly for the left z-half-plane (including the negative y-axis). Hence the image of thefull z-plane under “covers the w-plane twice” in the sense that every is theimage of two z-points; if is one, the other is For example, and are bothmapped onto

Now comes the crucial idea. We place those two copies of the cut w-plane upon eachother so that the upper sheet is the image of the right half z-plane R and the lower sheetis the image of the left half z-plane L. We join the two sheets crosswise along the cuts(along the negative u-axis) so that if z moves from R to L, its image can move from theupper to the lower sheet. The two origins are fastened together because is the imageof just one z-point, The surface obtained is called a Riemann surface (Fig. 395a).

is called a “winding point” or branch point. maps the full z-plane ontothis surface in a one-to-one manner.

By interchanging the roles of the variables z and w it follows that the double-valuedrelation

(2) (Sec. 13.2)

becomes single-valued on the Riemann surface in Fig. 395a, that is, a function in the usualsense. We can let the upper sheet correspond to the principal value of Its image isthe right w-half-plane. The other sheet is then mapped onto the left w-half-plane.

1z.

w � 1z

w � z2w � 0z � 0.

w � 0

w � �1.�iz � i�z1.z1

w � 0w � z2

z � 0,wr � 2z � 0,z � 0,

SEC. 17.5 Riemann Surfaces. Optional 755

(a) Riemann surface of z (b) Riemann surface of z3

Fig. 395. Riemann surfaces

Similarly, the triple-valued relation becomes single-valued on the three-sheetedRiemann surface in Fig. 395b, which also has a branch point at

The infinitely many-valued natural logarithm (Sec. 13.7)

becomes single-valued on a Riemann surface consisting of infinitely many sheets,corresponds to one of them. This sheet is cut along the negative x-axis and the

upper edge of the slit is joined to the lower edge of the next sheet, which corresponds tothe argument that is, to

The principal value Ln z maps its sheet onto the horizontal strip Thefunction maps its sheet onto the neighboring strip and soon. The mapping of the points of the Riemann surface onto the points of the w-planeis one-to-one. See also Example 5 in Sec. 17.1.

z � 0p v � 3p,w � Ln z � 2pi

�p v � p.

w � Ln z � 2pi.

p u � 3p,

w � Ln z

(n � 0, �1, �2, Á )w � ln z � Ln z � 2npi

z � 0.w � 1

3 z

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1. If z moves from twice around the circle what does do?

2. Show that the Riemann surface of has branch points at 1 and 2 sheets,

which we may cut and join crosswise from 1 to 2.Hint. Introduce polar coordinates and

so that

3. Make a sketch, similar to Fig. 395, of the Riemannsurface of w � 1

4 z � 1.

w � 1r1r2 ei(u1�u2)>2.z � 2 � r2eiu2,z � 1 � r1eiu1

1(z � 1)(z � 2)w �

w � 1zƒ z ƒ � 1

4 ,z � 14 4–10 RIEMANN SURFACES

Find the branch points and the number of sheets of theRiemann surface.

4. 5.

6. 7.

8. 9.

10. 2(4 � z2)(1 � z2)

2z3 � ze1z, 2ez

2n

z � z0ln (6z � 2i)

z2 � 23 4z � i1iz � 2 � i

P R O B L E M S E T 1 7 . 5

1. What is a conformal mapping? Why does it occur incomplex analysis?

2. At what points are and notconformal?

3. What happens to angles at under a mapping if

4. What is a linear fractional transformation? What canyou do with it? List special cases.

5. What is the extended complex plane? Ways of intro-ducing it?

6. What is a fixed point of a mapping? Its role in thischapter? Give examples.

7. How would you find the image of under?

8. Can you remember mapping properties of

9. What mapping gave the Joukowski airfoil? Explaindetails.

10. What is a Riemann surface? Its motivation? Its simplestexample.

11–16 MAPPING Find and sketch the image of the given region or curveunder

11.

12.

13. 14.

15. 16.

17–22 MAPPING w � 1/zFind and sketch the image of the given region or curveunder .

17.

18.

19. 20. 0 � arg z � p>42 ƒ z ƒ 3, y 0

ƒ z ƒ 1, 0 arg z p>2

ƒ z ƒ 1

w � 1>z

y � �2, 2x � �1, 1

0 y 2�4 xy 4

1>1p ƒ z ƒ 1p, 0 arg z p>2

1 ƒ z ƒ 2, ƒ arg z ƒ p>8

w � z2.

w � z2

w � ln z?

w � iz, z2, ez, 1>zx � Re z � 1

f r(z0) � 0, f s (z0) � 0, f t (z0) � 0?w � f (z)z0

w � cos (pz2)w � z5 � z

21.

22.

23–28 LINEAR FRACTIONALTRANSFORMATIONS (LFTs)

Find the LFT that maps

23. onto respectively






29–34 FIXED POINTSFind the fixed points of the mapping

29. 30.

31. 32.

33.

34.

35–40 GIVEN REGIONSFind an analytic function that maps

35. The infinite strip onto the upper half-plane

36. The quarter-disk onto the exteriorof the unit circle

37. The sector onto the region

38. The interior of the unit circle onto the exteriorof the circle

39. The region onto the strip

40. The semi-disk onto the exterior of thecircle ƒ w � p ƒ � p.

ƒ z ƒ 2, y 0

v 1.0 x 0, y 0, xy c

ƒ w � 2 ƒ � 2.ƒ z ƒ � 1

u 1.0 arg z p>2

ƒ w ƒ � 1.ƒ z ƒ 1, x 0, y 0

v 0.0 y p>4

w � f (z)

w � (iz � 5)>(5z � i)

w � z5 � 10z3 � 10z

(2iz � 1)>(z � 2i)w � (3z � 2)>(z � 1)

w � z4 � z � 64w � (2 � i)z

1 � i, 2, 0,�1, �i, i

�, 1, 0,0, 1, �

2i, 1 � 2i, 2 � 2i,0, 1, 2

i, �1, 1,1, i, �i

�, 12 , 14 ,0, 2, 4

4 � 3i, 5i>2, 4 � 3i,�1, 0, 1

z � 1 � iy (�� y �)

(x � 12 )2 � y2 � 1

4 , y 0

C H A P T E R 1 7 R E V I E W Q U E S T I O N S A N D P R O B L E M S

c17.qxd 11/1/10 7:36 PM Page 756

Summary of Chapter 17 757

A complex function gives a mapping of its domain of definition in thecomplex z-plane onto its range of values in the complex w-plane. If is analytic,this mapping is conformal, that is, angle-preserving: the images of any twointersecting curves make the same angle of intersection, in both magnitude and sense,as the curves themselves (Sec. 17.1). Exceptions are the points at which (“critical points,” e.g. for

For mapping properties of etc. see Secs. 17.1 and 17.4.Linear fractional transformations, also called Möbius transformations

(1) (Secs. 17.2, 17.3)

map the extended complex plane (Sec. 17.2) onto itself. They solvethe problems of mapping half-planes onto half-planes or disks, and disks onto disksor half-planes. Prescribing the images of three points determines (1) uniquely.

Riemann surfaces (Sec. 17.5) consist of several sheets connected at certain pointscalled branch points. On them, multivalued relations become single-valued, that is,functions in the usual sense. Examples. For we need two sheets (with branchpoint 0) since this relation is doubly-valued. For we need infinitely manysheets since this relation is infinitely many-valued (see Sec. 13.7).

w � ln zw � 1z

(ad � bc � 0)

w �az � b

cz � d

ez, cos z, sin zw � z2).z � 0

f r(z) � 0

f (z)w � f (z)

SUMMARY OF CHAPTER 17Conformal Mapping

c17.qxd 11/1/10 7:36 PM Page 757

758

C H A P T E R 1 8

Complex Analysisand Potential Theory

In Chapter 17 we developed the geometric approach of conformal mapping. This meantthat, for a complex analytic function defined in a domain D of the z-plane, weassociated with each point in D a corresponding point in the w-plane. This gave us aconformal mapping (angle-preserving), except at critical points where

Now, in this chapter, we shall apply conformal mappings to potential problems. Thiswill lead to boundary value problems and many engineering applications in electrostatics,heat flow, and fluid flow. More details are as follows.

Recall that Laplace’s equation is one of the most important PDEs inengineering mathematics because it occurs in gravitation (Secs. 9.7, 12.11), electrostatics(Sec. 9.7), steady-state heat conduction (Sec. 12.5), incompressible fluid flow, and otherareas. The theory of this equation is called potential theory (although “potential” is alsoused in a more general sense in connection with gradients (see Sec. 9.7)). Because wewant to treat this equation with complex analytic methods, we restrict our discussion tothe “two-dimensional case.” Then depends only on two Cartesian coordinates x and y,and Laplace’s equation becomes

An important idea then is that its solutions are closely related to complex analyticfunctions as shown in Sec. 13.4. (Remark: We use the notation to freeu and v, which will be needed in conformal mapping This important relation isthe main reason for using complex analysis in problems of physics and engineering.

We shall examine this connection between Laplace’s equation and complex analyticfunctions and illustrate it by modeling applications from electrostatics (Secs. 18.1,18.2), heat conduction (Sec. 18.3), and hydrodynamics (Sec. 18.4). This in turn willlead to boundary value problems in two-dimensional potential theory. As a result,some of the functions of Chap. 17 will be used to transform complicated regions intosimpler ones.

Section 18.5 will derive the important Poisson formula for potentials in a circular disk.Section 18.6 will deal with harmonic functions, which, as you recall, are solutions ofLaplace’s equation and have continuous second partial derivatives. In that section we willshow how results on analytic functions can be used to characterize properties of harmonicfunctions.

Prerequisite: Chaps. 13, 14, 17.References and Answers to Problems: App. 1 Part D, App. 2.

u � iv.)£ � i°£ � i°

£

�2£ � £xx � £yy � 0.

£

�2£ � 0

f r(z) � 0.

w � f (z)

c18.qxd 11/2/10 6:54 PM Page 758

18.1 Electrostatic FieldsThe electrical force of attraction or repulsion between charged particles is governed byCoulomb’s law (see Sec. 9.7). This force is the gradient of a function called theelectrostatic potential. At any points free of charges, is a solution of Laplace’s equation

The surfaces are called equipotential surfaces. At each point P at whichthe gradient of is not the zero vector, it is perpendicular to the surface through P; that is, the electrical force has the direction perpendicular to the equipotentialsurface. (See also Secs. 9.7 and 12.11.)

The problems we shall discuss in this entire chapter are two-dimensional (for thereason just given in the chapter opening), that is, they model physical systems that liein three-dimensional space (of course!), but are such that the potential is independentof one of the space coordinates, so that depends only on two coordinates, which wecall x and y. Then Laplace’s equation becomes

(1)

Equipotential surfaces now appear as equipotential lines (curves) in the xy-plane.Let us illustrate these ideas by a few simple examples.

E X A M P L E 1 Potential Between Parallel Plates

Find the potential of the field between two parallel conducting plates extending to infinity (Fig. 396), whichare kept at potentials and respectively.

Solution. From the shape of the plates it follows that depends only on x, and Laplace’s equation becomesBy integrating twice we obtain where the constants a and b are determined by the given

boundary values of on the plates. For example, if the plates correspond to and the solution is

The equipotential surfaces are parallel planes.

E X A M P L E 2 Potential Between Coaxial Cylinders

Find the potential between two coaxial conducting cylinders extending to infinity on both ends (Fig. 397)and kept at potentials and respectively.

Solution. Here depends only on for reasons of symmetry, and Laplace’s equation[(5), Sec. 12.10] with and becomes By separating

variables and integrating we obtain

and a and b are determined by the given values of on the cylinders. Although no infinitely extended conductorsexist, the field in our idealized conductor will approximate the field in a long finite conductor in that part whichis far away from the two ends of the cylinders. �

£

£s£ r � �

1

r , ln £ r � �ln r � a�, £ r �

a

r , £ � a ln r � b

r£s � £ r � 0.u � £uuu � 0r 2urr � rur � uuu � 0r � 2x2 � y2,£

£2,£1

£

�

£(x) � 12 (£2 � £1)x � 1

2 (£2 � £1) .

x � 1,x � �1£

£ � ax � b,£s � 0.£

£2,£1

£

�2£ �

02£

0x2�

02£

0y2� 0.

£

£

£ � const£

£ � const

�2£ � 0.

£

£,

SEC. 18.1 Electrostatic Fields 759

x

y

Fig. 396. Potentialin Example 1

c18.qxd 11/2/10 6:55 PM Page 759

E X A M P L E 3 Potential in an Angular Region

Find the potential between the conducting plates in Fig. 398, which are kept at potentials (the lower plate)and and make an angle where (In the figure we have

Solution. is constant on rays It is harmonic since it is the imaginarypart of an analytic function, (Sec. 13.7). Hence the solution is

with a and b determined from the two boundary conditions (given values on the plates)

Thus The answer is

Complex PotentialLet be harmonic in some domain D and a harmonic conjugate of in D.(Note the change of notation from u and v of Sec. 13.4 to and . From the next sectionon, we had to free u and v for use in conformal mapping. Then

(2)

is an analytic function of This function F is called the complex potentialcorresponding to the real potential Recall from Sec. 13.4 that for given a conjugate

is uniquely determined except for an additive real constant. Hence we may say thecomplex potential, without causing misunderstandings.

The use of F has two advantages, a technical one and a physical one. Technically, Fis easier to handle than real or imaginary parts, in connection with methods of complexanalysis. Physically, has a meaning. By conformality, the curves intersectthe equipotential lines in the xy-plane at right angles [except where Hence they have the direction of the electrical force and, therefore, are called linesof force. They are the paths of moving charged particles (electrons in an electronmicroscope, etc.).

F r(z) � 0].£ � const° � const°

°

£,£.z � x � iy.

F(z) � £(x, y) � i°(x, y)

°£

£°(x, y)£(x, y)

�u � arctan y

x .£(x, y) �

12

(£2 � £1) �1a

(£2 � £1) u,

a � (£2 � £1)>2, b � (£2 � £1)>a.

a � b(12 a) � £2 .a � b(�1

2 a) � £1,

£(x, y) � a � b Arg z

Ln zu � const.u � Arg z (z � x � iy � 0)

a � 120° � 2p>3.)0 � a � p.a,£2,£1£

760 CHAP. 18 Complex Analysis and Potential Theory

x

y

Fig. 398. Potentialin Example 3

x

y

Fig. 397. Potential in Example 2

c18.qxd 11/2/10 6:55 PM Page 760

E X A M P L E 4 Complex Potential

In Example 1, a conjugate is It follows that the complex potential is

and the lines of force are horizontal straight lines parallel to the x-axis.


In Example 2 we have A conjugate is Hence the complexpotential is

and the lines of force are straight lines through the origin. may also be interpreted as the complex potentialof a source line (a wire perpendicular to the xy-plane) whose trace in the xy-plane is the origin.


In Example 3 we get by noting that multiplying this by and adding a:

We see from this that the lines of force are concentric circles Can you sketch them?

SuperpositionMore complicated potentials can often be obtained by superposition.

E X A M P L E 7 Potential of a Pair of Source Lines (a Pair of Charged Wires)

Determine the potential of a pair of oppositely charged source lines of the same strength at the points andon the real axis.

Solution. From Examples 2 and 5 it follows that the potential of each of the source lines is

and

respectively. Here the real constant K measures the strength (amount of charge). These are the real parts of thecomplex potentials

and

Hence the complex potential of the combination of the two source lines is

(3)

The equipotential lines are the curves

thus

These are circles, as you may show by direct calculation. The lines of force are

We write this briefly (Fig. 399)

° � K(u1 � u2) � const.

° � Im F(z) � K[Arg (z � c) � Arg (z � c)] � const.

` z � c

z � c` � const.£ � Re F(z) � K ln ` z � c

z � c` � const,

F(z) � F1(z) � F2(z) � K [Ln (z � c) � Ln (z � c)] .

F2(z) � �K Ln (z � c).F1(z) � K Ln (z � c)

£2 � �K ln ƒ z � c ƒ ,£1 � K ln ƒ z � c ƒ

z � �cz � c

�ƒ z ƒ � const.

F(z) � a � ib Ln z � a � b Arg z � ib ln ƒ z ƒ .

�b,i Ln z � i ln ƒ z ƒ � Arg z,F(z)

�F(z)

F(z) � a Ln z � b

° � a Arg z.£ � a ln r � b � a ln ƒ z ƒ � b.

�y � const

F(z) � az � b � ax � b � iay,

° � ay.

SEC. 18.1 Electrostatic Fields 761

c18.qxd 11/2/10 6:55 PM Page 761

Now is the angle between the line segments from z to c and (Fig. 399). Hence the lines of forceare the curves along each of which the line segment S: appears under a constant angle. These curvesare the totality of circular arcs over S, as is (or should be) known from elementary geometry. Hence the linesof force are circles. Figure 400 shows some of them together with some equipotential lines.

In addition to the interpretation as the potential of two source lines, this potential could also be thought of asthe potential between two circular cylinders whose axes are parallel but do not coincide, or as the potentialbetween two equal cylinders that lie outside each other, or as the potential between a cylinder and a plane wall.Explain this using Fig. 400.

The idea of the complex potential as just explained is the key to a close relation of potentialtheory to complex analysis and will recur in heat flow and fluid flow.

�

�c � x � c�cu1 � u2


1–4 COAXIAL CYLINDERSFind and sketch the potential between two coaxial cylindersof radii and having potential and respectively.

1.

2.

3.

4. If and respectively, is the potential at equal to200 V? Less? More? Answer without calculation. Thencalculate and explain.

5–7 PARALLEL PLATESFind and sketch the potential between the parallel plateshaving potentials and . Find the complex potential.

5. Plates at potentials respectively.

6. Plates at and potentials respectively.

7. Plates at potentials respectively.

8. CAS EXPERIMENT. Complex Potentials. Graphthe equipotential lines and lines of force in (a)–(d) (four

U2 � 8 kV,20 kV,U1 �x2 � 24 cm,x1 � 12 cm,

U2 � 220 V,U1 � 0 V,y � x � k,y � x

U2 � 500 V,250 V,U1 �x2 � 5 cm,x1 � �5 cm,

U2U1

r � 4 cm100 V,U2 �U1 � 300 V,r1 � 2 cm, r2 � 6 cm

U2 � �10 kVr1 � 10 cm, r2 � 1 m, U1 � 10 kV,

r1 � 1 cm, r2 � 2 cm, U1 � 400 V, U2 � 0 V

U2 � 220 Vr1 � 2.5 mm, r2 � 4.0 cm, U1 � 0 V,

U2,U1r2r1

graphs, and on the same axes). Thenexplore further complex potentials of your choice withthe purpose of discovering configurations that mightbe of practical interest.

(a) (b)

(c) (d)

9. Argument. Show that arctanis harmonic in the upper half-plane and satisfies theboundary condition if and 0 if

and the corresponding complex potential is

10. Conformal mapping. Map the upper z-half-planeonto so that are mapped onto respectively. What are the boundary conditions on

resulting from the potential in Prob. 9? Whatis the potential at

11. Text Example 7. Verify, by calculation, that the equipo-tential lines are circles.

12–15 OTHER CONFIGURATIONS

12. Find and sketch the potential between the axes(potential 500 V) and the hyperbola (potential100 V).

xy � 4

w � 0?ƒ w ƒ � 1

1, i, �i,0, , �1ƒ w ƒ � 1

F(z) � �(i>p) Ln z.x 0,

x � 0£(x, 0) � 1

( y>x)£ � u>p � (1>p)

F(z) � i>zF(z) � 1>z

F(z) � iz2F(z) � z2

Im F(z)Re F(z)

P R O B L E M S E T 1 8 . 1

xc

z

–c

1 –

2

2 1

θ

θ

θ

θ

Fig. 399. Arguments in Example 7 Fig. 400. Equipotential lines and lines of force (dashed) in Example 7

c18.qxd 11/2/10 6:55 PM Page 762

SEC. 18.2 Use of Conformal Mapping. Modeling 763

14. Arccos. Show that in Prob. 13 gives the potentialsin Fig. 402.

F(z)

15. Sector. Find the real and complex potentials in thesector between the boundary

kept at 0 V, and the curve keptat 220 V.

x3 � 3xy2 � 1,�p>6,u ��p>6 � u � p>6

18.2 Use of Conformal Mapping. ModelingWe have just explored the close relation between potential theory and complex analysis.This relationship is so close because complex potentials can be modeled in complexanalysis. In this section we shall explore the close relation that results from the use ofconformal mapping in modeling and solving boundary value problems for the Laplaceequation. The process consists of finding a solution of the equation in some domain,assuming given values on the boundary (Dirichlet problem, see also Sec. 12.6). The keyidea is then to use conformal mapping to map a given domain onto one for which thesolution is known or can be found more easily. This solution thus obtained is then mappedback to the given domain. The reason this approach works is due to Theorem 1, whichasserts that harmonic functions remain harmonic under conformal mapping:

T H E O R E M 1 Harmonic Functions Under Conformal Mapping

Let be harmonic in a domain in the w-plane. Suppose that is analytic in a domain D in the z-plane and maps D conformally onto Thenthe function

(1)

is harmonic in D.

P R O O F The composite of analytic functions is analytic, as follows from the chain rule. Hence, takinga harmonic conjugate of as defined in Sec. 13.4, and forming the analyticfunction we conclude that is analytic in D.Hence its real part is harmonic in D. This completes the proof.

We mention without proof that if is simply connected (Sec. 14.2), then a harmonicconjugate of exists. Another proof of Theorem 1 without the use of a harmonicconjugate is given in App. 4. �

£*D*

£(x, y) � Re F(z)F(z) � F* ( f (z))F*(w) � £*(u, v) � i°*(u, v)

£*,°* (u, v)

£(x, y) � £* (u(x, y), v(x, y))

D*.w � u � iv � f (z)D*£*

13. Arccos. Show that (defined in ProblemSet 13.7) gives the potential of a slit in Fig. 401.

F(z) � arccos z

Fig. 401. Slit

y

x1–1

Fig. 402. Other apertures

y

xx

y

11

–1

c18.qxd 11/2/10 6:55 PM Page 763

E X A M P L E 1 Potential Between Noncoaxial Cylinders

Model the electrostatic potential between the cylinders and in Fig. 403. Then givethe solution for the case that is grounded, and has the potential

Solution. We map the unit disk onto the unit disk in such a way that is mapped ontosome cylinder By (3), Sec. 17.3, a linear fractional transformation mapping the unit disk onto theunit disk is

(2) w �z � b

bz � 1

C2*: ƒ w ƒ � r0.C2ƒ w ƒ � 1ƒ z ƒ � 1

U2 � 110 V.C2U1 � 0 V,C1

C2: ƒ z � 25 ƒ � 2

5 C1: ƒ z ƒ � 1


C1

U1 = 0

U2 = 110 V

y

xC2

Fig. 403. Example 1: z-plane

v

u

U1 = 0

U2 = 110 V

r0

Fig. 404. Example 1: w-plane

where we have chosen real without restriction. is of no immediate help here because centers of circlesdo not map onto centers of the images, in general. However, we now have two free constants b and and shallsucceed by imposing two reasonable conditions, namely, that 0 and (Fig. 403) should be mapped onto and

(Fig. 404), respectively. This gives by (2)

and with this,

a quadratic equation in with solutions (no good because and Hence our mappingfunction (2) with becomes that in Example 5 of Sec. 17.3,

(3)

From Example 5 in Sec. 18.1, writing w for z we have as the complex potential in the w-plane the functionand from this the real potential

This is our model. We now determine a and k from the boundary conditions. If then hence . If then hence Substitution of (3)now gives the desired solution in the given domain in the z-plane

The real potential is

Can we “see” this result? Well, if and only if that is, by (2) with These circles are images of circles in the z-plane because the inverse of a linear fractionaltransformation is linear fractional (see (4), Sec. 17.2), and any such mapping maps circles onto circles (or straightlines), by Theorem 1 in Sec. 17.2. Similarly for the rays Hence the equipotential lines

are circles, and the lines of force are circular arcs (dashed in Fig. 404). These two families ofcurves intersect orthogonally, that is, at right angles, as shown in Fig. 404. �£(x, y) � const

arg w � const.

b � 12 .

ƒ w ƒ � constƒ (2z � 1)>(z � 2) ƒ � const,£(x, y) � const

a � �158.7.£(x, y) � Re F(z) � a ln ` 2z � 1

z � 2` ,

F(z) � F* ( f (z)) � a Ln 2z � 1

z � 2 .

a � 110>ln (12 ) � �158.7.£* � a ln (1

2 ) � 110,ƒ w ƒ � r0 � 12 ,k � 0

£* � a ln 1 � k � 0,ƒ w ƒ � 1,

£* (u, v) � Re F* (w) � a ln ƒ w ƒ � k .

F*(w) � a Ln w � k

w � f (z) �2z � 1

z � 2 .

b � 12

r0 � 12 .r0 � 1)r0 � 2r0

�r0 �45 � b

4b>5 � 1�

45 � r0

4r0>5 � 1 ,r0 �

0 � b

0 � 1� b,

�r0

r045

r0

z0b � z0

c18.qxd 11/2/10 6:55 PM Page 764

E X A M P L E 2 Potential Between Two Semicircular Plates

Model the potential between two semicircular plates and in Fig. 405 having potentials and3000 V, respectively. Use Example 3 in Sec. 18.1 and conformal mapping.

Solution. Step 1. We map the unit disk in Fig. 405 onto the right half of the w-plane (Fig. 406) by usingthe linear fractional transformation in Example 3, Sec. 17.3:

w � f (z) �1 � z

1 � z .

�3000 VP2P1

SEC. 18.2 Use of Conformal Mapping. Modeling 765

y

x

P2: 3 kV

P1: –3 kV

2

1

0–1

–2

–3


v

u0

3 kV

–3 kV–2 kV

–1 kV

1 kV

2 kV


The boundary is mapped onto the boundary (the v-axis), with going onto respectively, and onto Hence the upper semicircle of is mapped onto the upper half,and the lower semicircle onto the lower half of the v-axis, so that the boundary conditions in the w-plane areas indicated in Fig. 406.

Step 2. We determine the potential in the right half-plane of the w-plane. Example 3 in Sec. 18.1 withand [with instead of yields

On the positive half of the imaginary axis this equals 3000 and on the negative half as itshould be. is the real part of the complex potential

Step 3. We substitute the mapping function into to get the complex potential in Fig. 405 in the form

The real part of this is the potential we wanted to determine:

As in Example 1 we conclude that the equipotential lines are circular arcs because theycorrespond to hence to Also, are rays from 0to the images of and respectively. Hence the equipotential lines all have and 1 (thepoints where the boundary potential jumps) as their endpoints (Fig. 405). The lines of force are circular arcs,too, and since they must be orthogonal to the equipotential lines, their centers can be obtained as intersectionsof tangents to the unit circle with the x-axis, (Explain!)

Further examples can easily be constructed. Just take any mapping in Chap. 17,a domain D in the z-plane, its image in the w-plane, and a potential in Then (1)gives a potential in D. Make up some examples of your own, involving, for instance,linear fractional transformations.

D*.£*D*w � f (z)

�

�1z � 1,z � �1,Arg w � constArg w � const.Arg [(1 � z)>(1 � z)] � const,

£(x, y) � const

£(x, y) � Re F(z) �6000p

Im Ln 1 � z1 � z

�6000p

Arg 1 � z1 � z

.

F(z) � F* ( f (z)) � � 6000 ip

Ln 1 � z1 � z

.

F(z)F*

F* (w) � � 6000 ip

Ln w.

£*�3000,(� � p>2),

� � arctan vu

.£* (u, v) �6000p

�,

£(x, y)]£* (u, v)U2 � 3000a � p, U1 � �3000,£* (u, v)

ƒ z ƒ � 1w � �i.z � �iw � 0, i, ,z � �1, i, 1u � 0ƒ z ƒ � 1

c18.qxd 11/2/10 6:55 PM Page 765


1. Derivation of (3) from (2). Verify the steps.

2. Second proof. Give the details of the steps given onp. A93 of the book. What is the point of that proof?

3–5 APPLICATION OF THEOREM 1

3. Find the potential in the region R in the first quadrantof the z-plane bounded by the axes (having potential

) and the hyperbola (having potential by mapping R onto a suitable infinite strip. Show that

is harmonic. What are its boundary values?

4. Let and Find What are its boundary values?

5. CAS PROJECT. Graphing Potential Fields. Graph equipotential lines (a) in Example 1 of the text,(b) if the complex potential is (c) Graph the equipotential surfaces for ascylinders in space.

6. Apply Theorem 1 to and any domain D, showing that the resulting

potential is harmonic.

7. Rectangle, Let D: the image of D under and What is the corresponding potential in D? What areits boundary values? Sketch D and

8. Conjugate potential. What happens in Prob. 7 if youreplace the potential by its conjugate harmonic?

9. Translation. What happens in Prob. 7 if we replaceby

10. Noncoaxial Cylinders. Find the potential betweenthe cylinders (potential and

(potential whereSketch or graph equipotential lines and

their orthogonal trajectories for Can you guesshow the graph changes if you increase c (� 1

2 )?c � 1

4 .0 � c � 1

2 .U2 � 220 V),C2: ƒ z � c ƒ � c

U1 � 0)C1: ƒ z ƒ � 1

cos z � sin (z � 12 p)?sin z

D*.£

£* � u2 � v2.w � sin z;0 � x � 1

2 p, 0 � y � 1; D*sin z.

£

f (z) � ez,w �£* (u, v) � u2 � v2,

F(z) � Ln zF(z) � z2, iz2, ez.

£.0 � y � p.D: x � 0,w � f (z) � ez,£* � 4uv,

£

U2)y � 1>xU1

£

11. On Example 2. Verify the calculations.

12. Show that in Example 2 the y-axis is mapped onto theunit circle in the w-plane.

13. At in Fig. 405 the tangents to the equipotentiallines as shown make equal angles. Why?

14. Figure 405 gives the impression that the potential onthe y-axis changes more rapidly near 0 than near Can you verify this?

15. Angular region. By applying a suitable conformalmapping, obtain from Fig. 406 the potential in thesector such that if

and if

16. Solve Prob. 15 if the sector is

17. Another extension of Example 2. Find the linearfractional transformation that maps onto with being mapped onto Show that is mapped onto and onto so that theequipotential lines of Example 2 look in asshown in Fig. 407.

ƒ Z ƒ � 1z � 1,Z2 � �0.6 � 0.8i

z � �1Z1 � 0.6 � 0.8iz � 0.Z � i>2ƒ z ƒ � 1

ƒ Z ƒ � 1z � g (Z )

� 18 p � Arg z � 1

8 p.

Arg z � 14 p.£ � 3 kVArg z � �1

4 p£ � �3 kV�1

4 p � Arg z � 14 p

£

�i.

z � �1

P R O B L E M S E T 1 8 . 2

Basic Comment on ModelingWe formulated the examples in this section as models on the electrostatic potential. It isquite important to realize that this is accidental. We could equally well have phrasedeverything in terms of (time-independent) heat flow; then instead of voltages we wouldhave had temperatures, the equipotential lines would have become isotherms ( lines ofconstant temperature), and the lines of the electrical force would have become lines alongwhich heat flows from higher to lower temperatures (more on this in the next section).Or we could have talked about fluid flow; then the electrostatic lines of force would havebecome streamlines (more on this in Sec. 18.4). What we again see here is the unifyingpower of mathematics: different phenomena and systems from different areas in physicshaving the same types of model can be treated by the same mathematical methods. Whatdiffers from area to area is just the kinds of problems that are of practical interest.

�

Y

X

Z2 Z

1

–3 kV

3 kV

2

10

Fig. 407. Problem 17

c18.qxd 11/2/10 6:55 PM Page 766

SEC. 18.3 Heat Problems 767

18.3 Heat ProblemsHeat conduction in a body of homogeneous material is modeled by the heat equation

where the function T is temperature, t is time, and is a positive constant(specific to the material of the body; see Sec. 12.6).

Now if a heat flow problem is steady, that is, independent of time, we have Ifit is also two-dimensional, then the heat equation reduces to

(1)

which is the two-dimensional Laplace equation. Thus we have shown that we can modela two-dimensional steady heat flow problem by Laplace’s equation.

Furthermore we can treat this heat flow problem by methods of complex analysis, sinceT (or ) is the real part of the complex heat potential

We call the heat potential. The curves are called isotherms, whichmeans lines of constant temperature. The curves are called heat flowlines because heat flows along them from higher temperatures to lower temperatures.

It follows that all the examples considered so far (Secs. 18.1, 18.2) can now bereinterpreted as problems on heat flow. The electrostatic equipotential lines now become isotherms and the lines of electrical force become lines ofheat flow, as in the following two problems.

E X A M P L E 1 Temperature Between Parallel Plates

Find the temperature between two parallel plates and in Fig. 408 having temperatures 0 and respectively.

Solution. As in Example 1 of Sec. 18.1 we conclude that From the boundary conditions,and The answer is

The corresponding complex potential is Heat flows horizontally in the negative x-directionalong the lines

E X A M P L E 2 Temperature Distribution Between a Wire and a Cylinder

Find the temperature field around a long thin wire of radius mm that is electrically heated to and is surrounded by a circular cylinder of radius which is kept at temperature bycooling it with air. See Fig. 409. (The wire is at the origin of the coordinate system.)

T2 � 60°Fr2 � 100 mm,T1 � 500°Fr1 � 1

�y � const.F(z) � (100>d)z.

T (x, y) �100

d x [°C].

a � 100>d.b � 0T (x, y) � ax � b.

100°C,x � dx � 0

T (x, y) � const,£(x, y) � const

° (x, y) � constT (x, y) � constT (x, y)

F(z) � T (x, y) � i°(x, y).

T (x, y)

�2T � Txx � Tyy � 0,

Tt � 0.

c2Tt � 0T>0t,

Tt � c2�2T

18. The equipotential lines in Prob. 17 are circles. Why?

19. Jump on the boundary. Find the complex and realpotentials in the upper half-plane with boundary values5 kV if and 0 if on the x-axis.x 2x � 2

20. Jumps. Do the same task as in Prob. 19 if the boundaryvalues on the x-axis are when and 0elsewhere.

�a � x � aV0

c18.qxd 11/2/10 6:55 PM Page 767

Solution. T depends only on r, for reasons of symmetry. Hence, as in Sec. 18.1 (Example 2),

The boundary conditions are

Hence (since and The answer is

The isotherms are concentric circles. Heat flows from the wire radially outward to the cylinder. Sketch T as afunction of r. Does it look physically reasonable? �

T (x, y) � 500 � 95.54 ln r [°F].

a � (60 � b)>ln 100 � �95.54.ln 1 � 0)b � 500

T1 � 500 � a ln 1 � b, T2 � 60 � a ln 100 � b.

T (x, y) � a ln r � b.


y

x

T = 60°F

Fig. 409. Example 2

y

x

T =

10

0°C

T =

0°C

Fig. 408. Example 1

y

x

T =

20

°C

T = 50°C 10

Insulated

Fig. 410. Example 3

Mathematically the calculations remain the same in the transition to another field ofapplication. Physically, new problems may arise, with boundary conditions that wouldmake no sense physically or would be of no practical interest. This is illustrated by thenext two examples.

E X A M P L E 3 A Mixed Boundary Value Problem

Find the temperature distribution in the region in Fig. 410 (cross section of a solid quarter-cylinder), whosevertical portion of the boundary is at the horizontal portion at and the circular portion is insulated.

Solution. The insulated portion of the boundary must be a heat flow line, since, by the insulation, heat isprevented from crossing such a curve, hence heat must flow along the curve. Thus the isotherms must meetsuch a curve at right angles. Since T is constant along an isotherm, this means that

(2) along an insulated portion of the boundary.

Here is the normal derivative of T, that is, the directional derivative (Sec. 9.7) in the direction normal(perpendicular) to the insulated boundary. Such a problem in which T is prescribed on one portion of the boundaryand on the other portion is called a mixed boundary value problem.

In our case, the normal direction to the insulated circular boundary curve is the radial direction toward theorigin. Hence (2) becomes meaning that along this curve the solution must not depend on r. Now

satisfies (1), as well as this condition, and is constant (0 and on the straight portions of theboundary. Hence the solution is of the form

The boundary conditions yield and This gives

u � arctan

y

x .T (x, y) � 50 �

60p

u,

a # 0 � b � 50.a # p>2 � b � 20

T (x, y) � au � b.

p>2)Arg z � u

0T>0r � 0,

0T>0n

0T>0n

0T

0n� 0

50°C,20°C,

c18.qxd 11/2/10 6:55 PM Page 768

The isotherms are portions of rays Heat flows from the x-axis along circles (dashed inFig. 410) to the y-axis. �

r � constu � const.

SEC. 18.3 Heat Problems 769

y

xT = 20°C1–1T = 0°C Insulated


u

T*

= 0

°C

T*

= 20

°C

π__2

π__2

–

v

Insulated


E X A M P L E 4 Another Mixed Boundary Value Problem in Heat Conduction

Find the temperature field in the upper half-plane when the x-axis is kept at for is insulatedfor and is kept at for (Fig. 411).

Solution. We map the half-plane in Fig. 411 onto the vertical strip in Fig. 412, find the temperature there, and map it back to get the temperature in the half-plane.

The idea of using that strip is suggested by Fig. 391 in Sec. 17.4 with the roles of and interchanged. The figure shows that maps our present strip onto our half-plane in Fig. 411. Hence theinverse function

maps that half-plane onto the strip in the w-plane. This is the mapping function that we need according toTheorem 1 in Sec. 18.2.

The insulated segment on the x-axis maps onto the segment on the u-axis.The rest of the x-axis maps onto the two vertical boundary portions and of the strip.This gives the transformed boundary conditions in Fig. 412 for where on the insulated horizontalboundary, because v is a coordinate normal to that segment.

Similarly to Example 1 we obtain

which satisfies all the boundary conditions. This is the real part of the complex potential Hence the complex potential in the z-plane is

and is the solution. The isotherms are in the strip and the hyperbolas in the z-plane,perpendicular to which heat flows along the dashed ellipses from the -portion to the cooler -portion of theboundary, a physically very reasonable result.

Sections 18.3 and 18.5 show some of the usefulness of conformal mappings and complexpotentials. Furthermore, complex potential models fluid flow in Sec. 18.4.

�0°20°

u � constT (x, y) � Re F(z)

F(z) � F* ( f (z)) � 10 �20p

arcsin z

F*(w) � 10 � (20>p) w.

T*(u, v) � 10 �20p

u

0T *>0n � 0T *>0v � 0T* (u, v),

p>2, v 0,u � �p>2�p>2 � u � p>2�1 � x � 1

w � f (z) � arcsin z

z � sin ww � u � ivz � x � iy

T (x, y)T* (u, v)

x 1T � 20°C�1 � x � 1,x � �1,T � 0°C

1. Parallel plates. Find the temperature between theplates and kept at and respec-tively. (i) Proceed directly. (ii) Use Example 1 and asuitable mapping.

100°C,20y � dy � 02. Infinite plate. Find the temperature and the complex

potential in an infinite plate with edges andkept at and respectively (Fig. 413).

In what case will this be an approximate model?40°C,�20y � x � 4

y � x � 4

P R O B L E M S E T 1 8 . 3

c18.qxd 11/2/10 6:55 PM Page 769


Fig. 413. Problem 2: Infinite plate

3. CAS PROJECT. Isotherms. Graph isotherms andlines of heat flow in Examples 2–4. Can you see fromthe graphs where the heat flow is very rapid?

4–18 TEMPERATURE T (x, y) IN PLATES

Find the temperature distribution and the complexpotential in the given thin metal plate whose facesare insulated and whose edges are kept at the indicatedtemperatures or are insulated as shown.

4. 5.

6. 7.

8.

9. y

xT = 0 T* = T1

T = 0

a b

v

xT* = T2

T* = T1

a

T = T1

T =

T2

y

xxT = 0°C

60°

0

T =

50°C

T = 2

0°C

T = –20°C

45°

45°

y

x

y = 10/x

T = 200°C

T = 0°CT =

20

0°C

F(z)T (x, y)

10.

11.

12.

Hint. Apply to Prob. 11.

13. 14.

15.

16.

17. First quadrant of the z-plane with y-axis kept at the segment of the x-axis insulated and thex-axis for kept at Hint. Use Example 4.

18. Figure 410,

19. Interpretation. Formulate Prob. 11 in terms of electro-statics.

20. Interpretation. Interpret Prob. 17 in Sec. 18.2 as a heatproblem, with boundary temperatures, say, on theupper part and on the lower.200°C

10°C

T (0, y) � �30°C, T (x, 0) � 100°C

200°C.x 10 � x � 1

100°C,

xT = 20°C

–20

T =

500

°C

Insulated

y = √3 x

xT = –20°C

45°T = 6

0°CInsulated

y

xT = 0°C

Insulated

T =

20

0°C

T = 0°CT = 100°C

0°C

10

0°C

y

x1

1

w � cosh z

y

x

T = 100°C

T = 0°C

T = 0°C0

π

y

xT = 0 T = 100°C T = 0

1–1

y

xT = T3

T = T2

T = T1

a b

x

y

T = 4

0°C

T = 2

0°C

4–4

c18.qxd 11/2/10 6:55 PM Page 770

18.4 Fluid FlowLaplace’s equation also plays a basic role in hydrodynamics, in steady nonviscous fluidflow under physical conditions discussed later in this section. For methods of complexanalysis to be applicable, our problems will be two-dimensional, so that the velocity vectorV by which the motion of the fluid can be given depends only on two space variables xand y, and the motion is the same in all planes parallel to the xy-plane.

Then we can use for the velocity vector V a complex function

(1)

giving the magnitude and direction of the velocity at each point Here and are the components of the velocity in the x and y directions. V is tangentialto the path of the moving particles, called a streamline of the motion (Fig. 414).

We show that under suitable assumptions (explained in detail following the examples),for a given flow there exists an analytic function

(2)

called the complex potential of the flow, such that the streamlines are given byand the velocity vector or, briefly, the velocity is given by

(3) V � V1 � iV2 � F r(z)

°(x, y) � const,

F(z) � £(x, y) � i°(x, y),

V2V1

z � x � iy.Arg Vƒ V ƒ

V � V1 � iV2

SEC. 18.4 Fluid Flow 771

Fig. 414. Velocity

y

x

V1

V2

V

Streamline

where the bar denotes the complex conjugate. is called the stream function. Thefunction is called the velocity potential. The curves are calledequipotential lines. The velocity vector V is the gradient of by definition, thismeans that

(4)

Indeed, for Eq. (4) in Sec. 13.4 is with bythe second Cauchy–Riemann equation. Together we obtain (3):

F r(z) � £x � i°x � £x � i£y � V1 � iV2 � V.

°x � �£yF r � £x � i°xF � £ � i°,

V1 �0£

0x , V2 �

0£

0y .

£;£(x, y) � const£

°

c18.qxd 11/2/10 6:55 PM Page 771

Furthermore, since is analytic, and satisfy Laplace’s equation

(5)

Whereas in electrostatics the boundaries (conducting plates) are equipotential lines, influid flow the boundaries across which fluid cannot flow must be streamlines. Hence influid flow the stream function is of particular importance.

Before discussing the conditions for the validity of the statements involving (2)–(5), letus consider two flows of practical interest, so that we first see what is going on from apractical point of view. Further flows are included in the problem set.

E X A M P L E 1 Flow Around a Corner

The complex potential models a flow with

Equipotential lines (Hyperbolas)

Streamlines (Hyperbolas).

From (3) we obtain the velocity vector

that is,

The speed (magnitude of the velocity) is

The flow may be interpreted as the flow in a channel bounded by the positive coordinates axes and a hyperbola,say, (Fig. 415). We note that the speed along a streamline S has a minimum at the point P where thecross section of the channel is large. �

xy � 1

ƒV ƒ � 2V12 � V2

2 � 22x2 � y2.

V1 � 2x, V2 � �2y.V � 2z � 2 (x � iy),

° � 2xy � const

£ � x2 � y2 � const

F(z) � z2 � x2 � y2 � 2ixy

�2£ �

02£

0x2�

02£

0y2� 0, �2

° �0

2°

0x2�

02°

0y2� 0.

°£F(z)


Fig. 415. Flow around a corner (Example 1)

y

x

S

0

P

E X A M P L E 2 Flow Around a Cylinder

Consider the complex potential

Using the polar form we obtain

Hence the streamlines are

°(x, y) � ar �1r

b sin u � const.

F(z) � reiu �1r e�iu � ar �

1r

b cos u � i ar �1r

b sin u.

z � reiu,

F(z) � £(x, y) � i°(x, y) � z �1z .

c18.qxd 11/2/10 6:55 PM Page 772

In particular, gives or Hence this streamline consists of the unit circlegives and the x-axis and For large the term in is small in absolute

value, so that for these z the flow is nearly uniform and parallel to the x-axis. Hence we can interpret this as aflow around a long circular cylinder of unit radius that is perpendicular to the z-plane and intersects it in theunit circle and whose axis corresponds to

The flow has two stagnation points (that is, points at which the velocity V is zero), at This followsfrom (3) and

hence (See Fig. 416.) �z2 � 1 � 0.F r(z) � 1 �1

z2 ,

z � �1.z � 0.ƒ z ƒ � 1

F(z)1>zƒ z ƒu � p).(u � 0r � 1)(r � 1>rsin u � 0.r � 1>r � 0°(x, y) � 0


Fig. 416. Flow around a cylinder (Example 2)

y

x

Fig. 417. Tangential component of the velocity with respect to a curve C

y

x

CV

α

Vt

Assumptions and Theory Underlying (2)–(5)

T H E O R E M 1 Complex Potential of a Flow

If the domain of flow is simply connected and the flow is irrotational andincompressible, then the statements involving (2)–(5) hold. In particular, then theflow has a complex potential which is an analytic function. (Explanation ofterms below.)

P R O O F We prove this theorem, along with a discussion of basic concepts related to fluid flow.

(a) First Assumption: Irrotational. Let C be any smooth curve in the z-plane givenby where s is the arc length of C. Let the real variable be thecomponent of the velocity V tangent to C (Fig. 417). Then the value of the real lineintegral

(6) �C

Vt ds

Vtz(s) � x(s) � iy(s),

F(z),

c18.qxd 11/2/10 6:55 PM Page 773

taken along C in the sense of increasing s is called the circulation of the fluid along C,a name that will be motivated as we proceed in this proof. Dividing the circulation by thelength of C, we obtain the mean velocity1 of the flow along the curve C. Now

(Fig. 417).

Hence is the dot product (Sec. 9.2) of V and the tangent vector of C (Sec. 17.1);thus in (6),

The circulation (6) along C now becomes

(7)

As the next idea, let C be a closed curve satisfying the assumption as in Green’s theorem(Sec. 10.4), and let C be the boundary of a simply connected domain D. Suppose furtherthat V has continuous partial derivatives in a domain containing D and C. Then we canuse Green’s theorem to represent the circulation around C by a double integral,

(8)

The integrand of this double integral is called the vorticity of the flow. The vorticitydivided by 2 is called the rotation

(9)

We assume the flow to be irrotational, that is, throughout the flow; thus,

(10)

To understand the physical meaning of vorticity and rotation, take for C in (8) a circle.Let r be the radius of C. Then the circulation divided by the length of C is the mean2pr

0V2

0x�

0V1

0y� 0.

v(x, y) � 0

v(x, y) �12

a 0V2

0x�

0V1

0yb .

�C

(V1 dx � V2 dy) � �D

�a 0V2

0x�

0V1

0y b dx dy.

�C

Vt ds � �C

(V1 dx � V2 dy).

Vt ds � aV1 dxds

� V2 dydsb ds � V1 dx � V2 dy.

dz>dsVt

Vt � ƒV ƒ cos a


1Definitions: mean value of f on the interval

mean value of f on C

mean value of f on D (A � area of D).1

A �

D

� f (x, y) dx dy �

(L � length of C ),1

L �

C

f (s) ds �

a � x � b,1

b � a �

b

a

f (x) dx �

c18.qxd 11/2/10 6:55 PM Page 774

velocity of the fluid along C. Hence by dividing this by r we obtain the mean angularvelocity of the fluid about the center of the circle:

If we now let the limit of is the value of at the center of C. Hence isthe limiting angular velocity of a circular element of the fluid as the circle shrinks to thepoint (x, y). Roughly speaking, if a spherical element of the fluid were suddenly solidifiedand the surrounding fluid simultaneously annihilated, the element would rotate with theangular velocity

(b) Second Assumption: Incompressible. Our second assumption is that the fluid isincompressible. (Fluids include liquids, which are incompressible, and gases, such as air,which are compressible.) Then

(11)

in every region that is free of sources or sinks, that is, points at which fluid is producedor disappears, respectively. The expression in (11) is called the divergence of V and isdenoted by div V. (See also (7) in Sec. 9.8.)

(c) Complex Velocity Potential. If the domain D of the flow is simply connected(Sec. 14.2) and the flow is irrotational, then (10) implies that the line integral (7) isindependent of path in D (by Theorem 3 in Sec. 10.2, where and z is the third coordinate in space and has nothing to do with our present z). Hence ifwe integrate from a fixed point (a, b) in D to a variable point (x, y) in D, the integralbecomes a function of the point (x, y), say,

(12)

We claim that the flow has a velocity potential which is given by (12). To provethis, all we have to do is to show that (4) holds. Now since the integral (7) isindependent of path, is exact (Sec. 10.2), namely, the differential of that is,

From this we see that and which gives (4).That is harmonic follows at once by substituting (4) into (11), which gives the first

Laplace equation in (5).We finally take a harmonic conjugate of Then the other equation in (5) holds.

Also, since the second partial derivatives of and are continuous, we see that thecomplex function

F(z) � £(x, y) � i°(x, y)

°£

£.°

£

V2 � 0£>0y,V1 � 0£>0x

V1 dx � V2 dy �0£

0x dx �

0£

0y dy.

£,V1 dx � V2 dy

£,

£(x, y) � �(x,y)

(a,b)

(V1 dx � V2 dy).

£(x, y):

F1 � V1, F2 � V2, F3 � 0,

0V1

0x�

0V2

0y� 0

v.

v(x, y)vv0r : 0,

v0 �1

2pr 2 �D

� a 0V2

0x�

0V1

0yb dx dy �

1pr 2 �

D

� v(x, y) dx dy.

v0


c18.qxd 11/2/10 6:55 PM Page 775

is analytic in D. Since the curves are perpendicular to the equipotentialcurves (except where we conclude that arethe streamlines. Hence is the stream function and is the complex potential of theflow. This completes the proof of Theorem 1 as well as our discussion of the importantrole of complex analysis in compressible fluid flow. �

F(z)°

°(x, y) � constF r(z) � 0),£(x, y) � const°(x, y) � const


1. Differentiability. Under what condition on the velocityvector V in (1) will in (2) be analytic?

2. Corner flow. Along what curves will the speed inExample 1 be constant? Is this obvious from Fig. 415?

3. Cylinder. Guess from physics and from Fig. 416 whereon the y-axis the speed is maximum. Then calculate.

4. Cylinder. Calculate the speed along the cylinder wallin Fig. 416, also confirming the answer to Prob. 3.

5. Irrotational flow. Show that the flow in Example 2 isirrotational.

6. Extension of Example 1. Sketch or graph andinterpret the flow in Example 1 on the whole upperhalf-plane.

7. Parallel flow. Sketch and interpret the flow withcomplex potential

8. Parallel flow. What is the complex potential of anupward parallel flow of speed in the directionof ? Sketch the flow.

9. Corner. What would be suitable in Example 1 ifthe angle of the corner were instead of ?

10. Corner. Show that also models a flowaround a corner. Sketch streamlines and equipotentiallines. Find V.

11. What flow do you obtain from K positivereal?

12. Conformal mapping. Obtain the flow in Example 1from that in Prob. 11 by a suitable conformal mapping.

13. - Sector. What would be suitable in Example 1if the angle at the corner were ?

14. Sketch or graph streamlines and equipotential lines of Find V. Find all points at which V ishorizontal.

15. Change in Example 2 slightly to obtain a flowaround a cylinder of radius that gives the flow inExample 2 if

16. Cylinder. What happens in Example 2 if you replacez by ? Sketch and interpret the resulting flow in thefirst quadrant.

17. Elliptic cylinder. Show that givesconfocal ellipses as streamlines, with foci at z � �1,

F(z) � arccos z

z2

r0 : 1.r0

F(z)

F(z) � iz3.

p>3F(z)60�

F(z) � �iKz,

F(z) � iz2

p>2p>4F(z)

y � xK 0

F(z) � z.

F(z)and that the flow circulates around an elliptic cylinderor a plate (the segment from to 1 in Fig. 418).�1

P R O B L E M S E T 1 8 . 4

Fig. 418. Flow around a plate in Prob. 17.

18. Aperture. Show that gives confocalhyperbolas as streamlines, with foci at and theflow may be interpreted as a flow through an aperture(Fig. 419).

z � �1,F(z) � arccosh z

19. Potential Show that the streamlines ofand circles through the origin with centers

on the y-axis.

20. TEAM PROJECT. Role of the Natural Logarithmin Modeling Flows. (a) Basic flows: Source and sink.Show that ln z with constant positivereal c gives a flow directed radially outward (Fig. 420),so that F models a point source at (that is, asource line in space) at which fluid isproduced. c is called the strength or discharge of thesource. If c is negative real, show that the flow isdirected radially inward, so that F models a sink at

a point at which fluid disappears. Note thatis the singular point of F(z).z � 0

z � 0,

x � 0, y � 0z � 0

F(z) � (c>2p)

F(z) � 1>zF(z) � 1>z.

–1 1

Fig. 419. Flow through an aperture in Prob. 18.

–1 1

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18.5 Poisson’s Integral Formula for PotentialsSo far in this chapter we have seen powerful methods based on conformal mappings andcomplex potentials. They were used for modeling and solving two-dimensional potentialproblems and demonstrated the importance of complex analysis.

Now we introduce a further method that results from complex integration. It will yieldthe very important Poisson integral formula (5) for potentials in a standard domain

SEC. 18.5 Poisson’s Integral Formula for Potentials 777

if they are at as K increases they move upon the unit circle until they unite at seeFig. 422), and if they lie on the imaginary axis(one lies in the field of flow and the other one lies insidethe cylinder and has no physical meaning).

K 4pz � i (K � 4p,

�1;K � 0

z �iK

4p�B

�K 2

16p2 � 1;

(d) Source and sink combined. Find the complexpotentials of a flow with a source of strength 1 at and of a flow with a sink of strength 1 at Addboth and sketch or graph the streamlines. Show that forsmall these lines look similar to those in Prob. 19.

(e) Flow with circulation around a cylinder. Addthe potential in (b) to that in Example 2. Show that thisgives a flow for which the cylinder wall is astreamline. Find the speed and show that the stagnationpoints are

ƒ z ƒ � 1

ƒ a ƒ

z � a.z � �a

(b) Basic flows: Vortex. Show that ln z with positive real K gives a flow circulating coun-terclockwise around (Fig. 421). is called avortex. Note that each time we travel around the vortex,the potential increases by K.

(c) Addition of flows. Show that addition of thevelocity vectors of two flows gives a flow whosecomplex potential is obtained by adding the complexpotentials of those flows.

z � 0z � 0

F(z) � �(Ki>2p)

Fig. 420. Point source

y

x

Fig. 422. Flow around a cylinder without circulation(K 0) and with circulation�

K = 0

K = 2.8π

K = 4π

K = 6π

Fig. 421. Vortex flow

y

x

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(a circular disk). In addition, from (5), we will derive a useful series (7) for these potentials.This allows us to solve problems for disks and then map solutions conformally onto otherdomains.

Derivation of Poisson’s Integral FormulaPoisson’s formula will follow from Cauchy’s integral formula (Sec. 14.3)

(1)

Here C is the circle (counterclockwise, and we assume that is analytic in a domain containing C and its full interior. Since we obtain from (1)

(2)

Now comes a little trick. If instead of z inside C we take a Z outside C, the integrals (1)and (2) are zero by Cauchy’s integral theorem (Sec. 14.2). We choose which is outside C because From (2) we thus have

and by straightforward simplification of the last expression on the right,

We subtract this from (2) and use the following formula that you can verify by directcalculation cancels):

(3)

We then have

(4)

From the polar representations of z and we see that the quotient in the integrand is realand equal to

R2 � r 2

(Reia � reiu)(Re�ia � re�iu)�

R2 � r 2

R2 � 2Rr cos (u � a) � r 2 .

z*

F(z) �1

2p �

2p

0

F(z*) z*z* � zz

(z* � z)( z* � z ) da.

z*

z* � z�

z

z � z*�

z*z* � zz

(z* � z)(z* � z) .

(zz*

0 �1

2p �

2p

0

F(z*) z

z � z* da.

0 �1

2p �

2p

0

F(z*) z*

z* � Z da �

12p

�2p

0

F(z*) z*

z* �z*z*

z

da

ƒ Z ƒ � R2> ƒ z ƒ � R2>r R.Z � z*z*>z � R2>z,

(z* � Reia, z � reiu).F(z) �1

2p �

2p

0

F(z*) z*

z* � z da

dz* � iReia da � iz* da,F(z*)0 � a � 2p),z* � Reia

F(z) �1

2pi �

C

F(z*)

z* � z dz*.


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conformal mapping

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