Z
.
Now consider integrating this function around the contour
C
=
∪
4
i
=1
C
i
, defined in Fig.
2.
As the contour
C
is traversed both
θ
and
φ
decrease
as the contour is traversed in a CCW fashion, and the phase of
f
varies
Mathematical Methods of Physics
v3.1
14 November 2015, 12:51 Noon
Autumn 2015
The University of Chicago, Department of Physics
Page 8 of 9
accordingly; we give some sample phases at AF in Table I.
From the foregoing considerations, in the limit where
2
,
4
→
0
, we can see that on the contour segment
C
1
, we
have arg
[
f
(
z
)] = 0
, so
Z
C
1
f
(
z
)
dz
=
Z
C
1

z


2
/
3

z

1


1
/
3
dz
→
Z
1
0
dx
(
x
2

x
3
)
1
/
3
=
I,
(53)
where I used that
z
=
x
here and

z

1

= (1

x
)
on this piece of contour. We chose the branch so that the overall
argument here is zero. Meanwhile, on the contour segment
C
3
, we have arg
[
f
(
z
)] = 2
π/
3
, so
Z
C
3
f
(
z
)
dz
=
Z
C
3

z


2
/
3

z

1


1
/
3
e
2
πi/
3
dz
→
e
2
πi/
3
Z
0
1
dx
(
x
2

x
3
)
1
/
3
=

e
2
πi/
3
Z
1
0
dx
(
x
2

x
3
)
1
/
3
=

e
2
πi/
3
I.
(54)
Meanwhile, on the circular contour segment
C
2
, we have
r
= 1+
O
(
2
)
and
ρ
=
2
, with
θ
= 0
and
φ
decreasing
from
π
to

π
, so the integral is (dropping subdominant powers of
2
)
Z
C
2
f
(
z
)
dz
∼
Z

π
π
(
i
2
e
iφ
dφ
)
e

iφ/
3+
iπ/
3

1
/
3
2
∼
2
/
3
2
→
0
,
(55)
the latter in the limit where
2
→
0
. Similarly, on the circular contour segment
C
4
, we have
r
=
4
and
ρ
= 1+
O
(
4
)
,
with
φ
=

π
and
θ
decreasing from
0
to

2
π
, so the integral is (dropping subdominant powers of
4
)
Z
C
4
f
(
z
)
dz
∼
Z

2
π
0
(
i
4
e
iθ
dθ
)

2
/
3
4
e

2
iθ/
3+2
πi/
3
∼
1
/
3
4
→
0
,
(56)
the latter in the limit where
4
→
0
. So in the limit
2
,
4
→
0
, we have
I
C
f
(
z
)
dz
= (1

e
2
πi/
3
)
I.
(57)
Now suppose I construct a single contour
˜
C
=
C
∪
P
∪
C
0
∪
(

P
)
where I start some point on
C
, move all the
way around
C
and back to just before this point, then pass to
C
0
by some path
P
(which does not cross or run near
the branch cut), then traverse all of
C
0
and finally then return to
C
in the opposite direction along a path which I
will call

P
which is infinitesimally displaced from
P
. Then Cauchy’s Theorem together with the observation that
f
(
z
)
is analytic in the interior of
˜
C
tells us that
H
˜
C
= 0 =
R
C
+
R
P
+
R
C
0
+
R

P
. On
P
and

P
, the argument of
f
must differ by the amount of phase
f
(
z
)
picks up on
C
or
C
0
, but since
P
and

P
are not running near branch
cuts, the phase shift
must
be a multiple of
2
π
(this is almost by definition and construction: if it did not, there
TABLE I
Phase of the integrand at selected points AF as defined in Fig. 2. Note that a full traversal of the
contour (change in
Δ
θ
= Δ
φ
=

2
π
) yields
Δ
arg
f
[
z
] = +2
π
as well, which is important:
f
is singlevalued.
Point
θ
φ
arg[
f
(
z
)]
A
0
π
0
B
0
π
0
C
0
0
π/
3
D
0

π
2
π/
3
E
0

π
2
π/
3
F

π

π
4
π/
3
A (after traversing
C
once)

2
π

π
2
π
Mathematical Methods of Physics
v3.1
14 November 2015, 12:51 Noon
Autumn 2015
The University of Chicago, Department of Physics
Page 9 of 9
would be a branch cut between
P
and

P
); this is confirmed by Table I. Therefore, the integrals over
P
and

P
will cancel out in the limit where the infinitesimal displacement between them vanishes, and we have
I
C
f
(
z
)
dz
=

I
C
0
f
(
z
)
dz
⇒
I
C
0
f
(
z
)
dz
= (
e
2
πi/
3

1)
I
= 2
ie
iπ/
3
sin(
π/
3)
I
=
i
√
3
e
iπ/
3
I.
(58)
Now consider that in the limit where the radius
R
of the contour
C
0
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 Physics, University of Chicago, Methods of contour integration, Complex Plane, Pole