Graphics Reference
In-Depth Information
1
2
f
()
z
()
n
()
=
Ú
f
z
d
z
.
p
i
n
+
1
g
(
)
z
-
z
Proof.
One way to prove this theorem is to use Theorem E.3.2 and show that dif-
ferentiation commutes with integration.
E.3.4. Corollary.
An analytic function is infinitely differentiable.
E.4
More on Complex Series
The first theorem is the complex version of the Taylor series expansion for a function.
E.4.1. Theorem.
If f(
z
) is an analytic function on an open disk of radius r about a
point
z
0
, then for every point
z
in the disk
()
¢¢
()
-
n
()
-
f
z
f
z
2
n
0
0
()
=
()
+¢
()
-
(
)
+
(
)
(
)
(E.4)
f
zz zzz
f
f
zz
+ ◊◊◊+
zz
+ ◊◊◊
.
0
0
0
0
0
2!
n
!
In particular, the series on the right side of equation (E.4) converges.
Proof.
See [Ahlf66].
Let f(
z
) be a function that is analytic in the neighborhood of a point
a
except pos-
sibly at the point
a
itself. More precisely, assume that there exists a d>0 and that f(
z
)
is analytic for all
z
satisfying 0 < |
z
-
a
| <d.
Definition.
The point
a
is called an
isolated singularity
of f(
z
). It is called a
remov-
able singularity
of f if some definition of f(
a
) will make f analytic at
a
. The point
a
is
called a
pole
if
()
=•
lim
f
z
,
za
Æ
and in that case we set f(
a
) =•.
Assume that f(
z
) has a pole at a point
a
. If
1
g
()
=
z
,
()
f
z
then g(
z
) has a removable singularity at
a
and we can remove the singularity by defin-
ing g(
a
) to be
0
, making g into a function which is analytic in a neighborhood of
a
.
Because
a
is a zero of g, it follows that
m
zza z
()
=-
(
)
( )
,
g
h
for some m > 0 and h(
z
) an analytic function at
a
with h(
a
) π
0
.
