Page 701 - Advanced_Engineering_Mathematics o'neil

19.2 Complex Functions 681

u(x + h, y) − u(x, y) v(x + h, y) − v(x, y)
= lim + i
h→0 h h
∂u ∂v
= + i .
∂x ∂x
Path 2 Along the Imaginary Axis.
Now h = ik with k real and z + h = z + ik = x + i(y + k). Then
u(x, y + k) + iv(x, y + k) − u(x, y) − iv(x, y)
f (z) = lim

k→0 ik

1 u(x, y + k) − u(x, y) v(x, y + k) − v(x, y)
= lim +
k→0 i k k
∂u ∂v
=−i + .
∂y ∂y
Here we used the fact that 1/i =−i. We conclude that
∂u ∂v ∂u ∂v

f (z) = + i =−i + . (19.1)
∂x ∂x ∂y ∂y
Equating real parts and then imaginary parts of opposite sides of this equation, we obtain
∂u ∂v ∂v ∂u
= and =− .
∂x ∂y ∂x ∂y
The Cauchy-Riemann equations are a necessary condition for a complex function to be
differentiable. Given f , u, and v are uniquely determined and f cannot be differentiable at any
point where u and v do not satisfy the Cauchy-Riemann equations.

EXAMPLE 19.10
We already know that f (z) = z is not differentiable. To illustrate the theorem, we will use the
Cauchy-Riemann equations to show this. Write
f (z) = z = x − iy = u(x, y) + iv(x, y),
so u(x, y) = x and v(x, y) =−y. Then
∂u ∂v
= 1 and =−1,
∂x ∂y
so the Cauchy-Riemann equations fail to hold and f is not differentiable at any z.

EXAMPLE 19.11
Let f (z) = z Re(z). Then
2
f (z) = f (x + iy) = (x + iy)x = x + ixy,
2
so u(x, y) = x and v(x, y) = xy. Compute
∂u ∂v
= 2x, = x
∂x ∂y
and
∂u ∂v
= 0, = 0.
∂y ∂x
If z = 0, the Cauchy-Riemann equations do not hold, so f cannot be differentiable at any
z = 0.

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October 15, 2010 18:5 THM/NEIL Page-681 27410_19_ch19_p667-694