when
Iz+il
< i·
11.3 ANALYTIC CONTINUATION 303
A consequence
of
this theorem is the following corollary.
11.3.2.
Corollary.
The behavior01afunction that is analytic in a region S c C is
completely determined by its behavior in a (small) neighborhood 01an arbitrary
point in that region.
analytic
continuation
This process
of
detetmining the behavior
of
an analytic function outside the
regionin whichit was originallydefinedis called
analytic
continuation.
Although
there are infinitely many ways
of
analytically continuing beyond regions
of
defi-
nition, the values
of
all functions obtained as a result
of
diverse continuations are
the sarne at any given point. This follows from Theorem 11.3.1.
Let
It,Iz :C
-+
C be analyticin regions
8t
and 82, respectively. Supposethat
II
and fzhave differentfunctional forms in theirrespective regions
of
analyticity.
If
there is an overlap between 81 and 82 and
if
II
= fzwithin that overlap, then
the (unique) analyticcontinuation
of
II
into 82
must
be
h.
and vice versa.
In
fact,
we may regard
!J
and h as a single function I :
iC
-+
C such that
I(z)
=
{!J(Z)
when ZE 81,
I:
(z) when Z E 82.
Clearly,
I is analytic for the combined region 8 = 81 U 82. We then say that II
and fzare analytic continuations
of
one
another.
11.3.3.
Example.
Let us consider the function
11
(z)
= L~o
t",
which is analytic
for lel
< 1. We have seen that it converges to
1(1
- z) for
[z]
< 1. Thus, we have
II
(z) =1((1 - z) when
[z]
< I,
and!J
is not definedfor
Izi
> 1.
Now let us consider a second function,
h(z)
= L;:'=o
(~)n+1
(z + i)n, which
converges for [z +
~I
<
~.
To see what it converges to, we note that
h(z)
=
~
L~o
[~(z
+
i)]",
Thus,
~
I
h(z)
=
=-
I-~(z+~)
I-z
Weobservethat although
II
(z) and h (z) havedifferentseriesrepresentations in the two
overlapping regions(seeFigure
11.7),theyrepresentthe samefunction,
I(z)
=
1(1-
z).
Wecan
therefore
write
I
{
fJ(z) when
Izl
< I,
(z) -
-
h(z)
when lz+
~I
< i,
and
I:
and h areanalyticcontinuations of ooe
another.
In fact,
I(z)
=
1(1
- z) is the
analyticcontinuation of both
fJ and Iz for all of iCexcept z= 1. Figure 11.7sbowsSi,
theregion of definitionof Ii.for i = 1, 2.
III
11.3.4.
Example.
ThefunctionfJ(z) =1
0
00
e-
zt
dt existsonly
if
Re(z) > 0, in which
case
II
(z) =I(z.ItsregionofdefinitionSI is showninFigure H.S andissimplytheright
half-plane.