Complex Analysis - Maths KU

286 Chapter 5 Integration in the Complex Plane
(i) If F(x, y) = P (x, y)i + Q(x, y)j is a vector field for which
div F = 0 and curl F = 0 in D, and if f(z) =P (x, y)+iQ(x, y)
is the complex representation of F, then the function g(z) =f(z) =
P (x, y) − iQ(x, y) is analytic in D.
(ii) Conversely, if g(z) =u(x, y)+iv(x, y) is analytic in D, then the function
f(z) =g(z) =u(x, y) − iv(x, y) is the complex representation
of a vector field F(x, y) =P (x, y)i + Q(x, y)j for which div F =0
and curl F = 0 in D.
Proof (i) If we let u(x, y) and v(x, y) denote the real and imaginary parts
of g(z) =f(z) =P (x, y) − iQ(x, y), then P = u and Q = −v. Because div F
= 0 and curl F = 0, the equations in (2) become, respectively,
That is,
∂u
= −∂(−v)
∂x ∂y
∂u ∂v
=
∂x ∂y
and
and
∂u
∂y
∂u
∂y
= ∂(−v)
∂x .
∂v
= − . (3)
∂x
The equations in (3) are the usual Cauchy-Riemann equations, and so by
Theorem3.5 we conclude that g(z) =f(z) =P (x, y) − iQ(x, y) is analytic
in D.
(ii) We now let P (x, y) and Q(x, y) denote the real and imaginary parts
of f(z) =g(z) =u(x, y) − iv(x, y). Since u = P and v = −Q, the Cauchy-
Riemann equations become
∂P
∂x
= ∂(−Q)
∂y
= −∂Q
∂y
and
∂P
∂y
= −∂(−Q)
∂x
∂Q
= . (4)
∂x
These are the equations in (2) and so div F = 0 and curl F = 0. ✎
EXAMPLE 1 Vector Field Gives an Analytic Function
The two-dimensional vector field
F(x, y) = K
�
�
2π
,K >0,
y − y0
(x − x0) 2 2 i −
+(y − y0)
x − x0
(x − x0) 2 2 j
+(y − y0)
can be interpreted as the velocity field of the flow of an ideal fluid in a domain
D of the xy-plane not containing (x0, y0). It is easily verified that the fluid is
incompressible (div F = 0) and irrotational (curl F = 0) inD. The complex
representation of F is
f(z) = K
�
y − y0
2π (x − x0) 2 x − x0
2 − i
+(y − y0) (x − x0) 2 +(y − y0) 2
�
.