Complex Analysis - Maths KU

3.4 Applications 165
tangent is the negative reciprocal of the slope of the other by showing that the
product of the two slopes is −1. We begin by differentiating u(x, y) =u0 and
v(x, y) =v0 with respect to x using the chain rule of partial differentiation:
∂u
∂x
∂u dy
+ = 0 and
∂y dx
∂v ∂v dy
+
∂x ∂y dx =0.
We then solve each of the foregoing equations for dy/dx:
slope of a tangent to curve u(x, y) =u0 � �� �
dy
= −∂u/∂x
dx ∂u/∂y ,
slope of a tangent to curve v(x, y) =v0 � �� �
dy
dx
= −∂v/∂x , (3)
∂v/∂y
At (x0, y0) we see from (3), the Cauchy-Reimann equations ux = vy,
uy = −vx, and from f ′ (z0) �= 0, that the product of the two slope functions is
�
− ∂u/∂x
∂u/∂y
��
− ∂v/∂x
∂v/∂y
�
=
� ∂v/∂y
∂v/∂x
��
− ∂v/∂x
∂v/∂y
�
= −1. (4)
EXAMPLE 1 Orthogonal Families
For f(z) =z2 = x2−y 2 +2xyi we identify u(x, y) =x2−y2 and v(x, y) =2xy.
For this function, the families of level curves x2 − y2 = c1 and 2xy = c2 are
two families of hyperbolas. Since f is analytic for all z, these families are
orthogonal. At a specific point, say, z0 =2+i we find 22 − 12 =3=c1 and
2(2)(1) = 4 = c2 and two corresponding orthogonal curves are x2 − y2 =3
and xy = 2. Inspection of Figure 3.4(a) shows x2 −y 2 = 3 in color and xy =2
in black; the curves are orthogonal at z0 =2+i (and at −2 − i, by symmetry
of the curves). In Figure 3.4(b) both families are superimposed on the same
coordinate axes, the curves in the family x2 − y2 = c1 are drawn in color
whereas the curves in family 2xy = c2 are in black.
(–2, –1)
y
(2, 1)
(a) Curves are orthogonal at points
of intersection
Figure 3.4 Orthogonal families
x
y
(b) Families x 2 – y 2 = c 1 and
2xy = c 2
Gradient Vector In vector calculus, if f(x, y) is a differentiable scalar
function, then the gradient of f, written either grad f or ∇f (the symbol
x