1. 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

2. 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.

3. Fig 20.1

4. 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.

5. Fig 20.2

6. Example 2
The complex function f = 1/z has domain z  0 and

7. Example 2 (2)
Likewise v(x, y) = b, b  0 can be written as See Fig 20.3.

8. Fig 20.3

9. Translation and Rotation
The function f(z) = z + z0 is interpreted as a translation. The function is interpreted as a rotation. See Fig 20.4.

10. Example 3 Find a complex function that maps −1  y  1 onto 2  x  4.
Solution See Fig We find that −1  y  1 is first rotated through 90 and shifted 3 units to the right. Thus the mapping is

11. Fig 20.5

12. 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 then the vector is rotated through 0, magnified by a factor r0, and then translated using b.

13. 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).

14. Power Functions
A complex function f(z) = z where  is a fixed positive number, is called a real power function. See Fig If z = rei, then w = f(z) = rei.

15. 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.

16. Successive Mapping
See Fig If  = f(z) maps R onto R, and w = g() maps R onto R, w = g(f(z)) maps R onto R.

17. Fig 20.7

18. 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.

19. 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.

20. 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

21. Fig 20.10

22. Referring to Fig 20.10, we have Likewise

23. If f(z) is analytic in the domain D and f’(z)  0, then
THEOREM 20.1
If f(z) is analytic in the domain D and f’(z)  0, then
f is conformal at z = z0.
Conformal Mapping
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 If C1 and C2 intersect at z = z0, then

24. THEOREM 20.1 proof
Since f (z0)  0, we can use (2) to obtain

25. 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.

26. 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.

27. Example 2 (2)
Since cosh2 t − sinh2 t = 1, then 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, ).

28. 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 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

29. Fig 20.12

30. 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 half-plane v  0. This follows from (3) since 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.

31. Conformal Mappings Using Tables
Conformal mappings are given in Appendix IV. We may use this table to solve the transformations.

32. 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 x-axis onto the interval (0, 1) on the u-axis.

33. 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.

34. Example 5 (2)
Therefore 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.

35. If f be an analytic function that maps a domain D onto
THEOREM 20.2
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.
Transformation Theorem for
Harmonic Functions

36. 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.

37. Solving Dirichlet Problems Using Conformal Mapping
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.
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().

38. Fig 20.13

39. Solve the Dirichlet problem in R
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.
The solution to the original Dirichlet problems is u(x, y) = U(f(z)).

40. 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).

41. Fig 20.14

42. 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.

43. Example 7
From C-1 in Appendix IV, the analytic function maps the region outside the two open disks |z| < 1 and |z – 5/2| < ½ onto the circular region r0  |w|  1, See Fig 20.15(a) and (b).

44. Fig 20.15

45. Example 7 (2)
In problem 10 of Exercise 14.1, we found is the solution to the new problem. From Theorem 20.2 we conclude that the solution to the original problem is

46. 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

47. Fig 20.16

48. It follows that the solution in R to the Dirichlet problem with is the harmonic function U(x, y) = (c0/)(Arg(z – b) – Arg(z – a))

49. 20.3 Linear Fractional Transformations
Linear Fractional Transformation If a, b, c, d are complex constants with ad – bc  0, then the function

50. T is conformal at z provided.  = ad – bc  0 and z  −d/c
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

51. Example 1 If T(z) = (2z + 1)/(z – i), compute T(0), T(), T(i).
Solution

52. 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,

54. It is easy to show that all points w that satisfy 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|.

55. THEOREM 20.3
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.
Circle-Preserving Property

56. 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.

57. 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,

58. Example 2 (2)
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

59. Fig 20.33

60. Matrix Methods
We associate the matrix

61. where

62. Example 3
Solution

63. Triples to Triples
The linear fractional transformation 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

64. Likewise, the linear fractional transformation 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

65. 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 Solving for w, we get w = −i(z – i)/(z + i).

66. 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 cross-product are replaced by 1. Then

67. Example 5 (2)
If we use the matrix method to find w = S-1(T(z)),

68. 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.

69. Fig 20.35(a)

70. 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 Since , 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).

71. 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.

72. Example 6 (4)
The level curves u(x, y) = c can be written as and are therefore circles that pass through z = 1. See Fig

73. Fig 20.36

74. 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).

75. The mapping is the composite of  = z – x1 and w = /
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).

76. Next assume that f(z) is a function that is analytic in the upper half-plane and that jas the derivative (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 (2)

77. 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 See Fig

78. Fig 20.41

79. THEOREM 20.4
Let f(z) be a function that is analytic in the upper half-plane y > 0 and that has the derivative (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.
Schwarz-Christoffel Formula

80. 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
(iii) If the polygonal region is bounded only n – 1 of the n interior angles should be included in the Schwarz-Christoffel formula.

81. 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

82. 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

83. Example 1 (3)

84. 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).

85. 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

86. Example 2 (3)

87. 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).

88. 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

89. Example 3 (3)

90. 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

91. 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 Note that as u0 approaches −, 1 and 2 approach 2 and 0, respectively.

92. 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.

93. 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 half-plane 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.

94. 20.5 Poisson Integral Formulas
Introduction It would be helpful if a general solution could be found for Dirichlet problem in either the upper half-plane 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.

95. 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

96. The solution of the corresponding Dirichlet problem in the upper half-plane is (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 (2)

97. The superposition principle can be used to solve the more general Dirichlet problem in Fig 20.56.

98. If u(x, 0) = ui for xi-1  x  xi, and u(x, 0) = 0 outside the interval [a, b], then from (1) (3) 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).

100. on the upper half-plane y  0.
THEOREM 20.5
Let u(x, 0) be a piecewise-continuous function on every finite interval and bounded on － < x < . Then the
function defined by is the solution of the corresponding Dirichlet problem
on the upper half-plane y  0.
Poisson Integral Formula for the
Upper Half-Plane

101. 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

102. Example 1 (2)

103. 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).

104. Fig 20.57

105. 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

106. Example 2 (3)

107. THEOREM 20.6
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 (5)
Poisson Integral Formula for the
Unit Disk

108. Geometric Interpretation
Fig shows a thin membrane (as a soap film) that has been stretched across a frame defined by u = u(ei).

109. Fig 20.58

110. The displacement u in the direction perpendicular to the z-plane satisfies the two-dimensional wave equation 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 [−, ].

111. 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),

112. Example 3 (2)
When (x, y) = (0, 0), we get 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) = , u(0.5, 0) =

113. 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

115. 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.

116. 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

117. Fig 20.59

118. 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 (1)

119. THEOREM 20.7
(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 is analytic in D.
(ii) Conversely, if g(z) is analytic in D, then F(x, y) = defined a vector field in D for which div F = 0 and curl F = 0.
Vector Fields and Analyticity

120. 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 Equations in (2) are the Cauchy-Riemann equations for analyticity.

121. 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

122. Example 2
The complex function g(z) = Az, A > 0, is analytic in the first quadrant and therefore gives rise to the vector field

123. 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 (4) in D, which is called a complex potential for the vector filed F.

124. Therefore F =  , and the harmonic function  is called a (real) potential function of F.

125. 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 (a). Determine a complex potential, the equipotential lines, and the field F.

126. Fig

127. 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.

128. 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

129. 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

130. 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

131. Here special importance is placed on the level curves (x, y) = c
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

132. The function (x, y) is called a stream function and the level curves (x, y) = c are streamlines for the flow.

133. 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.

134. Fig 20.69(a)

135. Example 5
The analytic function G(z) = z2 gives rise to the vector field 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).

136. Fig 20.69(b)

137. is placed is placed inside R, its path z = z(t) remains in R.
THEOREM 20.8
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) = 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.
Streamline

138. 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 shows the streamlines. The velocity field is given by

139. 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.

140. Fig 20.70

141. 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 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.

142. 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 See Fig

143. Fig 20.71

144. Example 8
The analytic function f(w) = w + ew + 1 maps the strip 0  v   onto the region R shown in Fig 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.

145. 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 See Fig

146. Fig 20.72