dg
dz
0
¼ lim
z!0
g 0 þ z
ð
ÞÀg 0
ð Þ
z
¼ lim
z!0
z À 0
z
¼ 1:
As a result, the derivative takes the same value 1, regardless of what straight lines
the differentiation is taken along.
Though simple, the above example gives us a heuristic method. As before, let f (z)
be a function described as (6.44) such that
f z
ð Þ ¼ f x, y
ð Þ ¼ u x, y
ð Þþiv x, y
ð Þ,
ð6:48Þ
where both u(x, y) and v(x, y) possess the first-order partial derivatives with respect to
x and y. Then we have a derivative df (z)/dz such that
df z
ð Þ
dz
¼ lim
Δz!0
f z þ Δz
ð
ÞÀf z
ð Þ
Δz
¼
lim
Δx!0, Δy!0
u x þ Δx, y þ Δy
ð
Þ À u x, y
ð Þþi v x þ Δx, y þ Δy
ð
Þ À v x, y
ð Þ
½
Š
Δx þ iΔy
:
ð6:49Þ
We wish to seek the condition that
df z
ð Þ
dz gives the same result regardless of the
order of taking the limit of Δx ! 0 and Δy ! 0. That is, taking the limit of Δy ! 0
first, we have
df z
ð Þ
dz
¼ lim
Δx!0
u x þ Δx, y
ð
ÞÀu x, y
ð Þþi v x þ Δx, y
ð
ÞÀv x, y
ð Þ
½
Š
Δx
¼
∂u x, y
ð Þ
∂x
þ i
∂v x, y
ð Þ
∂x
:
ð6:50Þ
Next, taking the limit of Δx ! 0, we get
df z
ð Þ
dz
¼ lim
Δy!0
u x, y þ Δy
ð
ÞÀu x, y
ð Þþi v x, y þ Δy
ð
ÞÀv x, y
ð Þ
½
Š
iΔy
¼ Ài
∂u x, y
ð Þ
∂y
þ
∂v x, y
ð Þ
∂y
:
ð6:51Þ
Consequently, by equating the real and imaginary parts of (6.50) and (6.51) we
must have
∂u x, y
ð Þ
∂x
¼
∂v x, y
ð Þ
∂y
,
ð6:52Þ
202
6 Theory of Analytic Functions