Mathematical Methods for Physicists: A concise introduction - Site Map

DIFFERENTIAL CALCULUS
y. Suppose we ®rst take the x route, so y is ®xed as we change x, that is, y ˆ 0
and x ! 0, and we have
f 0 u…x ‡ x; y†u…x; y† v…x ‡ x; y†v…x; y†
…z† ˆ lim
‡ i ˆ @u
x!0 x
x
@x ‡ i @v
@x :
We next take the y route, and we have
f 0 u…x; y ‡ y†u…x; y† v…x; y ‡ y†v…x; y†
…z† ˆ lim
‡ i
y!0 iy
iy
ˆi @u
@y ‡ @v
@y :
Now f …z† cannot possibly be analytic unless the two derivatives are identical.
Thus a necessary condition for f …z† to be analytic is
from which we obtain
@u
@x ‡ i @v @u ˆi
@x @y ‡ @v
@y ;
@u
@x ˆ @v
@y
and
@u
@y ˆ@v @x :
…6:11†
These are the Cauchy±Riemann conditions, named after the French mathematician
A. L. Cauchy (1789±1857) who discovered them, and the German mathematician
Riemann who made them fundamental in his development of the theory of
analytic functions. Thus if the function f …z† ˆu…x; y†‡iv…x; y† is analytic in a
region R, then u…x; y† and v…x; y† satisfy the Cauchy±Riemann conditions at all
points of R.
Example 6.9
If f …z† ˆz 2 ˆ x 2 y 2 ‡ 2ixy, then f 0 …z† exists for all z: f 0 …z† ˆ2z, and
@u @v ˆ 2x ˆ
@x @y ; and @u ˆ2y ˆ@v
@y @x :
Thus, the Cauchy±Riemann equations (6.11) hold in this example at all points z.
We can also ®nd examples in which u…x; y† and v…x; y† satisfy the Cauchy±
Riemann conditions (6.11) at z ˆ z 0 , but f 0 …z 0 † doesn't exist. One such example
is the following:
(
f …z† ˆu…x; y†‡iv…x; y† ˆ z5 =jzj 4 if z 6ˆ 0
:
0 if z ˆ 0
The reader can show that u…x; y† and v…x; y† satisfy the Cauchy±Riemann conditions
(6.11) at z ˆ 0, but that f 0 …0† does not exist. Thus f …z† is not analytic at
z ˆ 0. The proof is straightforward, but very tedious.
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