Contents
20.1 Complex Functions as Mappings
20.2 Conformal Mappings
20.3 Linear Fractional Transformations
20.4 Schwarz-Christoffel Transformations
20.5 Poisson Integral Formulas
20.6 Applications
Ch20_1
20.1 Complex Functions as Mappings
Introduction
The complex function w = f(z) = u(x, y) + iv(x, y)
may be considered as the planar transformation. We
also call w = f(z) is the image of z under f. See Fig
20.1.
Ch20_2
Fig 20.1
Ch20_3
Example 1
Consider the function f(z) = ez. If z = a + it, 0  t  ,
w = f(z) = eaeit. Thus this is a semicircle with center
w = 0 and radius r = ea. If z = t + ib, −  t  , w =
f(z) = eteib. Thus this is a ray with Arg w = b, |w| = et.
See Fig 20.2.
Ch20_4
Fig 20.2
Ch20_5
Example 2
The complex function f = 1/z has domain z  0 and
real part : u ( x , y ) 
imaginary
x
x  y
2
part : v ( x , y ) 
2
 y
x  y
2
2
When a  0 , u ( x , y )  a can be written as
x 
2
1
a
x  y  0, ( x 
2
1
2a
)  y (
2
2
1
)
2
2a
Ch20_6
Example 2 (2)
Likewise v(x, y) = b, b  0 can be written as
x  (y 
2
1
2b
) (
2
1
)
2
2b
See Fig 20.3.
Ch20_7
Fig 20.3
Ch20_8
Translation and Rotation
The function f(z) = z + z0 is interpreted as a
translation. The function g ( z )  e i  0 z is interpreted
as a rotation. See Fig 20.4.
Ch20_9
Example 3
Find a complex function that maps −1  y  1 onto 2  x
 4.
Solution
See Fig 20.5. We find that −1  y  1 is first rotated
through 90 and shifted 3 units to the right. Thus the
mapping is
h( z)  e
i / 2
z  3  iz  3
Ch20_10
Fig 20.5
Ch20_11
Magnification
A magnification is the function f(z) = z, where  is a
fixed positive real number. Note that |w| = |z| = |z|.
If g(z) = az + b and a  r0 e i 0 then the vector is
rotated through 0, magnified by a factor r0, and then
translated using b.
Ch20_12
Example 4
Find a complex function that maps the disk |z|  1 onto
the disk |w – (1 + i)|  ½.
Solution
Magnified by ½ and translated to 1 + i, we can have the
desired function as w = f(z) = ½z + (1 + i).
Ch20_13
Power Functions
A complex function f(z) = z where  is a fixed
positive number, is called a real power function. See
Fig 20.6. If z = rei, then w = f(z) = rei.
Ch20_14
Example 5
Find a complex function that maps the upper half-plane
y  0 onto the wedge 0  Arg w  /4.
Solution
The upper half-plane can also be described by 0  Arg w
 . Thus f(z) = z1/4 will map the upper half-plane onto
the wedge 0  Arg w  /4.
Ch20_15
Successive Mapping
See Fig 20.7. If  = f(z) maps R onto R, and w =
g() maps R onto R, w = g(f(z)) maps R onto R.
Ch20_16
Fig 20.7
Ch20_17
Example 6
Find a complex function that maps 0  y   onto the
wedge 0  Arg w  /4.
Solution
We have shown that f(z) = ez maps 0  y   onto to 0 
Arg    and g() =  1/4 maps 0  Arg    onto 0 
Arg w  /4. Thus the desired mapping is w = g(f(z)) =
g(ez) = ez/4.
Ch20_18
Example 7
Find a complex function that maps /4  Arg z  3/4
onto the upper half-plane v  0.
Solution
First rotate /4  Arg z  3/4 by  = f(z) = e-i/4z. Then
magnify it by 2, w = g() =  2. Thus the desired
mapping is w = g(f(z)) = (e-i/4z)2 = -iz2.
Ch20_19
20.2 Conformal Mappings
Angle –Preserving Mappings
A complex mapping w = f(z) defined on a domain D
is called conformal at z = z0 in D when f preserves
that angle between two curves in D that intersect at z0.
See Fig 20.10.
Ch20_20
Fig 20.10
Ch20_21
Referring to Fig 20.10, we have
'
z1

'
z2
2

'
z1
2

'
z2
2
 2 z1 z 2 cos 
'
'
 z' 2  z' 2  z'  z'
2
1
2
1  1
or   cos
'
'

2 z1 z 2

2




(1)




(2)
Likewise
 w' 2  w' 2  w'  w'
1
2
1
2
1 
  cos
'
'

2 w1 w 2

2
Ch20_22
THEOREM 20.1
Conformal Mapping
If f(z) is analytic in the domain D and f’(z)  0, then
f is conformal at z = z0.
Proof
If a curve C in D is defined by z = z(t), then w = f(z(t)) is
the image curve in the w-plane. We have
w  f ( z ( t )), w '  f ' ( z ( t )) z ' ( t )
If C1 and C2 intersect at z = z0, then
w1  f ' ( z 0 ) z1 , w 2  f ' ( z 0 ) z 2
'
'
'
'
Ch20_23
THEOREM 20.1 proof
Since f (z0)  0, we can use (2) to obtain
 f '( z ) z ' 2  f '( z ) z ' 2  f '( z ) z '  f '( z ) z '
0
1
0
2
0
1
0
2
1 
  cos
'
'

2 f ' ( z 0 ) z1 f ' ( z 0 ) z 2

 z' 2  z' 2  z'  z'
2
1
2
1  1
 cos
'
'

2 z1 z 2

2
2





Ch20_24




Example 1
(a) The analytic function f(z) = ez is conformal at all
points, since f (z) = ez is never zero.
(b) The analytic function g(z) = z2 is conformal at all
points except z = 0, since g(z) = 2z  0, for z  0.
Ch20_25
Example 2
The vertical strip −/2  x  /2 is called the
fundamental region of the trigonometric function
w = sin z. A vertical line x = a in the interior of the
region can be described by z = a + it, −  t  . We
find that
sin z = sin x cosh y + i cos x sinh y
and so
u + iv = sin (a + it)
= sin a cosh t + i cos a sinh t.
Ch20_26
Example 2 (2)
Since cosh2 t − sinh2 t = 1, then
u
2
2
sin a

v
2
2
1
cos a
The image of the vertical line x = a is a hyperbola with
 sin a as u-intercepts and since −/2 < a < /2, the
hyperbola crosses the u-axis between u = −1 and u = 1.
Note if a = −/2, then w = − cosh t, the line x = − /2 is
mapped onto the interval (−, −1]. Likewise, the line x
= /2 is mapped onto the interval [1, ).
Ch20_27
Example 3
The complex function f(z) = z + 1/z is conformal at all
points except z = 1 and z = 0. In particular, the function
is conformal at all points in the upper half-plane
satisfying |z| > 1. If z = rei, then
w = rei + (1/r)e-i, and so
1
1
u  ( r  ) cos  , v  ( r  ) sin 
r
r
(3)
Note if r = 1, then v = 0 and u = 2 cos  . Thus the
semicircle z = eit, 0  t  , is mapped onto [−2, 2] on
the u-axis. If r > 1, the semicircle z = reit, 0  t  , is
mapped onto the upper half of the ellipse u2/a2 + v2/b2 =
1, where a = r + 1/r, b = r − 1/r. See Fig 20.12.
Ch20_28
Fig 20.12
Ch20_29
Example 3 (2)
For a fixed value of , the ray tei, for t  1, is mapped
to the point u2/cos2 − v2/sin2 = 4 in the upper halfplane v  0. This follows from (3) since
2
2
2
2
1
1



 t    t    4
2
2
t
t
cos  sin  

u
v
Since f is conformal for |z| > 1 and a ray  = 0
intersects a circle |z| = r at a right angle, the hyperbolas
and ellipses in the w-plane are orthogonal.
Ch20_30
Conformal Mappings Using Tables
Conformal mappings are given in Appendix IV. We
may use this table to solve the transformations.
Ch20_31
Example 4
Use the conformal mappings in Appendix IV to find a
conformal mapping between the strip 0  y  2 and the
upper half-plane v  0.What is the image of the negative
x-axis?
Solution
From H-2, letting a = 2 then f(z) = ez/2 and noting the
positions E, D, E and D. We can map the negative xaxis onto the interval (0, 1) on the u-axis.
Ch20_32
Example 5
Use the conformal mappings in Appendix IV to find a
conformal mapping between the strip 0  y  2 and the
disk |w|  1. What is the image of the negative x-axis?
Solution
Appendix IV des not have an entry that maps 0  y  2
directly onto the disk. In Example 4, the strip was
mapped onto f(z) = ez/2 the upper half-plane, and from
C-4, the complex mapping w = (i – )/ (i + ) maps the
upper half-plane to the disk |w|  1.
Ch20_33
Example 5 (2)
Therefore
w  g ( f ( z )) 
ie
z / 2
ie
z / 2
maps the strip 0  y  2 onto the disk |w|  1.
The negative x-axis is first mapped to the interval
(0, 1) in the -plane and from the position of points C
and C in C-4, this interval is mapped to the circular arc
w = ei, 0 <  < /2 in the w-plane.
Ch20_34
THEOREM 20.2
Transformation Theorem for
Harmonic Functions
If f be an analytic function that maps a domain D onto
a domain D. If U is harmonic in D, then the real-valued
function u(x, y) = U(f(z)) is harmonic in D.
Ch20_35
THEOREM 20.2
Proof
We will give a special proof for the special case in
which D is simply connected. If U has a harmonic
conjugate V in D, then H = U + iV is analytic in D, and
so the composite function H(f(z)) = U(f(z)) + iV(f(z)) is
analytic in D. It follow that the real part U(f(z)) is
harmonic in D.
Ch20_36
Solving Dirichlet Problems Using Conformal
Mapping
 Solving Dirichlet Problems Using Conformal
Mapping
1. Find a conformal mapping w = f(z) that transform
s the original region R onto the image R. The
region R may be a region for which many
explicit solutions to Dirichlet problems are
known.
2. Transfer the boundary conditions from the R to
the boundary conditions of R. The value of u at a
boundary point  of R is assigned as the value of
U at the corresponding boundary point f().
Ch20_37
Fig 20.13
Ch20_38
3. Solve the Dirichlet problem in R. The solution
may be apparent from the simplicity of the
problem in R or may be found using Fourier or
integral transform methods.
4. The solution to the original Dirichlet problems is
u(x, y) = U(f(z)).
Ch20_39
Example 6
The function U(u, v) = (1/) Arg w is harmonic in the
upper half-plane v > 0 since it is the imaginary part of
the analytic function g(w) = (1/) Ln w. Use this
function to solve the Dirichlet problem in Fig 20.14(a).
Ch20_40
Fig 20.14
Ch20_41
Example 6 (2)
Solution
The analytic function f(z) = sin z maps the original
region to the upper half-plane v  0 and maps the
boundary segments to the segments shown in Fig
20.14(b). The harmonic function U(u, v) = (1/) Arg w
satisfies the transferred boundary conditions U(u, 0) = 0
for u > 0 and U(u, 0) = 1 for u < 0.
u ( x, y ) 
1

tan
1 
cos x sinh y 


 sin x cosh y 
Ch20_42
Example 7
From C-1 in Appendix IV, the analytic function
f (z) 
za
az  1
, a 
72 6
5
maps the region outside the two open disks |z| < 1 and
|z – 5/2| < ½ onto the circular region r0  |w|  1,
where r0  5  2 6 . See Fig 20.15(a) and (b).
Ch20_43
Fig 20.15
Ch20_44
Example 7 (2)
In problem 10 of Exercise 14.1, we found
U ( r ,  )  (log e r ) /(log
e r0 ) z )
is the solution to the new problem. From Theorem 20.2
we conclude that the solution to the original problem is
u ( x, y ) 
1
log e ( 5  2 6 )
log
z  (7  2 6 ) / 5
e
(7  2 6 ) z / 5  1
Ch20_45
A favorite image region R for a simply connected
region R is the upper half-plane y  0. For any real
number a, the complex function
Ln(z – a) = loge|z – a| + i Arg (z – a)
is analytic in R and is a solution to the Dirichlet
problem shown in Fig 20.16.
Ch20_46
Fig 20.16
Ch20_47
It follows that the solution in R to the Dirichlet
problem with
 c0 , a  x  b
U ( x, 0)  
 0 , otherwise
is the harmonic function
U(x, y) = (c0/)(Arg(z – b) – Arg(z – a))
Ch20_48
20.3 Linear Fractional Transformations
Linear Fractional Transformation
If a, b, c, d are complex constants with ad – bc  0,
then the function
T (z) 
az  b
cz  d
is called a llinear fractional
T '( z) 
transform ation. Since
ad  bc
( cz  d )
2
Ch20_49
T is conformal at z provided
 = ad – bc  0 and z  −d/c.
Note when c  0, T(z) has a simple zero at z0 = −d/c,
and so
lim T ( z )   ,
z  z0
We will write T ( z 0 )   . In addition,
lim T ( z )  lim
z 
z 
a b/z
cd/z

a
if c  0 , then
,
c
and we write T (  )  a / c .
Ch20_50
Example 1
If T(z) = (2z + 1)/(z – i), compute T(0), T(), T(i).
Solution
T ( 0 )  1 /(  i )  i , T (  )  lim T ( z )  2 ,
z 
T ( i )  lim T ( z )   , T ( i )  
zi
Ch20_51
Circle Preserving Property
If c = 0, the transformation reduces to a linear
function T(z) = Az + B. This is a composition of a
rotation, magnification, and translation. As such, a
linear function will map a circle in the z-plane to a
circle in the w-plane. When c  0,
w
az  b
cz  d

bc  ad
1
c
cz  d

c
(1)
a
Ch20_52
Letting A 
bc  ad
1
z1
, w  Az 2  B
Note that if z  z1  r , w 
1
w

1
w1

w  w1
w w1
, T ( z ) can be written as
c
c
z1  cz  d , z 2 
,B 
a
1
(2)
, then
z
 r or w  w1  ( r w1 ) w  0
(3)
Ch20_53
It is easy to show that all points w that satisfy
w  w1   w  w 2
(4)
is a line when  = 1 and is a circle when  > 0 and 
 1. It follows from (3) that the image of the circle
|z – z1| = r under the inversion w = 1/z is a circle
except when r = 1/|w1| = |z1|.
Ch20_54
THEOREM 20.3
Circle-Preserving Property
A linear fractional transformation maps a circle in the
z-plane to either a line or a circle in the w-plane. The
image is a line if and only if the original circle passes
through a pole of the linear fractional transformation.
Ch20_55
Example 2
Find the images of the circles |z| = 1 and |z| = 2 under
T(z) = (z + 2)/(z – 1). What are the images of the
interiors of these circles?
Solution
The circle |z| = 1 passes through the pole z0 = 1 of the
linear transformation and so the image is a line. Since
T(−1) = −½ and T(i) = −(1/2) – (3/2)i, we conclude that
the line is u = −½.
Ch20_56
Example 2 (2)
The image of the interior |z| = 1 is either the half-plane
u < −½ or the half-plane u > −½. Using z = 0 as a test
point, T(0) = −2 and so the image is the half-plane u <
−½.
The circle |z| = 2 does not pass through the pole so the
image is a circle. For |z| = 2,
z  2,T ( z ) 
z2
z 1

z2
z 1
 T (z)
Ch20_57
Example 2 (2)
Therefore T ( z ) is a point on the image circle
and the image circle is symmetric
w.r. t . the u - axis.
Since T(−2) = 0 and T(2) = 4 the center of the circle is w
= 2 and the image is the circle |w – 2| = 2. The interior
of |z| = 2 is either the interior or the exterior of the
image |w – 2| = 2. Since T(0) = −2, we conclude that the
image is |w – 2| > 2. See Fig 20.33.
Ch20_58
Fig 20.33
Ch20_59
Matrix Methods
We associate the matrix
a
A 
c
If T1 ( z ) 
b
az  b
 with T ( z ) 
cz  d
d
a1 z  b1
c1 z  d 1
, T2 ( z ) 
a 2 z  b2
c2 z  d 2
then T 2 (T1 ( z )) is given by T ( z ) 
,
az  b
cz  d
Ch20_60
where
a

c
b   a2

d   c2
If w  T ( z ) 
that is, T
matrix is
1
b 2   a1

d 2   c1
az  b
cz  d
(w) 
b1 

d1 
, then z 
dw  b
 cw  a
 d
adj A  
 c
(5 )
dw  b
 cw  a
, and the associated
 b

a 
(6)
Ch20_61
Example 3
If T ( z ) 
2z 1
z2
Solution
Let S (T ( z )) 
-1
a

c
b
1
  adj 
d
i
and S ( z ) 
az  b
cz  d
S
1
(T ( z )) 
iz  1
-1
, find S (T ( z )).
, where
 i 2
 
 1  1
  1 i  2


  i 1  1
zi
 1

2 
 1   2  i

2   1  2i
 1  2i 
 , then
2i 
( 2  i) z  1  2i
(1  2 i ) z  2  i
Ch20_62
Triples to Triples
The linear fractional transformation
T (z) 
z  z1 z 2  z 3
z  z 3 z 2  z1
has a zero at z = z1, a pole at z = z3 and T(z2) = 1. Thus
T(z) maps three distinct complex numbers z1, z2, z3 to
0, 1, and , respectively. The term
The term
z  z1 z 2  z 3
z  z 3 z 2  z1
is called the cross - ratio of
z , z1 , z 2 , z 3 .
Ch20_63
Likewise, the linear fractional transformation
S (w) 
w  w1 w 2  w 3
w  w 3 w 2  w1
sends w1, w2, w3 to 0, 1, and , and so S-1maps 0, 1,
and  to w1, w2, w3. It follows that w = S-1(T(z)) maps
the triple z1, z2, z3 to the triples w1, w2, w3. From w =
S-1(T(z)), we have S(w) = T(z) and
w  w1 w 2  w 3
w  w 3 w 2  w1

z  z1 z 2  z 3
z  z 3 z 2  z1
(7)
Ch20_64
Example 4
Construct a linear fractional transformation that maps
the points 1, i, −1 on the circle |z| = 1 to the points −1, 0
and 1 on the real x-axis.
Solution
From (7) we get
w 1 0 1
w  1 0  (  1)

z 1i 1
z 1i 1
or 
w 1
w 1
 i
z 1
z 1
Solving for w, we get w = −i(z – i)/(z + i).
Ch20_65
Example 5
Construct a linear fractional transformation that maps
the points , 0, 1 on the real x-axis to the points 1, i, −1
on the circle |w| = 1.
Solution
Since z1 = , the terms z − z1 and z2 − z1 in the crossproduct are replaced by 1. Then
w 1i 1
w 1i 1

1
0 1
z 1 1
or S ( w )   i
w 1
w 1

1
z 1
 T (z)
Ch20_66
Example 5 (2)
If we use the matrix method to find w = S-1(T(z)),
a

c
b
 i
  adj 
d
 1
and so w 
i 0
 
1  1
 iz  1  i
 iz  1  i

 1   i

 1   i
z 1 i
z 1 i
1 i

1 i 
.
Ch20_67
Example 6
Solve the Dirichlet problem in Fig 20.35(a) using
conformal mapping by constructing a linear fractional
transformation that maps the given region into the upper
half-plane.
Ch20_68
Fig 20.35(a)
Ch20_69
Example 6 (2)
Solution
The boundary circles |z| = 1 and |z – ½| = ½ each pass
through z = 1. We can map each boundary circle to a
line by selecting a linear fractional transformation that
has a pole at z = 1. If we require T(i) = 0 and T(-1) = 1,
then
T (z) 
z  i 11
z 11 i
 (1  i )
zi
z 1
Since T ( 0 )  1  i , T ( 12  12 i )   1  i , T maps the interior
of |z| = 1 onto the upper half-plane and maps |z – ½| = ½
onto the line v = 1. See Fig 20.35(b).
Ch20_70
Example 6 (3)
The harmonic function U(u, v) = v is the solution to the
simplified Dirichlet problem in the w-plane, and so u(x,
y) = U(T(z)) is the solution to the original Dirichlet
problem in the z-plane.
Since the imaginary
1 x  y
2
2
( x  1)  y
2
part of T ( z )  (1  i )
zi
z 1
1 x  y
2
2
, the solution is u ( x , y ) 
is
2
( x  1)  y
2
2
Ch20_71
Example 6 (4)
The level curves u(x, y) = c can be written as
2


 1 
2
x
  y 

1 c

1  c 
c
2
and are therefore circles that pass through z = 1. See Fig
20.36.
Ch20_72
Fig 20.36
Ch20_73
20.4 Schwarz-Christoffel Transformations
Special Cases
First we examine the effect of f(z) = (z – x1)/, 0 < 
< 2, on the upper half-plane y  0 shown in Fig
20.40(a).
Ch20_74
The mapping is the composite of  = z – x1 and w =
/. Since w = / changes the angle in a wedge by a
factor of /, the interior angle in the image is (/)
= . See Fig 20.40(b).
Ch20_75
 Note that
f'(z)  A ( z  x1 )
(  /  ) 1
, A   /
Next assume that f(z) is a function that is analytic in
the upper half-plane and that jas the derivative
(  1 /  ) 1
2

f ( z )  A ( z  x1 )
( z  x 2 )(  /  )  1
(1)
where x1 < x2. We use the fact that a curve w = w(t) is
a line segment when the argument of its tangent
vector w(t) is constant. From (1) we get
arg f  ( t )
 1

 Arg A  
 1  Arg ( t  x1 ) 


(2)
2


1

 Arg ( t  x 2 )
 

Ch20_76
Since Arg(t – x) =  for t < x, we can find the
variation of f (t) along the x-axis. They are shown in
the following table.
Interval
(  , x1 )
arg f ' ( t )
Change in arg.
Arg A  ( α1  π )  ( α 2  π )
0
( x1 , x 2 )
Arg A  ( α 2  π )
π  α1
( x2 ,  )
Arg A
π  α2
See Fig 20.41.
Ch20_77
Fig 20.41
Ch20_78
THEOREM 20.4
Schwarz-Christoffel Formula
Let f(z) be a function that is analytic in the upper
half-plane y > 0 and that has the derivative
f ( z )
 A ( z  x1 )
(  1 /  ) 1
( z  x2 )
(  2 /  ) 1
 ( z  xn )
(  n /  ) 1
(3)
where x1 < x2 < … < xn and each i satisfies
0 < i < 2. Then f(z) maps the upper half-plane y  0
to a polygonal region with interior angles 1, 2, …,
n.
Ch20_79
Comments
(i) One can select the location of three of the points xk
on the x-axis.
(ii) A general form for f(z) is
f (z)  A
 ( z  x )
(  1 /  ) 1
1
and can be considered
g (z) 
 ( z  x1 )
(  1 /  ) 1
...( z  x n )
(  n /  ) 1
as the compisite
...( z  x n )
(  n /  ) 1

dz  B
of
dz
and w  Az  B .
(iii) If the polygonal region is bounded only n – 1 of the
n interior angles should be included in the SchwarzChristoffel formula.
Ch20_80
Example 1
Use the Schwarz-Christoffel formula to construct a
conformal mapping from the upper half-plane to the
strip |v| 1, u  0. See Fig 20.42.
Ch20_81
Example 1 (2)
Solution
We may select x1 = −1, x2 = 1 on the x-axis, and we will
construct a mapping f with f(−1) = −i, f(1) = i. Since 1
= 2 = /2, (3) gives
f ' ( z )  A ( z  1)
 A
1 / 2
( z  1)
1
( z  1)
2
1/ 2

A
1 / 2
1
i (1  z )
2 1/ 2
.
Ch20_82
Example 1 (3)
Thus f(z)   Ai sin
1
z  B.
Since f (  1)   i , f (1)  i , then
 i  Ai

 B , i   Ai
2
then
f (z) 

 B imply B  0 , A   2 / 
2
2

i sin
1
z.
Ch20_83
Example 2
Use the Schwarz-Christoffel formula to construct a
conformal mapping from the upper half-plane to the
region shown in Fig 20.43(b).
Ch20_84
Example 2 (2)
Solution
We may select x1 = −1, x2 = 1 on the x-axis, and we will
require f(−1) = ai, f(1) = 0. Since 1 = 3/2
2 = /2, (3) gives
f ' ( z )  A ( z  1)
1/ 2
( z  1)
1 / 2
. If we write


z
1
f '( z)  A  2
 2
, then
1/ 2
1/ 2 
( z  1) 
 ( z  1)

f ( z )  A ( z  1)
2
1/ 2
 cosh
1

z B
Ch20_85
Example 2 (3)
Notice that cosh
1
(  1)   i and cosh
1
1  0,
and so ai  f (  1)  A ( i )  B , 0  f (1)  B .
Therefore
f (z) 
a

( z
2
 1)
1/ 2
 cosh
1
z

Ch20_86
Example 3
Use the Schwarz-Christoffel formula to construct a
conformal mapping from the upper half-plane to the
region shown in Fig 20.44(b).
Ch20_87
Example 3 (2)
Solution
Since the region is bounded, only two of the 60 interior
angles should be included. If x1 = 0, x2 = 1, we obtain
f ' ( z )  Az
2 / 3
( z  1)
2 / 3
We can use Theorem 18.8 to get the
antideriva tive
f (z)  A
1
z
0
s
2/3
( s  1)
2/3
ds  B .
Ch20_88
Example 3 (3)
We require
1  A
1
1
0
Thus
f ( 0 )  0 , f (1)  1, then B  0 and
x
2/3
( x  1)
f (z) 
2/3
1
dx   (1 / 3 )
z
1
ds

2
/
3
2
/
3
 (1 / 3 ) 0 s ( s  1)
Ch20_89
Example 4
Use the Schwarz-Christoffel formula to construct a
conformal mapping from the upper half-plane to the
upper half-plane with the horizontal line v = , u  0,
deleted. See Fig 20.45.
Ch20_90
Example 4 (2)
Solution
The nonpolygonal target region can be approximated by
a polygonal region by adjoining a line segment from w
= i to a pint u0 on the negative u-axis. See Fig 20.45(b).
If we require f(−1) = i, f(0) = u0, then
f ' ( z )  A ( z  1)
(  1 /  ) 1
z
(  2 /  ) 1
Note that as u0 approaches −, 1 and 2 approach 2
and 0, respectively.
Ch20_91
Example 4 (3)
This suggests we examine the mappings that satisfy w
= A(z + 1)1z-1 = A(1 + 1/z) or w = A(z + Ln z) + B.
First we determine the image of the upper half-plane
under g(z) = z + Ln z and then translate the image region
if needed. For t real
g(t) = t + loge |t| + i Arg t
If t < 0, Arg t =  and u(t) = t + loge |t| varies from − to
−1. It follows that w = g(t) moves along the line v = 
from − to −1.
Ch20_92
Example 4 (4)
When t > 0, Arg t = 0 and u(t) varies from − to .
Therefore g maps the positive x-axis onto the u-axis. We
can conclude that g(z) = z + Ln z maps the upper halfplane onto the upper half-plane with the horizontal line
v = , u  −1, deleted. Therefore w = z + Ln z + 1 maps
the upper half-plane onto the original target region.
Ch20_93
20.5 Poisson Integral Formulas
Introduction
It would be helpful if a general solution could be
found for Dirichlet problem in either the upper halfplane y  0 or the unit disk |z| = 1. The Poisson
formula fro the upper half-plane provides such a
solution expressing the value of a harmonic function
u(x, y) at a point in the interior of the upper half-plane
in terms of its values on the boundary y = 0.
Ch20_94
Formula for the Upper Half-Plane
Assume that the boundary function is given by u(x, 0)
= f(x), where f(x) is the step function indicated in Fig
20.55.
Ch20_95
The solution of the corresponding Dirichlet problem
in the upper half-plane is
u ( x, y ) 
ui

[ Arg ( z  b )  Arg ( z  a )]
(1)
Since Arg(z – b) is an exterior angle formed by z, a
and b, Arg(z – b) = (z) + Arg(z – a), where 0 <  < ,
and we can write
zb
u ( x , y )   ( z )  Arg 



za
ui
ui
(2)
Ch20_96
The superposition principle can be used to solve the
more general Dirichlet problem in Fig 20.56.
Ch20_97
If u(x, 0) = ui for xi-1  x  xi, and u(x, 0) = 0 outside
the interval [a, b], then from (1)
n
u ( x, y ) 

ui
1
n
i 1



[ Arg ( z  x i )  Arg ( z  x i 1 )]
 u i i ( z )
(3)
i 1
Note that Arg(z – t) = tan-1(y/(x – t)), where tan-1 is
selected between 0 and , and therefore
(d/dt) Arg(z – t) = y/((x – t)2 + y2).
Ch20_98
From (3),
u ( x, y ) 

1

n

xi
 x
i 1
1
i 1
n
xi
 x
i 1
i 1
ui
dt
Arg ( z  t ) dt
d
ui y
(x  t)  y
2
2
dt
Since u ( x , 0 )  0 outside the interval [ a,b ],
we have
u ( x, y ) 
y


u (t ,0 )
  ( x  t ) 2 
y
2
dt
(4)
Ch20_99
THEOREM 20.5
Poisson Integral Formula for the
Upper Half-Plane
Let u(x, 0) be a piecewise-continuous function on every
finite interval and bounded on － < x < . Then the
function defined by
u ( x, y ) 
y

u (t , 0 )
   ( x  t ) 2 
y
2
dt
is the solution of the corresponding Dirichlet problem
on the upper half-plane y  0.
Ch20_100
Example 1
Find the solution of the Dirichlet problem in the upper
half-plane that satisfies the boundary condition u(x, 0) =
x, where |x| < 1, and u(x, 0) = 0 otherwise.
Solution
By the Poisson integral formula
u ( x, y ) 
y

1
t
1 ( x  t ) 2 
y
2
dt .
Ch20_101
Example 1 (2)
Using s  x  t , then
1 y
2
2
u ( x , y )   log e (( x  t )  y )  x tan
 2
 ( x  1) 2  y 2  x 

log e 
  tan
2
2
2
 ( x  1)  y   
y
x  t  t  1


 y  t  1
1 
x  1

  tan
 y 
1 
x  1 


 y 
1 
Ch20_102
Example 2
The conformal mapping f(z) = z + 1/z maps the region in
the upper half-plane and outside the circle |z| = 1 onto
the upper half-plane v  0. Use the mapping and the
Poisson integral formula to solve the Dirichlet problem
shown in Fig 20.57(a).
Ch20_103
Fig 20.57
Ch20_104
Example 2 (2)
Solution
Using the result of Example 4 of Sec 20.2, we can
transfer the boundary conditions to the w-plane. See Fig
20.57(b). Since U(u, 0) is a step function, we will use
the integrated solution (3) rather than the Poisson
integral. The solution to the new Dirichlet problem is
U (u , v ) 
1

Arg ( w  2 ) 
1
1

Arg (
w2
w2
1

[  Arg ( w  2 )]
)
Ch20_105
Example 2 (3)
and therefore
u ( x, y )  U ( z 
1
z
) 1
1

 z  1
u ( x , y )  1  Arg 


 z  1
1
Arg (
z 1 z  2
z 1 z  2
)
2
Ch20_106
THEOREM 20.6
Poisson Integral Formula for the
Unit Disk
Let u(ei) be a bounded and piecewise continuous for
－    . Then the solution to the corresponding
Dirichlet Problem on the open units disk |z| < 1 is given
by
u ( x, y ) 
1


2 
it
u (e )
1 | z |
2
|e  z|
it
2
dt
(5)
Ch20_107
Geometric Interpretation
Fig 20.58 shows a thin membrane (as a soap film)
that has been stretched across a frame defined by u =
u(ei).
Ch20_108
Fig 20.58
Ch20_109
The displacement u in the direction perpendicular to
the z-plane satisfies the two-dimensional wave
equation
2
2
2



u

u

u
2

a  2 
2 
2
y  t
 x
and so at equilibrium, the displacement function u =
u(x, y) is harmonic. Formula (5) provides an explicit
solution for u and has the advantage that the integral
is over the finite interval [−, ].
Ch20_110
Example 3
A frame for a membrane is defined by u(ei) = | | for
−    . Estimate the equilibrium displacement of
the membrane at (−0.5, 0), (0, 0) and (0.5, 0).
Solution
Using (5),
u ( x, y ) 
1
2

  t
1 z
2
e z
it
2
dt
Ch20_111
Example 3 (2)
When (x, y) = (0, 0), we get
u ( x, y ) 
1
2

  t dt


2
For the other two values of (x, y), the integral is not
elementary and must be estimated using a numerical
solver. We have u(−0.5, 0) = 2.2269, u(0.5, 0) =
0.9147.
Ch20_112
Fourier Series Form
Note that un(r,) = rn cos n and vn(r,) = rn sin n
are each harmonic, since these functions are the real
and imaginary parts of zn. If a0, an, bn are chosen to be
the Fourier coefficients of u(ei) for − <  < , then
u (r ,  ) 
a0
2


 a n r
n
cos n   b n r sin n 
n

(6)
n 1
Ch20_113
We find (6) is harmonic and
u (1,  ) 
a0


2
From
i


a
cos
n


b
sin
n


u
(
e
)
 n
n
n 1
(5) we have
u ( r , ) 

1
2
a0
2

 
1 r
it
u (e )
e  re
it
2
i
2
dt


n
n

a
r
cos
n


b
r
sin n  
 n
n
n 1
Ch20_114
Example 4
Find the solution of the Dirichlet problem in the unit
disk satisfying the boundary condition u(ei) = sin 4 .
Sketch the level curve u = 0.
Solution
Rather than using (5), we use (6) which reduces to u(r,
) = r4 sin 4 . Therefore u = 0 if and only if
sin 4 = 0. This implies u = 0 on the lines x = 0, y = 0
and y = x.
Ch20_115
Example 4 (2)
If we switch to rectangular coordinates, u(x, y) =
4xy(x2 – y2). The surface of u(x, y) = 4xy(x2 – y2), the
frame u(ei) = sin 4, and the system of level curves
were sketched in Fig 20.59.
Ch20_116
Fig 20.59
Ch20_117
20.6 Applications
Vector Fields
A vector field F(x, y) = P(x, y)i + Q(x, y)j in a domain
D can also be expressed in the complex form
F(x, y) = P(x, y) + iQ(x, y)
Recall that div F = P/x +Q/y and curl F = (Q/x
−P/y)k. If we require both of them are zeros, then
P
x

Q
y
and
P
y

Q
x
(1)
Ch20_118
THEOREM 20.7
Vector Fields and Analyticity
(i) Suppose that F(x, y) = P(x, y) + Q(x, y) is a vector
field in a domain D and P(x, y) and Q(x, y) are
continuous and have continuous first partial
derivatives in D. If div F = 0 and curl F = 0, then
complex function
g ( z )  P ( x , y )  iQ ( x , y )
is analytic in D.
(ii) Conversely, if g(z) is analytic in D, then F(x, y) =
g ( z ) defined a vector field in D for which div F = 0
and curl F = 0.
Ch20_119
THEOREM 20.7
Proof
If u(x, y) and v(x, y) denote the real and imaginary parts
of g(z), then u = P and v = −Q. Then
u
 ( v ) u  ( v )
 
,

; that is,
x
y
y
x
u
v u
v

,
 
x y y
x
(2)
Equations in (2) are the Cauchy-Riemann equations for
analyticity.
Ch20_120
Example 1
The vector field F(x, y) = (−kq/|z − z0|2)(z − z0) may be
interpreted as the electric field by a wire that is
perpendicular to the z-plane at z = z0 and carries a
charge of q coulombs per unit length. The
corresponding complex function is
g (z) 
 kq
z  z0
2
( z  z0 ) 
 kq
z  z0
Since g ( z ) is analytic for z  z 0 , div F  0, curl F  0 .
Ch20_121
Example 2
The complex function g(z) = Az, A > 0, is analytic in the
first quadrant and therefore gives rise to the vector field
V ( x , y )  g ( z )  Ax  iAy
which satisfies div V  0 , curl V  0 .
Ch20_122
Potential Functions
Suppose that F(x, y) is a vector field in a simply
connected domain D with div F = 0 and curl F = 0.
By Theorem 18.8, the analytic function g(z) = P(x, y)
− iQ(x, y) has an antiderivative
G ( z )   ( x , y )  i ( x , y )
(4)
in D, which is called a complex potential for the
vector filed F.
Ch20_123
 Note that g ( z )  G ' ( z )    ( x , y )  i   ( x , y )
x


x
( x, y )  i
and so

x

y
x
( x, y )
 P,

y
Q
(5)
Therefore F =  , and the harmonic function  is
called a (real) potential function of F.
Ch20_124
Example 3
The potential  in the half-plane x  0 satisfies the
boundary conditions (0, y) = 0 and (x, 0) = 1 for x
 1. See Fig. 20.68(a). Determine a complex
potential, the equipotential lines, and the field F.
Ch20_125
Fig. 20.68
Ch20_126
Example 3 (2)
Solution
We knew the analytic function z = sin w maps the strip 0
 u  /2 in the w-plane to the region R in question.
Therefore f(z) = sin-1z maps R onto the strip, and Fig
20.68(b) shows the transferred boundary conditions.
The simplified Dirichlet problem has the solution U(u, v)
= (2/)u, and so (x, y) = U(sin-1z) = Re((2/) sin-1z) is
the potential function on D, and G(z) = (2/)u sin-1z is a
complex potential for F.
Ch20_127
Example 3 (3)
Note that the equipotential lines  = c are the images of
the equipotential lines U = c in the w-plane under the
inverse mapping z = sin w. We found that the vertical
lines u = a is mapped onto a branch of the hyperbola
x
2
2
sin a

y
2
2
1
cos a
Ch20_128
Example 3 (4)
Since the equipotential lines U = c, 0 < c < 1 is the
vertical line u = /2c, it follows that the equipotential
lines  = c is the right branch of the hyperbola
x
2
sin ( c / 2 )
2

y
cos ( c / 2 )
2
Since F  G' ( z ) and ( d / dz ) sin
then F 
2
1
 (1  z )
2 1/ 2

2
2
1
1
z  1 /( 1  z )
2 1/ 2
,
1
 (1  z 2 )1 / 2
Ch20_129
Steady-State Fluid Flow
The vector V(x, y) = P(x, y) + iQ(x, y) may also be
expressed as the velocity vector of a two-dimensional
steady-state fluid flow at a point (x, y) in a domain D.
If div V = 0 and curl V = 0, V has a complex velocity
potential
G(z) = (x, y) + (x, y)
that satisfies
G' ( z )  V
Ch20_130
Here special importance is placed on the level curves
(x, y) = c. If z(t) = x(t) + iy(t) is the path of a
particle, then
dx
dt
Hence
 P ( x , y ),
dy
 Q ( x, y )
(6)
dt
dy / dx  Q ( x , y ) / P ( x , y ) or
 Q ( x,y ) dx  P ( x , y ) dy  0 .
Ch20_131
 Since div V  0 implies
 (Q )
y

P
x
and by the Cauchy - Riemann equations

x
 

y
and all solutions
  Q and

y


x
 P
of (6) satisfy ψ ( x , y )  c .
The function (x, y) is called a stream function and
the level curves (x, y) = c are streamlines for the
flow.
Ch20_132
Example 4
The uniform flow in the upper half-plane is defined
by V(x, y) = A(1, 0), where A is a fixed positive
constant. Note that |V| = A, and so a particle in the
fluid moves at a constant speed. A complex potential
for the vector field is G(z) = Az = Ax + iAy, and so the
streamlines are the horizontal lines Ay = c. See Fig
20.69(a). Note that the boundary y = 0 of the region is
itself streamline.
Ch20_133
Fig 20.69(a)
Ch20_134
Example 5
The analytic function G(z) = z2 gives rise to the vector
field V ( x , y )  G' ( z )  ( 2 x ,  2 y ) in the first quadrant.
Since z2 = x2 − y2 + i(2xy), the stream function is (x,
y) = 2xy and the streamlines are the hyperbolas 2xy =
c. See Fig 20.69(b).
Ch20_135
Fig 20.69(b)
Ch20_136
THEOREM 20.8
Streamline
Suppose that G(x) = (x, y) + i(x, y) is analytic in a
region R and (x, y) is continuous on the boundary of
R. Then V(x, y) = G  ( z ) defined an irrotational and
incompressible fluid flow in R. Moreover, if a particle
is placed is placed inside R, its path z = z(t) remains in
R.
Ch20_137
Example 6
The analytic function G(z) = z + 1/z maps the region R
in the upper half-plane and outside the circle |z| = 1onto
the upper half-plane v  0. The boundary of R is
mapped onto the u-axis, and so v = (x, y) = y – y/(x2 +
y2) is zero on the boundary of R. Fig 20.70 shows the
streamlines. The velocity field is given by
G' ( z )  1  1 / z , and so
2
G' ( re
i
) 1
1
r
2
e
2 i
Ch20_138
Example 6 (2)
It follows that V  (1, 0) for large values of r, and so the
flow is approximately uniform at large distance from the
circle |z| = 1. The resulting flow in R is called flow
around a cylinder.
Ch20_139
Fig 20.70
Ch20_140
Example 7
The analytic function f(w) = w + Ln w + 1 maps the
upper half-plane v  0 to the upper half-plane y  0,
with the horizontal line y = , x  0, deleted. See
Example 4 in Sec 20.4. If G(z) = f -1(z) = (x, y) + i(x,
y), then G(z) maps R onto the upper half-plane and maps
the boundary of R onto the u-axis. Therefore (x, y) = 0
on the boundary of R.
Ch20_141
Example 7 (2)
It is not easy to find an explicit formula for (x, y). The
streamlines are the images of the horizontal lines v = c
under z = f(w). If we write w = t + ic, c > 0, then the
streamlines can be
z  f ( t  ic )  t  ic  Ln( t  ic )  1, that is,
x  t 1
1
2
log e ( t  c ), y  c  Arg ( t  ic )
2
2
See Fig 20.71.
Ch20_142
Fig 20.71
Ch20_143
Example 8
The analytic function f(w) = w + ew + 1 maps the strip 0
 v   onto the region R shown in Fig 20.71.
Therefore G(z) = f -1(z) = (x, y) + i(x, y) maps R back
to the strip and from M-1 in the Appendix IV, maps the
boundary line y = 0 onto the u-axis and maps the
boundary line y =  onto the horizontal line v = .
Therefore (x, y) is constant on the boundary of R.
Ch20_144
Example 8 (2)
The streamlines are the images of the horizontal lines v
= c, 0 < c < , under z = f(w). If we write w = t + ic,
then the streamlines can be
z  f ( t  ic )  t  ic  e
( t  ic )
 1, that is,
x  t  1  e cos c , y  c  e sin c
t
t
See Fig 20.72.
Ch20_145
Fig 20.72
Ch20_146