1549312215-Complex_Analysis_for_Mathematics_and_Engineering_5th_edition__Mathews

502 CHAPTER 11 • APPLICATIONS OF HARMONIC FUNCTIONS

Next, we use the change of variable h = r tan t and dh = r sec^2 t dt and the

· · ·d 't^2 t r

2

+ h

2
tngonometnc 1 ent1 y sec = b. th. I t. t al
r^2 to o tam e equ1va en 1n egr :

J

~(q/2)costr^2 +h^2 qjlf q

IE(x,y)I = 2 f2 dt =

2

cost dt = -.

.::;.r+i r r;;;. 1'

Multiplying this magnitude ~ by the unit vector l: I establishes Formula (11-41).

If q > 0, then the field is directed away from zo = 0 and, if q < 0, then it is
directed toward zo = 0. An electrical field located at zo :/= 0 is given by

E x _ q(z - zo) _ q

(,y)- (^1) z-zo 12 - =--=· z -zo
and the corresponding complex potential is
F(z) = -qlog(z-zo).

• EXAMPLE 11 .30 (Source and sink of equal strength) Let a source and sink
  of unit strength be located at the points +1 and - 1, respectively. The complex
  potential for a fluid ftowing from the source at + 1 to the sink at - 1 is

z-1

F (z) = log(z - 1) - log(z + 1) =log--.

z+l

The velocity potential and stream function are

z - 11

z+ 1

z-l

and 'I/! (x , y) = arg --

1

,

z+

respectively. Solving for the streamline 'ljJ (x, y) = c, we start with

c = arg --= arg = arctan ~-"=---

z + 1 (x+l )^2 +y2 x^2 +y^2 - l

and obtain the equation (tanc) (x^2 + y^2 - 1) = Zy. A straightforward calcula-

tion shows that points on the streamline must satisfy the equat ion

whicli is the equation of a circle with center at (0, cot c) that passes through the

points (±1, 0). Several streamlines are indicated in Figure ll.95(a).