These fundamental equations pertain to a complex valued function of a complex variable. Recall the
complex numbers should be considered as points in the plane. Thus a complex number is of the form x + iy
where i^{2} = −1. The complex conjugate is defined by
------
x+ iy ≡ x − iy
and for z a complex number,
∘ -------
|z| ≡ (zz)1∕2 = x2 +y2.
Thus when x + iy is considered an ordered pair
(x,y)
∈ ℝ^{2} the magnitude of a complex number is nothing
more than the usual norm of the ordered pair. Also for z = x + iy,w = u + iv,
∘ ----------------
|z − w| = (x − u )2 +(y − v)2
so in terms of all topological considerations, ℝ^{2} is the same as ℂ. Thus to say z → f
(z)
is continuous, is
the same as saying
(x,y) → u (x,y) , (x,y) → v(x,y)
are continuous where f
(z)
≡ u
(x,y)
+ iv
(x,y)
with u and v being called the real and imaginary parts of
f. The only new thing is that writing an ordered pair
(x,y)
as x + iy with the convention
i^{2} = −1 makes ℂ into a field. Now here is the definition of what it means for a function to be
analytic.
Definition 15.2.1Let U be an open subset of ℂ (ℝ^{2}) and let f : U → ℂ be a function. Then f is said tobe analytic on U if for every z ∈ U,
f-(z-+-Δz)−-f-(z) ′
Δlzim→0 Δz ≡ f (z)
exists and is a continuous function of z ∈ U. For a function having values in ℂ denote by u
(x,y)
the realpart of f and v
(x,y)
the imaginary part. Both u and v have real values and
f (x + iy) ≡ f (z) ≡ u (x,y)+ iv(x,y)
Definition 15.2.2More generally, suppose f has values in X, a complex normed complete vector space.Then define f to be analyticon an open subset U of ℂ if
f (z + h)− f (z) ′
lhim→0-------h------ ≡ f (z)
exists and f^{′}is a continuous function of z ∈ U. This time, the limit takes place in terms of the norm on X.Meaning the usual thing: For every ε > 0 there exists δ > 0 such that
∥∥ ′ f(z+h)−-f(z)∥∥
∥f (z)− h ∥
< ε whenever
0 <
|h|
< δ. Here there are no real and imaginary parts to consider.
First are some simple results in the case that f has values in ℂ.
Proposition 15.2.3Let U be an open subset of ℂ . Then f : U → ℂ is analytic if and only iffor
f (x+ iy) ≡ u(x,y)+ iv(x,y)
u
(x,y)
,v
(x,y)
being the real and imaginary parts of f, it follows
ux(x,y) = vy(x,y),uy(x,y) = − vx(x,y)
and all these partial derivatives, u_{x},u_{y},v_{x},v_{y}are continuous on U. (The above equations are called theCauchy Riemann equations.)
Proof: First suppose f is analytic. First let Δz = ih and take the limit of the difference quotient as
h → 0 in the definition. Thus from the definition,
′ f (z +-ih)−-f-(z)
f (z) ≡ lhim→0 ih
u(x,y+ h)+ iv(x,y+ h) − (u (x,y) +iv(x,y))
= lhim→0 -------------------ih--------------------
1
= lhim→0 i (uy(x,y)+ ivy(x,y)) = − iuy(x,y)+ vy(x,y)
Next let Δz = h and take the limit of the difference quotient as h → 0.
f (z + h) − f (z)
f ′(z) ≡ lhim→0 --------------
h
= lim u(x+-h,y)+-iv(x-+-h,y)-−-(u-(x,y)-+iv(x,y))
h→0 h
= ux(x,y)+ ivx(x,y).
Therefore, equating real and imaginary parts,
ux = vy,vx = − uy (15.6)
(15.6)
and this yields the Cauchy Riemann equations. Since z → f^{′}
(z)
is continuous, it follows the real and
imaginary parts of this function must also be continuous. Thus from the above formulas for f^{′}
(z)
, it
follows from the continuity of z → f^{′}
(z)
all the partial derivatives of the real and imaginary parts are
continuous.
Next suppose the Cauchy Riemann equations hold and these partial derivatives are all continuous. For
Δz = h + ik,
f (z + Δz )− f (z) = u(x + h,y+ k)+ iv(x+ h,y+ k) − (u (x,y)+ iv(x,y))
= ux (x,y)h + uy(x,y)k+ i(vx(x,y)h+ vy(x,y)k)+ o ((h,k))
= ux (x,y)h + uy(x,y)k+ i(vx(x,y)h+ vy(x,y)k)+ o (Δz )
This follows since C^{1} implies differentiable along with the definition of the norm (absolute value) in ℂ. By
the Cauchy Riemann equations this equals
For functions of a real variable, it is perfectly possible for the derivative to exist and not be continuous.
For example, consider
{ ( )
x2sin 1x if x ⁄= 0
f (x) ≡ 0 if x = 0
You can verify that f^{′}
(x)
exists for all x but at 0 this derivative is not continuous. This will NEVER
happen with functions of a complex variable. This will be shown later when it is more convenient. For now
make continuity of f^{′} part of the requirement for f to be analytic.