Complex Analysis - Maths KU

3.2 Cauchy-Riemann Equations 153
We now let ∆z → 0 along a vertical line. With ∆x = 0 and ∆z = i∆y,
(3) becomes
f ′ u(x, y +∆y) − u(x, y)
(z) = lim
∆y→0 i∆y
v(x, y +∆y) − v(x, y)
+ i lim
. (6)
∆y→0 i∆y
In this case (6) shows us that ∂u/∂y and ∂v/∂y exist at z and that
f ′ (z) =−i ∂u
∂y
∂v
+ . (7)
∂y
By equating the real and imaginary parts of (5) and (7) we obtain the pair of
equations in (1). ✎
Because Theorem 3.4 states that the Cauchy-Riemann equations (1) hold
at z as a necessary consequence of f being differentiable at z, we cannot use
the theorem to help us determine where f is differentiable. But it is important
to realize that Theorem 3.4 can tell us where a function f does not possess a
derivative. If the equations in (1) are not satisfied at a point z, then f cannot
be differentiable at z. We have already seen in Example 3 of Section 3.1 that
f(z) =x +4iy is not differentiable at any point z. If we identify u = x and
v =4y, then ∂u/∂x =1,∂v/∂y=4, ∂u/∂y =0, and ∂v/∂x =0. In view of
∂u ∂v
=1�=
∂x ∂y =4
the two equations in (1) cannot be simultaneously satisfied at any point z. In
other words, f is nowhere differentiable.
It also follows from Theorem 3.4 that if a complex function f(z) =
u(x, y)+iv(x, y) is analytic throughout a domain D, then the real functions u
and v satisfy the Cauchy-Riemann equations (1) at every point
in D.
EXAMPLE 1 Verifying Theorem 3.4
The polynomial function f(z) =z 2 +z is analytic for all z and can be written as
f(z) =x 2 −y 2 +x+i(2xy+y). Thus, u(x, y) =x 2 −y 2 +x and v(x, y) =2xy+y.
Foranypoint(x, y) in the complex plane we see that the Cauchy-Riemann
equations are satisfied:
∂u
∂v
=2x +1=
∂x ∂y
and
∂u
∂y
= −2y = − ∂v
∂x .