chapter 9 complex functions masatsugu sei suzuki...
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Chapter 9 Complex functions
Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton
(Date: October 23, 2010)
Baron Augustin-Louis Cauchy (21 August 1789 – 23 May 1857; French pronunciation was a French mathematician who was an early pioneer of analysis. He started the project of formulating and proving the theorems of infinitesimal calculus in a rigorous manner. He also gave several important theorems in complex analysis and initiated the study of permutation groups in abstract algebra. A profound mathematician, Cauchy exercised a great influence over his contemporaries and successors. His writings cover the entire range of mathematics and mathematical physics.
http://en.wikipedia.org/wiki/Augustin-Louis_Cauchy 9.1 Function of a complex
iyxz . z is a complex number and both x and y are real. Suppose we write f(z) - that is a function of z - by this we mean that for each value of z - f(z) can take a complex value. f(z) is a function of two real variables x and y.
),(),()( yxivyxuzf where ),( yxu and ),( yxv are real functions.
)3()3(
)()(3)(3
)()(
3223
3223
33
yyxixyx
iyiyxiyxx
iyxzzf
32
23
3
3
yyxv
xyxu
(a)
In our study of complex functions, we will consider those sorts of functions which turn up in physical problems. -i.e., our functions will have a “smoothness requirement”. -i.e., they are differentiable. (1) How do we define the derivative of a complex function? It seems like it should be
z
zfzzf
z
fz
)()(lim
0.
But z has two paths - i.e., yixz -thus 0z can assume in an infinite number of ways.
irez and 0z is obtained when 0r but can have any value.
((Example)) Say iyxzf 2)( and let us compute directly
z
zfzzfz
)()(lim
0 for z = 0,
and two different values: (a) 0 , rz
1)2(}2){(
lim)()(
lim00
r
iyxiyrx
z
zfzzfrz
(b) 2/ , rirez i 2/ and 0x .
2)2()}(2{
lim)()(
lim00
ri
iyxryix
z
zfzzfrz
Thus for iyxzf 2)( , we do not get a unique result. We say that this )(zf is not differentiable.
For a function of z to be differentiable at 0zz , we need z
zfzzfz
)()(lim 00
0 to
exist (i.e. finite) and the result should be independent of any z . 9.2 Cauchy-Riemann conditions
Let’s see what restrictions are imposed on )(zf of it is to be differentiable.
),(),()( yxivyxuzf
yix
yxvyyxxviyxuyyxxu
z
zfzzf
dz
df
yx
z
)],(),([)],(),([lim
)()(lim
00
0
We get unique result no matter what is in irez . For = 0, 0x , 0y
x
vi
x
u
dz
df
(1)
For = /2, 0x , 0y
y
v
y
ui
dz
df
(2)
And both results must be the same
y
v
x
u
, y
u
x
v
(Cauchy-Riemann condition)
or
yx
yx
uv
vu
The Cauchy-Riemann conditions are necessary for the existence of a derivative of )(zf .
If dzdf / exists, the Cauchy-Riemann conditions must hold. Conversely, if the Cauchy-Riemann conditions are satisfied and the partial derivatives of u and v are continuous, then dzdf / exists.
yy
vi
y
ux
x
vi
x
uf
x
yi
x
y
y
vi
y
u
x
vi
x
u
yix
yy
vi
y
ux
x
vi
x
u
z
f
1
Note that
x
vi
x
ui
x
v
x
ui
y
vi
y
u,
using the Cauchy-Riemann conditions. Then we have
x
vi
x
u
z
f
(1)
On the other hand
x
vi
x
u
x
ui
x
vi
y
vi
y
u
iyi
yy
vi
y
u
z
f
xy
)(1
00
(2)
x
vi
x
u
z
f
yx
00
, (3)
Eqs. (1) ,(2), and (3) show that z
fz
0lim
is independent of the direction of approach in the
complex plane as long as the partial derivatives are continuous. ((Note)) Cauchy-Riemann conditions
y
v
x
u
, y
u
x
v
,
2
22
2
2
2
y
u
yx
v
xy
v
x
u
02
2
2
2
uyx
Laplace eq. for 2D
Similarly
2
22
2
2
2
y
v
yx
u
xy
u
x
v
02
2
2
2
vyx
Laplace eq. for 2D
Therefore, both u and v are harmonic function that satisfy the Laplace’s equation. Furthermore, comparing the 2D gradients
),(),(
),(),(
x
u
y
u
y
v
x
vv
x
v
y
v
y
u
x
uu
,
Then we have
0 vu .
The lines of constant u (level curves) are orthogonal to the lines of constant v, anywhere that 0)(' zf . If u represents a potential function, then v represents the corresponding stream function (lines of force), or vice versa. ((Example))
xyiyxivuzzf 2)( 222 or
22 yxu , v = 2 xy. We make a ContourPlot of u = const, a StreamPlot of u , a ContourPlot of v = const and a StreamPlot of v by using the Mathematica.
x
y
O
u = x2 - y2
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
Fig. A ContourPlot of u = x2 - y2 = constant and a StreamPlot of u .
x
y
v = 2xy
-3 -2 -1 0 1 2 3
-3
-2
-1
0
1
2
3
Fig. A ContourPlot of v = 2xy = constant and a StreamPlot of v . 9.3 Example James J. KellyGraduate Mathematical Physics
Level curves of f(z)=z2=u+Âv are shown.
f x y3 Expand
x3 3 x2 y 3 x y2 y3
u SimplifyRef, x Reals, y Realsx3 3 x y2
v SimplifyImf, x Reals, y Reals3 x2 y y3
ContourPlotEvaluateTableu k, k, 5, 5, 0.1,
x, 2, 2, y, 2, 2
-2 -1 0 1 2
-2
-1
0
1
2
ContourPlotEvaluateTablev k, k, 5, 5, 0.1,
x, 2, 2, y, 2, 2
-2 -1 0 1 2
-2
-1
0
1
2
______________________________________________________________________ 9.4 Example
)()()('
iy
vi
iy
u
y
v
y
ui
x
vi
x
uzf
)3()3()()( 322333 yyxixyxiyxzzf
)3( 23 xyxu , )3( 32 yyxv
xyiyxx
vi
x
uzf 633)(' 22
or
xyiyxyxxyiy
v
y
uizf 63333)6()(' 2222
((Laplace equation))
xx
u6
2
2
, xy
u6
2
2
02
2
2
2
uyx
Can we write )(' zf as a derivative? The answer is yes.
ixyyx
ixyyxiyxzzf
6)(3
)2(3)(32)('
22
2222
9.5 Mathematica Cauchy-Riemann condition
Cauchy-Rieman conditions
SuperStar : expr_ : expr . Complexa_, b_ Complexa, bRCf_ : Moduleu, v, w, uy, vx, ux, vy, w ComplexExpandfx y;
u w w 2 Simplify; v w w 2 Simplify;
uy Du, y Simplify; vx Dv, x Simplify;
ux Du, x Simplify; vy Dv, y Simplify;
uxx Dux, x Simplify; uyy Duy, y Simplify;
vxx Dvx, x Simplify; vyy Dvy, y Simplify;
List"u", u, "v", v, "uy", uy, "vx", vx, "ux", ux,
"vy", vy, "uyvx", uy vx, "uxvy", ux vy,
"uxxuyy", uxx uyy, "vxxvyy", vxx vyy
f1 Functionz, z4Functionz, z4
RCf1 Simplify TableForm
u x4 6 x2 y2 y4
v 4 x y x2 y2uy 4 y 3 x2 y2vx 12 x2 y 4 y3
ux 4 x3 3 x y2vy 4 x3 3 x y2uyvx 0uxvy 0uxxuyy 0vxxvyy 0
f2 Functionz, z3Functionz, z3
RCf2 Simplify TableForm
u x3 3 x y2
v 3 x2 y y3
uy 6 x yvx 6 x y
ux 3 x2 y2vy 3 x2 y2uyvx 0uxvy 0uxxuyy 0vxxvyy 0
f3 Functionz, z5Functionz, z5
RCf3 Simplify TableForm
u x5 10 x3 y2 5 x y4
v 5 x4 y 10 x2 y3 y5
uy 20 x y x2 y2vx 20 x y x2 y2ux 5 x4 6 x2 y2 y4vy 5 x4 6 x2 y2 y4uyvx 0uxvy 0uxxuyy 0vxxvyy 0
f4 Functionz, 1zFunctionz,
1
z
RCf4 Simplify TableForm
u xx2y2
v yx2y2
uy 2 x y
x2y22
vx 2 x y
x2y22
ux x2y2
x2y22
vy x2y2
x2y22
uyvx 0uxvy 0uxxuyy 0vxxvyy 0
f5 Functionz, z zFunctionz, z z
RCf5 Simplify TableForm
u x2 y2
v 0uy 2 yvx 0ux 2 xvy 0uyvx 2 yuxvy 2 xuxxuyy 4vxxvyy 0
f6 Functionz, zFunctionz, z
RCf6 Simplify TableForm
u xv yuy 0vx 0ux 1vy 1uyvx 0uxvy 2uxxuyy 0vxxvyy 0
((Definition)) Analytic
If f(z) is differentiable at z = z0 and in some small region around z0, we say that f(z) is analytic at z = z0. If f(z) is analytic everywhere in the (finite) complex plane, we call it an entire function. ((Example)) Arfken 6-2-8
Using ),(),()( rii erRref , in which ),( rR and ),( r are differentiable real
functions of r and , show that the Cauchy-Riemann condition in polar coordinates become
(a)
),(),(),( r
r
rR
r
rR, (b)
r
rrR
rR
r
),(),(
),(1
Hint: set up the derivative first with dz radial (r → r+dr) and then with dz tangential (r → + d). ((Mathematica))
Arfken 6 - 2 - 8(a) Derivative with dr while q fixed
Clear"Global`"
eq1 Rr r, rr, Rr, r,
r r r Simplify
r, Rr, rr, Rr r, r
We apply the L' Hospital theorem to calculate.
eq11 D r, Rr, rr, Rr r, , r . r 0
Dr, r . r 0
r, R1,0r, r, Rr, 1,0r,
(b) Derivative with dq while r fixed
eq2 Rr, r, Rr, r,
r r Simplify
r, Rr, r, Rr, 1 r
We apply the L' Hospital theorem to calculate.
eq22 D r, Rr, r, Rr, , . 0
D1 r, . 0 Expand
r, R0,1r,
r r, Rr, 0,1r,
r
eq3 r eq11 eq22 r, FullSimplify
R0,1r, r R1,0r, Rr, 0,1r, r 1,0r,
The real part and the imaginary part of eq3=0,
Rr, 0,1r, r R1,0r, =0
R0,1r, r Rr, 1,0r, =0
________________________________________________________________________ 9.6 Cauchy’s integral theorem
Integral of a complex variable over a contour in the complex plane. We divide the contour 00 zz into n intervals by picking n-1 intermediate points (i) on the contour.
Consider the sum
n
jjjjn zzfS
11 ))(( ,
where j is a point on the curve between zj and zj-1. Now let n with 01 jj zz
for all j.
0
0
)())((lim1
1
z
z
n
jjjj
ndzzfzzf ,
(contour integral) (exists and is independent of the details of choosing points zj and j.)
22
11
22
11
22
11
2
1
)()())(()(yx
yx
yx
yx
yx
yx
z
z
udyvdxivdyudxidydxivudzzf .
________________________________________________________________________ 9.7 Cauchy’s integral theorem
If a function )(zf is analytic and its partial derivatives are continuous throughout some simple connected region R, for every closed path C in R, the line integral of )(zf around C is zero:
0)()( dzzfdzzfC
.
C
R
________________________________________________________________________ (A) Stroke’s theorem proof
idydxdz ,
),(),()( yxivuxuzf ,
)()()( udyvdxivdyudxdzzf .
We consider the Stroke’s theorem
rAaA dd
with
za ˆdxdyd
S
xy
S
z
C
yx dxdyy
A
x
AdxdydyAdxA )()( A
Let uAx , vAy .
First term
SC
dxdyy
u
x
vvdyudx )(
Let vAx , uAy .
Second term
SC
dxdyy
v
x
uudyvdx )(
Thus
0
)()()(
SS
dxdyy
v
x
uidxdy
x
v
y
u
udyvdxivdyudxdzzf
If )(zf is analytic, the Cauchy-Riemann condition is satisfied. (B) Cauchy-Goursat Proof
Suppose that )(zf is analytic and its partial derivatives are continuous throughout the region R.
0)( C
dzzf
C
R
We subdivide the region inside the contour C into a network of small squares. Then
j CC j
dzzfdzzf )()( .
All integrals along interior lines cancel out. Now we consider
jC
dzzf )( ,
with the contour path Cj,
C jz j
zj: an interior point of the j-th subregion. We construct the function
jzzj
jjj dz
df
zz
zfzfzz
)()(),( .
We may make for 0 ,
),( jj zz
where e is an arbitrary chosen small positive quantity.
)()()(),()( jzz
jjjj zzdz
dfzfzfzzzz
j
j jjj C
jzz
j
C
jjj
C
dzzzdz
dfzfdzzzzzdzzf )}()({),()()(
Since 0jC
dz and 0jC
zdz ,
0),()()( Adzzzzzdzzfjj C
jjj
C
.
If a function )(zf is analytic on and within a closed path C,
0)( C
dzzf .
((Note)) (1)
001
11
1 )())((0
0
zzzzzzfdzn
jjj
n
jjjj
zz
z
n
.
(2)
)(2
1
)(2
1)(
2))((
20
20
1
21
2
11
1
11
0
0
zz
zzzzzz
zzfzdzn
jjj
n
jjj
jjn
jjjj
zz
z
n
.
Thus we have
0C
dz and 0C
zdz
____________________________________________________________ ((Multiply connected region))
We consider the multiply connected region in which )(zf is not defined (not analytic) in the interior R.
RA
BC
FE
D
We can construct a contour for which the theorem holds. The new contour never crosses the interior R. From the Cauchy’s theorem, we have
0)()()()()( FADEFCDABC
dzzfdzzfdzzfdzzfdzzf .
Since
CDFA
dzzfdzzf )()( ,
then
FEDDEFABC
dzzfdzzfdzzf )()()( .
Therefore we obtain
FEDABC
dzzfdzzf )()( .
where the contours C1 (= A-B-C) and C2 (= F-E-D) are the counterclockwise. ________________________________________________________________________ 9.8 Cauchy’s integral formula
)(2)(
00
zifdzzz
zf Cauchy’s integral
z = z0
AB
C
FE
D
We consider a function )(zf that is analytic on a closed contour A-B-C and within the
interior by the path A-B-C. )/()( 0zzzf is not analytic at 0zz . We apply the
Cauchy’s integral theorem to the contour ABCDEF. In this region )/()( 0zzzf is
analytic.
0)()()()(
0000
FADEFCDABC
dzzz
zfdz
zz
zfdz
zz
zfdz
zz
zf,
or
0)()(
00
FEDABC
dzzz
zfdz
zz
zf.
Let irezz 0 , r is small and will eventually be made to approach zero.
We have
FED
i
FED
ii
i
FED
idrezfidrere
rezfdz
zz
zf
)()()(
00
0
.
Taking the limit as 0r , we obtain
)(2)()(
0
2
0
00
zifdzifdzzz
zf
ABC
(Residue)
f z0 is exterior to C1,
0)(
1 0
C
dzzz
zf.
We have therefore
exterior: 0
interior: )()(
2
1
0
00
01z
zzfdz
zz
zf
i C.
9.9 Derivatives Suppose that )(zf is analytic. By the definition of derivative, we have
dzzz
zf
i
dzzzzzz
zfz
zi
zz
dzzf
zzz
dzzf
zi
z
zfzzfzf
z
z
z
20
000
0
00
00000
0
000
00
)(
)(
2
1
))((
)(
2
1lim
)(
)(
)(
2
1lim
)()(lim)(
0
0
0
Similarly,
dz
zz
zf
i
nzf
nn
10
0)(
)(
)(
2
!)(
220
01
01
000
100
10
0 )(
))(1(
)()(
)()(lim
0
k
k
kk
kk
z zz
zzk
zzzzzz
zzzzz
Thus
dzzz
zf
i
k
dzzfzz
zzk
i
k
z
zfzzf
k
k
kkk
z
20
220
0
0
0)(
00)(
0
)(
)(
2
)!1(
)()(
))(1(
2
!)()(lim
0
The requirement that )(zf be analytic not only gurantees a first derivative but also derivatives of all orderes as well. Goursat’s theorem
The existence of )(zf shows that )(zf is continuous.”
)(zf is analytic
)(zf is analytic
)(zf is analytic
. . .
)()( zf n is analytic
All the derivatives of )(zf are analytic within R. Morera’s theorem
This is the converse of Cauchy’s integral theorem.
“If a function )(zf is continuous in a simple connected region R and 0)( C
dzzf for
every closed contour C within R, then )(zf is analytic throughout R. ________________________________________________________________________ 9.10 Taylor expansion and Laurent expansion (a) Taylor expansion
)(zf is analytic on and within C. 1zz is the nearest point for which )(zf is not
analytic. C is a circle centered at z0 with radius )( 010 zzzz .
C
z0
z1
z
z'
Cauchy integral formula
nn
n
n Cn
n
n C
n
n
C
C
C
zzzfn
zz
zdzf
izz
zz
zdzz
zzzf
i
zz
zzzz
zdzf
i
zzzz
zdzf
i
zz
zdzf
izf
))((!
1
)(
)(
2
1)(
)(
)(
)()(
2
1
)(
)(1)(
)(
2
1
)()(
)(
2
1
)(
2
1)(
00)(
0
01
00
0 0
0
0
0
00
00
(1)
where z is a point on the contour C, and z is any point interior to C.
01
1
n
ntt
for 1t (even for complex number t).
C nn dz
zz
zf
i
nzf '
'
)'(
2
!)( 1
0
0)(
Taylor expansion: )(zf is analytic at 0zz .
(b) Laurentz expansion f(z) is not analytic in the regions denoted by green.
z0A
BC
FE
D
R
r
z
z'
z'
According to the Cauchy theorem, we have
])()(
[2
1
])()()()(
[2
1)(
2
1)(
FEDABC
FADEFCDABCABCDEFA
zz
zdzf
zz
zdzf
i
zz
zdzf
zz
zdzf
zz
zdzf
zz
zdzf
izz
zdzf
izf
On the path (ABC), we have
|||'| 00 zzzz
On the path (FED), we have
|||'| 00 zzzz
Then
00
01
00
0 0
0
0
0
00
00
)(
)(
)(
2
1)(
)(
)(
)()(
2
1
)(
)(1)(
)(
2
1
)()(
)(
2
1)(
2
1
n
nn
n Cn
n
n C
n
n
ABC
ABCABC
zza
zz
zdzf
izz
zz
zdzz
zzzf
i
zz
zzzz
zdzf
i
zzzz
zdzf
izz
zdzf
i
and
1
100
01
0
0
0 0
0
0
0
00
00
)')(()(2
1
)(
)')((
2
1
)(
)(
)'()(
2
1
)(
)'(1)(
)(
2
1
)()(
)(
2
1)(
2
1
n FED
nn
n FEDn
n
n FED
n
n
FED
FEDFED
zdzzzfzzi
zz
zdzzzf
i
zz
zdzz
zzzf
i
zz
zzzz
zdzf
i
zzzz
zdzf
izz
zdzf
i
These two series are combined into one series (Laurent series)
1
00
0 )()()(n
nn
n
nn zzbzzazf ,
with
ABC nn dzzz
zf
ia '
'
)'(
2
11
0.
FED
nn zdzzzf
ib 1
0 )')((2
1
_____________________________________________________________________ (1) Analytic at z = z0
If f(z) is analytic in the vicinity of z0, we can perform a Taylor series expansion.
0
0 )()(n
nn zzazf .
(2) Isolated singular point at z = z0
We say we have an isolated singular point at z = z0.
.......)()(
02
0
2
0
1
00
n
n
n
nn
zz
b
zz
b
zz
bzzazf
(i) As long as one of the b’s are nonzero, z0 is a singular point. (ii) If bn does not vanish, but all bj with j>n vanish, then f(z) has a pole of order n. (iii) If b1 ≠0 and all other bj’s = 0 the pole is a single pole. (iv) Isolated essential singularity (all the bj’s) normally does not occur in physics. 9.11 Examples of Taylor and Lorentz expansions (Mathematica) 9.11.1 Example-1
Clear"Global`"SeriesExpz, z, 0, 10
1 z z2
2
z3
6
z4
24
z5
120
z6
720
z7
5040
z8
40 320
z9
362 880
z10
3 628 800 Oz11
SeriesCosz, z, 0, 10
1 z2
2
z4
24
z6
720
z8
40 320
z10
3 628 800 Oz11
SeriesSinz, z, 0, 10
z z3
6
z5
120
z7
5040
z9
362 880 Oz11
SeriesCotz, z, 0, 10 Normal
1
z
z
3
z3
45
2 z5
945
z7
4725
2 z9
93 555
SeriesTanz, z, 0, 10 Normal
z z3
3
2 z5
15
17 z7
315
62 z9
2835
SeriesCscz2 , z, 0, 10 Normal
1
3
1
z2
z2
15
2 z4
189
z6
675
2 z8
10 395
1382 z10
58 046 625
SeriesSecz2 , z, 0, 10 Normal
1 z2 2 z4
3
17 z6
45
62 z8
315
1382 z10
14 175
Series 1
Sinz, z, 0, 10 Normal
1
z
z
6
7 z3
360
31 z5
15 120
127 z7
604 800
73 z9
3 421 440
Series z
Sinz, z, 0, 10 Normal
1 z2
6
7 z4
360
31 z6
15 120
127 z8
604 800
73 z10
3 421 440
Seriesz3 Sinz, z, 0, 10 Normal
1
6
1
z2
z2
120
z4
5040
z6
362 880
z8
39 916 800
z10
6 227 020 800
Series 1
Cosz, z, 0, 10 Normal
1 z2
2
5 z4
24
61 z6
720
277 z8
8064
50 521 z10
3 628 800
SeriesCscz2 Log1 z, z, 0, 10 Normal
1
2
1
z
2 z
3
5 z2
12
17 z3
45
17 z4
60
229 z5
945
1531 z6
7560
502 z7
2835
5903 z8
37 800
21 872 z9
155 925
158 707 z10
1 247 400
Series z
Sinz Tanz, z, 0, 10 Normal
1
2
2
z2
11 z2
120
157 z4
15 120
641 z6
604 800
1417 z8
13 305 600
1 402 631 z10
130 767 436 800
SeriesSinhz, z, 0, 10
z z3
6
z5
120
z7
5040
z9
362 880 Oz11
SeriesCoshz, z, 0, 10
1 z2
2
z4
24
z6
720
z8
40 320
z10
3 628 800 Oz11
SeriesCsczz
, z, 0, 10
1
z2
1
6
7 z2
360
31 z4
15 120
127 z6
604 800
73 z8
3 421 440
1 414 477 z10
653 837 184 000 Oz11
SeriesSeczz
, z, 0, 10
1
z
z
2
5 z3
24
61 z5
720
277 z7
8064
50 521 z9
3 628 800 Oz11
9.11.2 Example-2
Clear"Global`"
Series 1
1 z, z, 0, 10
1 z z2 z3 z4 z5 z6 z7 z8 z9 z10 Oz11
SeriesLog1 z, z, 0, 10
z z2
2
z3
3
z4
4
z5
5
z6
6
z7
7
z8
8
z9
9
z10
10 Oz11
Series 1
z z 23, z, 2, 10 Normal
1
16
1
2 2 z3
1
4 2 z2
1
8 2 z 1
322 z
1
642 z2
1
1282 z3
1
2562 z4
1
5122 z5
2 z6
10242 z7
20482 z8
40962 z9
81922 z10
16 384
SeriesExp zz2 12
, z, , 10 Normal
9
16
1
4 z2
2 z
23 z48
67 z2
192
37 z3
160
1663 z4
11 520
6983 z5
80 640
5431 z6
107 520
5237 z7
181 440
942 659 z8
58 060 800
213 359 z9
23 654 400
38 020 573 z10
7 664 025 600
Series 1
z4 a4, z,
1
2a, 5 Normal
PowerExpand
3
8 a4
14
4
2 a3 1 a2
z 5
16 5
16 1 a
2 z
2 a5
5 1 a2
z2
32 a6 1
64
64 1 a
2 z3
2 a7
7 1 a2
z4
128 a8 15
256 15
256 1 a
2 z5
2 a9
Series1 z , z, 0, 5 Normal Simplify
1
1212 12 z 6 6 z2 6 2 z3 5 z4 4 z5
Series z4
z2 c24, z, c, 5 Normal
3
256 c4
1
16 c z4
1
8 c c z3
1
32 c2 c z2
1
32 c3 c z c z2
256 c6
c z3
256 c7
11 c z4
4096 c8
3 c z5
2048 c9
Series z4
z2 c24, z, c, 5 Normal
3
256 c4
1
16 c z4
1
8 c c z3
1
32 c2 c z2
1
32 c3 c z c z2
256 c6c z3
256 c7
11 c z4
4096 c8
3 c z5
2048 c9
SeriesCot zz a2
, z, a, 1 Normal
2 Cota Cota a z2
2 Cota 3 Cota 2
a z
a z 3
3
4
33 Cota 2 3 Cota 4
9.11.3 Example-3
Clear"Global`"
Series z
Expz 1, z, 0, 6
1 z
2
z2
12
z4
720
z6
30 240 Oz7
Series 1
z 1, z, 0, 10
1
z
1
2
z
12
z3
720
z5
30 240
z7
1 209 600
z9
47 900 160 Oz11
Series1z, z, , 101
1
z
1
2 z2
1
6 z3
1
24 z4
1
120 z5
1
720 z6
1
5040 z7
1
40 320 z8
1
362 880 z9
1
3 628 800 z10 O1
z11
________________________________________________________________________ 9.12 Calculation of residue
....)(Re)(Re)([Re2
......)()()()(
321
321
zzszzszzsi
dzzfdzzfdzzfdzzfCCCC
......)()( 2
0
2
0
1010
zz
b
zz
bzzaazf
Using Cauchy’s theorem,
CC
dzzz
b
zz
bdzzzaadzzf ..][...])([)( 2
0
2
0
1010
or
dzzz
b
zz
bdzzf
C
..][)( 20
2
0
1
We now calculate
dzzzI n0
irezz 0
ieidrdz )(
1,1
2
0
)1(12
0
2)( nnninini irdeireidrreI
Then we have
)(Re22)()( 010
1 zzsiibdzzz
bdzzf
C
The problem is reduced to finding the value of b1. (i) f(z) has a simple pole.
)(Re)()(lim
...)()()()(
...)()(
010
12
01000
0
1010
0
zzsbzfzz
bzzazzazfzz
zz
bzzaazf
zz
(ii) A pole of order n
10
1
01
110
1
202
10101000
02
0
2
0
1010
)]()[(
)!1(
1lim][Re
)!1()()(
......])([)()()(
......)()(
0
n
nn
zz
n
nn
nnnnn
nn
dz
zfzzd
nzzsb
nbdz
zfzzd
bzzbzzbzzaazzzfzz
zz
b
zz
b
zz
bzzaazf
((Note)) Mathematica Residue[expr, {x, x0}];
to find the residue of expression when x equals x0.
The residue is defined as the coefficient of 1/(z-z0) in the Laurent expansion of expr.
NResidue[expr, {x, x0}];
to find numerically the residue of expression when x equals x0. Series[f, {x, x0, n}];
to generate a power series expansion for f about the point x=x0 to order (x - x0)n.
________________________________________________________________________ 9.13 Mathematica 9.13.1 Example-1 (a) Find the poles of
1
1)(
8
zzf
Clear"Global`";
hz_ 1
1 z8;
eq1 NSolveDenominatorhz 0, z;
list1 TableRez . eq1i, Imz . eq1i,
i, Lengtheq11., 0, 0.707107, 0.707107,0.707107, 0.707107, 0., 1., 0., 1.,0.707107, 0.707107, 0.707107, 0.707107, 1., 0
ListPlotlist1, PlotStyle Red, Thick, PointSize0.02,
AspectRatio 1, Background LightGray,
AxesLabel "x", "y"
-1.0 -0.5 0.5 1.0x
-1.0
-0.5
0.5
1.0
y
(b) Find the poles of
1
1)( 8
zzf
Clear"Global`";
hz_ 1
1 z8;
eq1 NSolveDenominatorhz 0, z;
list1 TableRez . eq1i, Imz . eq1i,
i, Lengtheq10.92388, 0.382683, 0.92388, 0.382683,
0.382683, 0.92388, 0.382683, 0.92388,
0.382683, 0.92388, 0.382683, 0.92388,0.92388, 0.382683, 0.92388, 0.382683
ListPlotlist1, PlotStyle Blue, Thick, PointSize0.02,
AspectRatio 1, Background LightGray,
AxesLabel "x", "y"
-0.5 0.5x
-0.5
0.5
y
_______________________________________________________________________ 9.13.2 Example-2 ((Mathematica))
The residue is the coefficient of the term z- z0-1 in the Laurent expansion of a function about the point z0. (For a very detailed survey of applications of residues, In Mathematica, we get the residue of a function at a given point via Residue. Find the residue of the function
1
11)(
zn ezzf
at z = 0. (a) n = 4. (b) n = 0, 1, ..., 10.
Residue[expr, {x, x0}] the residue of expr when x equals x0
Clear"Global`";
Fz_, n_ :1
zn
1
z 1;
SeriesFz, 4, z, 0, 10 Normal
1
z5
1
2 z4
1
12 z3
1
720 z
z
30 240
z3
1 209 600
z5
47 900 160
691 z7
1 307 674 368 000
z9
74 724 249 600
ResidueFz, 4, z, 0
1
720
Tablen, ResidueFz, n, z, 0, n, 0, 10 TableForm
0 1
1 12
2 112
3 0
4 1720
5 0
6 130 240
7 0
8 11 209 600
9 0
10 147 900 160
________________________________________________________________________ 9.13.3 Example-3
Find the residues of the function given by
z
z
e
ezzf
2
2
1)(
Clear"Global`";
f1 z2 z
1 2 z;
eq1 SolveDenominatorf1 0, z;
z1 z . eq11
2
z2 z . eq12
2
Seriesf1, z, z1, 3 Normal
2
2
8 2
z 1
48 24 2
2 z
1
12
2 z2
240 7 2
2 z3
2880
Seriesf1, z, z2, 3 Normal
2
2
8 2
z 1
48 24 2
2 z
1
12
2 z2
240 7 2
2 z3
2880
Residuef1, z, z1
2
8
Residuef1, z, z2 2
8
9.13.4 Examples 4
Find the residue of fz Cot zz2 at z0 0.
Clear"Global`";
fz_ Cot z
z2;
Seriesfz, z, 0, 101
z3
2
3 z4 z
45
2 6 z3
9458 z5
4725
2 10 z7
93 555
1382 12 z9
638 512 875 Oz11
Residuefz, z, 0
2
3 9.13.5 Example 5
Show that C 2 zz2 2
z 4 , where C is the circle C : z 2 taken with positive
orientation.
fz_ 2 z
2 z2;
eq1 SolveDenominatorfz 0, zz 2 , z 2
data1 TableRez, Imz . eq1, i, 1, Lengtheq10, 2 , 0, 2 , 0, 2 , 0, 2
f1 ListPlotdata1, PlotStyle PointSize0.02, Red,
AspectRatio Automatic;
g1 GraphicsBlue, Thick, Circle0, 0, 2;
Showf1, g1
-2 -1 1 2
-2
-1
1
2
r1 Residuefz, z, 2 1
r2 Residuefz, z, 2 1
sol 2 r1 2 r2
4
9.13.6 Example 6
Use residues to integrate C 1z4 z3 2 z2 z around the circle C : z 3.
fz_ 1
2 z2 z3 z4;
eq1 SolveDenominatorfz 0, zz 2, z 0, z 0, z 1
Lengtheq14
data1 TableRez, Imz . eq1, i, 1, Lengtheq12, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 1, 0,2, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 0, 0, 0, 1, 0
f1 ListPlotdata1, PlotStyle PointSize0.04, Red,
AspectRatio Automatic; g1 GraphicsBlue, Thick, Circle0, 0, 3;
Showf1, g1
-3 -2 -1 1 2 3
-3
-2
-1
1
2
3
Tablez . eq1i, ResiduefZ, Z, z . eq1i, i, 1, Lengtheq12,
1
12, 0,
1
4, 0,
1
4, 1,
1
3
Sum2 ResiduefZ, Z, z . eq2i, i, 1, Lengtheq20
________________________________________________________________________9.14 Calculus of residue
9.4.1 Jordan’s lemma Jordan's lemma is a result frequently used in conjunction with the residue theorem to
evaluate contour integrals and improper integrals. The theorem is named after the French mathematician Camille Jordan.
We want to calculate the integral given by
dxexf iax)( ,
where a>0 or a<0, and
0)(lim||
zfz
.
In order to do that, we consider the path integral around the contour C1 for a>0 (the upper half plane) and the contour C2 for a<0 (the lower half plane) in the complex plane.
x
y
C1
O
a>0
G1
-¶ ¶
(i) a>0
RR
iax
C
iaz Idxexfdzezf lim)()(1
If the integral reduces to zero in the upper semi-circle (1, R→∞)
0)(limlim1
dzzfeI iaz
RR
R.
Note: iaaiiaiaz )( . When >0, the integral on the large semi-circle becomes zero. Then we have
planehalfupperC
iaziax siduesidzezfdxexf Re2)()(1
where f(z) has poles inside the contour C1. (ii) a<0
x
y
C1
O
a<0
G2
-¶ ¶
planehalflolwerC
iazR
R
iax siduesidzezfIdxexf Re2)(lim)(2
when
0)(limlim2
dzzfeI iaz
RR
R
((Note))
iaaiiaiaz )( . When <0, the integral on the large semi-circle becomes zero. Then we have
planehalflolwerC
iaziax siduesidzezfdxexf Re2)()(2
In summary, For a>0,
planehalfupperC
iaziax siduesidzezfdxexf Re2)()(1
.
For a<0,
planehalflowerC
iaziax siduesidzezfdxexf Re2)()(2
.
(2) Integral along the contour of half-circle
Suppose that f(z) has a simple pole on the real axis. We now consider the contour integral along the half circle centered at the pole (z = z0),
1
)(C
dzzf
For simplicity we suppose that f(z) has a single pole at z = z0.
)()()(0
1
0
1 zgzz
bzg
zz
azf
,
where a-1 = b1, g(z) is analytic.
Now we have the integral around the contour C1 (upper-half circle with the radius located at the point z = z0)
1
0
1
0
10
1
0
1
1
)]([)(
1
11
ibidbidee
bdzzz
b
dzzgzz
bdzzfI
ii
C
CC
.
z0
C1
e
or
)(Re)()( 0
1
zzsidzzfIC
[clock-wise (CW) rotation]
where ).(Re 01 zzsa
Next we have the integral around the contour C2 (lower-half circle with the radius located at the point z = z0)
1
2
1
2
10
1
0
1
1
)]([)(
2
22
ibidbidee
bdzzz
b
dzzgzz
bdzzfI
ii
C
CC
.
z0
C2
e
or
)(Re)( 0
2
zzsidzzfIC
[counter clock-wise(CCW) rotation]
____________________________________________________________________ 9.15 Calculation of residue Residue theorem
n
nn zzazf )()( 0
,
C
nn dzzzaI )( 0 ,
irezz 0 ,
r
z0
0,11
2
0
)1(12
0
2)( n
nn
ninn
inin riaderiaidrereaI
.
Then we have
][Re)(2
101 zzsadzzf
i
.
A set of isolated singularities;
C1
C2
C3
C4
C
z1
z2
z3
z4
0...)()()()(321
CCCC
dzzfdzzfdzzfdzzf .
Here C is a path with CCW and C1, C2, … are paths with CW.
...])(Re)(Re)([Re2
)()()()(
321
321
zszszsi
dzzfdzzfdzzfdzzfCCCC
((Note))
nn
n
nn
zz
b
zz
b
zz
b
zz
bzzazf
03
0
32
0
2
0
1
00 ...)()(
(a) As long as one of the b’s are non zero, z0 is a singular point. (b) If bn does not vanish, but all bj with j>n does vanish, then f(z) has a pole of order n. (c) If b1≠0, and all other bj’s = 0, then the pole is a simple pole. (d) If f(z) has a pole of order n, 1/f(z) has a zero of order n. (e) A function that is analytic in a region except for isolated singular points is (f) Isolated essential singularity (all the bj’s) normally does not occur in physics. _____________________________________________________________________ 9.16 Mathematica 9.16.1 Application I
dx
bx
eI
iax
22 (b>0).
22
1)(
bzzf
x
y
C1
C2
G
b-b
Note that
01
|)(| 2 R
zf ( R ).
For a>0,
1
)(C
iaz zfdzeI
For a<0,
2
)(C
iaz zfdzeI
since iaz = ia ( + i ) = ia - a). We need to choose the path (upper half plane) for a>0 and the path (lower half plane) for a<0 (Jordan's lemma). There are poles on the real axis at z = ± b. We must specify how to go around. We consider the four cases (Cases I - IV). ________________________________________________________________ ((Case I))
x
y
C1
C2
G
b-b
For a>0 (upper half-plane),
)(Re2
)(Re)(Re221
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
The second term of the right-hand side clock-wise The third term of the right hand side counter clock-wise Then we have
)sin()](Re)([Re22
abb
bzsbzsidxbx
eP
iax
For a<0 (lower half-plane),
)(Re2
)(Re)(Re222
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
or
)sin()](Re)([Re22
abb
bzsbzsidxbx
eP
iax
____________________________________________________________________ ((Case II))
x
y
C1
C2
G
b-b
For a>0 (upper half-plane)
)(Re2
)(Re)(Re221
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
The second term of the right-hand side counter clock-wise The third term of the right hand side clock-wise Then we have
)sin()](Re)([Re22
abb
bzsbzsidxbx
eP
iax
For a<0 (lower half-plane)
)(Re2
)(Re)(Re222
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
or
)sin()](Re)([Re22
abb
bzsbzsidxbx
eP
iax
________________________________________________________________ ((Case-III))
x
y
C1
C2
G
b-b
(a) For a>0 (upper half-plane), we have
)(Re2)(Re2
)(Re)(Re221
22
bzsibzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
since there are two poles inside the contour. Then we have
)sin(
)](Re)([Re22
abb
bzsbzsidxbx
eP
iax
(for a>0).
(b) For a<0 (lower half-plane), we have
0
)(Re)(Re222
22
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
since there is no pole in the contour.
)sin(
)](Re)([Re22
abb
bzsbzsidxbx
eP
iax
((Case-IV))
x
y
C1
C2
G
b-b
For a>0 (upper half-plane)
)(Re2
)(Re)(Re221
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
C
iaz
or
)sin(
)](Re)([Re22
abb
bzsbzsidxbx
eP
iax
(a>0)
For a<0 (lower half plane)
)(Re2
)(Re)(Re2231
22
bzsi
bzsibzsidxbx
ePdz
bz
e iax
CC
iaz
or
)sin(
)](Re)([Re22
abb
bzsbzsidxbx
eP
iax
(a<0)
9.16.2 Application II
x
y
C1
C2
G
ib
-ib
a>0
a<0
dx
bx
edx
bx
axdx
bx
axI
iax
22220
22 2
1)cos(
2
1)cos( (b>0).
The last equality holds because of
0)sin(22
dxbx
ax (the integrand is an odd function of x)
So we calculate
dx
bx
eI
iax
22.
For a>0 (upper-half plane),
b
eibzsidx
bx
edz
bz
edz
bz
e abiax
C
iaz
C
iaz
)(Re2221
221
22
since there is one single pole at z = ib and the contour integral around the path C1 is zero (Jordan's lemma). For a<0 (lower half-plane),
b
eibzsidx
bx
edz
bz
edz
bz
e abiax
C
iaz
C
iaz
)(Re2222
222
22
since there is one single pole at z = -ib and the contour integral around the path C2 is zero (Jordan's lemma). 9.16.3 Application III
We consider the derivation of familiar integral
2
sin
0
dxx
x
((Case-I))
It can be found by integrating
dzz
eI
CC
iz
21
around the contour of Fig. The integrand has a simple pole at z = 0. This pole is avoided by placing a semicircular path C2 (the radius r →0) around it. There are no poles inside the contour. So I = 0.
x
y
C1
C2
O
G G
02121
C
iziz
C
iz
CC
iz
dzz
edz
z
edz
z
edz
z
eI
The first term is equal to zero (the radius of C1 R→∞, Jordan's lemma). Then we have
0)0(Re
zsidxx
eP
ix
or
idxx
xidx
x
eP
ix
0
sin2
or
2
sin
0
dxx
x
((Case-II))
For the calculation of the integral, we can choose the different contour shown in Fig.
x
y
C1
C2
G G
)0(Re22121
zsidzz
edz
z
edz
z
edz
z
eI
C
iziz
C
iz
CC
iz
The integrand has a simple pole at z = 0 in this contour. Using the Jordan's lemma for the contour integral around the contour C1, we have
izsidxx
eP
ix
)0(Re .
9.16.4 Application IV Unit step function
dxx
esI
isx
)(
x
y
C1
C2
C3
C4
G G
For s>0,
03131
C
iszisx
C
isz
CC
isz
dzz
edx
x
ePdz
z
edz
z
e
since there is no pole inside the contour. From the Jordan's lemma, the first term (the contour integral around C1) is equal to zero. Then we have
izsidzz
edx
x
eP
C
iszisx
)0(Re)(3
For s<0
04242
C
iszisx
C
isz
CC
isz
dzz
edx
x
ePdz
z
edz
z
e
since there is no pole inside the contour. From the Jordan's lemma, the first term (the contour integral around C2) is equal to zero. Then we have
izsidzz
edx
x
eP
C
iszisx
)0(Re)(4
In summary, we obtain
)0(
)0()(
s
s
i
isI
For s = 0,
01
dxx
We consider
)0(
)0(
,0
,1
2
1
2
1)(
s
sdx
x
eP
isu
isx
and
u(s) = 1/2 at s = 0.
-1 1 2 3s
0.2
0.4
0.6
0.8
1.0
us
9.16.5 The i prescription
We derive the formula
)(11
xix
Pix
where (0) is a positive infinitesimally small quantity. (1) Case-I
dxix
xfI
)(lim
01
x
y
C1
G1 G2z=ie
Since the only singularity near the real axis is z = i, we make the following deformation of the contour without changing the value of I. The contour runs along the real axis (the path ) and goes around counterclockwise, below the origin in a semicircle (C1), and resumes along the real axis (the path ).
)0()(
)0(Re)(
)()()(
211
1
ifdxx
xfP
zsidxx
xfP
dxx
xfdz
z
zfdx
x
xfI
C
or
)(11
xix
Pix
.
(2) Case-II
dxix
xfI
)(lim
02
x
y
C2G1 G2
z=-ie
Since the only singularity near the real axis is z = -i, we make the following deformation of the contour without changing the value of I2 The contour runs along the real axis (the path 1) and goes around clockwise, above the origin in a semicircle (C2), and resumes along the real axis (the path 2).
)0()(
)0(Re)(
)()()(
221
2
ifdxx
xfP
zsidxx
xfP
dxx
xfdz
z
zfdx
x
xfI
C
or
)(11
xix
Pix
____________________________________________________________________ 9.16.7 Fresnel integral
22
1)sin()cos(
0
2
0
2
dxxdxx
O A
B
0222
BO
z
AB
z
OA
z dzedzedze
OA
20
1
22
dxedzeI x
OA
z
BO
OB
z
BO
z dzedzeI22
2
4/irez , dredz i 4/ , 22/22 irerz i .
0
22
0
4/2 )]sin()cos([
2
)1(2
drriri
dreeI iri
AB
iz Re , iddz iRe , 222 ieRz .
4/
0
223 Re)exp(
2
ideRdzeI ii
AB
z
2/
0
22/
0
24/
0
23 )]sin(exp[
2)]cos(exp[
2)]2cos(exp[
dRR
dRR
RdRI
or
04
)1()]sin(exp[
2
22/
0
23
R
edR
RI
R
as R→∞. Note that we use the inequality
2sin for 0≤≤/2.
p8
p4
3 p8
p2
f
-0.20
-0.15
-0.10
-0.05
f f
f f= 2
pf - Sinf
Since 021 II , we have
0
22 )]sin()[cos(2
1
2drrir
i
or
0
22 )]sin()[cos()1(22
1drriri
or
22
1)sin()cos(
0
2
0
2
drrdrr
9.16.8 Arfken 7-1-15 Use the large square contour with R0 to prove that
-R R-r r
iR
x
y
C1
C2
C3
G
dxx
xsin
((Solution))
dzz
edz
z
edz
z
edz
z
edx
x
edx
x
edz
z
e iz
C
iz
C
iz
C
izR
r
ixr
R
ixiz
321
0
0)1(1
00
)(
1
RR yR iyRi
C
iz
eRR
edy
iyR
eidydz
z
e.
021)(
2
RR
R
RR
R
iRxi
C
iz
edxeRiRx
edxdz
z
e
0)1(11
0
0 )(
3
RR
y
R
iyRi
C
iz
eR
dyeRiyR
eidydz
z
e
Then
321
0C
iz
C
iz
C
iz
dzz
edz
z
edz
z
e.
Since
izsidzz
eiz
)0(Re
we have
idxx
eP
ix
or
dxx
xP
sin
9.17 Cut line for the multivalued function
Single valued function: a function which is mostly analytical with some singularities We can generalize discussion to multi-valued functions using a geometrical construction known as Riemann surfaces. We consider logarithmic functions
zizz arglnln
Argz is uniquely defined only up to multiple of 2.
nArgz 2 (i) To see what this means, we consider a closed path C enclosing z = 0. Start at z =
z0 and follow the value of lnz as it changes continuously as we go along the contour C. Whatever we had for argz0 initially we find that argz0 has increased by 2 when we return to z0.
izz initialfinal 2)(ln)(ln 00 .
((Branch point))
A point in complex plane that has this property, i.e.,
initialfinal zfzf ))()( 00
for any loop around it, is called a “branch point,” of the function.
lnz has a branch point at z = 0. (ii) Let z = 1/z’, then we have
'arg'
1ln'lnln zi
zzz
It is easy to see that lnz’ has a branch point at z’ = 0 and z = ∞. In summary, lnz has two branch points at |z| = 0 and |z| = ∞. (iii) If we draw line or curve joining the two branch points, this line is referred to as a
branch cut. So the complex plane is cut from branch point to another. ((Example))
zzf ln)( We cut the plane along the positive half of the real axis.
irzf ln)(0
where
irez where r>0 and 20 . This is really a single valued function. Similarly we can define
)2(ln)(1 irzf where r>0 and 20 . and
)2(ln)(1 irzf where r>0 and 20 .
and a whole inifinite sequence for functions
f0, f±1, f±2, f±2,……………….., f±n, which are single value and can be used to replace the multivalued function lnz. (iv) Each fn(z) suffers a discontinuity across the cut – i.e., 2i- but the values of fn(z)
above cut is the same as fn+1(z) below cut. Thus it is suggestive to say that we have an infinite series of cut planes on top
of one another. The plane that fn+1(z) is defined on, lied right above the fn(z). The adjacent planes are connected across the cut. The lower lip pf the n+1 plane is connected to the upper lip of the n-th plane. So when we cross a cut, we are going from one-cut-plane to another cut-plane. Each plane is called a Rieman sheet. Supposition of all planes in the helix array is called Rieman surface.
(v) What has all of this achieved?
Started with lnz. We have one single valued function defined on Rieman surface instead of multivalued function.
Branch points: when we go around them, we go to the different sheet. We can define the order of branch points. We have an n-th order branch point if
we have to go around it n+1 times to some back to a function original value [minimum n].
So z = 0 is a branch point of infinite order for lnz. ______________________________________________________________________ 9.19 Cut-line
The cut line runs along some curve from the z = 0 out to infinity. We consider
(a) z Cut along the negative real axis
irez ( ) The angle is restricted so as not to cross the cut line.
(i) Cut along the negative real axis as the cut line
2/ierz for
point a ( ia rez ) → rier i 2/
point b ( ib rez ) → rier i 2/
There is thus a discontinuity across the cut. The function is not analytic on the cut (ii) Cut along the positive real axis as the cut line
2/ierz for 20
point a ( 0ia rez )→ rer i 0
point b ( 2ib rez ) → rer i
There is thus a discontinuity across the cut. The function is not analytic on the cut line
(b) 'zaz
azz ' z’=0 (or z = a) and infinity (z = ∞) are branch points.
(c) )1)(1( zz
Check on the possibility of taking the line segment joining z+1 and z-1 as a cut line
irez 1 iez 1
where 20 and 20
2/)()( ierzf
(+)/2 1 0 0 0 2 0 3 0 4 5 2 6 2 7 2
(i) The phase at points 5 and 6 is not the same as the points 2 and 3. This behavior
can be expected at a branch point cut line. (ii) The phase at point 7 exceeds that at 1 by 2. So the function f(z) is therefore
single valued for the contour shown encircling both branch points.
(d) ))()(( czbzaz
There are three branch points. So there are two cut lines.
(d) ))()()(( dzczbzaz
There are four branch points. So there are two cut lines.
((Note)) (i) A function with a branch point and a required cut line will not be a continuous
across the cut line. (ii) In general, there will be a phase difference on opposite sides of this cut line. (iii) Thus the line integrals on opposite sides of the cut line will not generally cancel
each other. 9.20 Examples of cut line (Mathematica)
9.20.1 z
Plot3D of Im[ z ]there is a cut line on the negative x axis (x<0).
9.20.2 )1)(1( zz
Plot3D of Im[ )1)(1( zz ]there is a cut line between z = 1 and z = -1.
9.20.3 )1()1( zzz
Plot3D of Im[ )1()1( zzz ]A cut line between z = 1 and z = 0, and a cut line
between z = -1 and the negative x axis (x<-1).
9.20.4 )2)(1()1( zzzz
Plot3D of Im[ )2)(1()1( zzzz ]A cut line between z = -1 and z = 0, and a cut line
between z = 1 and z = 2.
9.20.5 ln(z) Plot3D of Im[ln(z)]there is a cut line on the negative x axis.
___________________________________________________________________
9.21 Applications 9.21.1 Arfken 7-1-18
Show that
)sin()1(02 a
adx
x
xa
where |a|<1. We note that z = 0 is a branch point and the positive x axis is a cut line.
x
y
B
D
H
F
A
EGC
2)1()(
z
zzf
a
.
Since there is one pole at z = ei, we have
)(Re2)()()()( i
HGFEFABCDEHA
ezsidzzfdzzfdzzfdzzf .
HA
0ixez , 0iaaa exz
0idxedz
02
0
02
.
1 )1()1()( dx
x
xdxe
x
xdzzfI
ai
a
HA
EF
ixez 2 , iaaa exz 2
idxedz 2
12
02
20
22
.
2 )1()1()( Iedx
x
xedxe
x
xdzzfI ia
aiaia
a
EF
ABCD (radius R)
iz Re , iaaa eRz
iddz iRe
0)2(Re)1(Re
)( 12
02
12
02
a
ai
i
iaa
ABCDE
RdR
Rid
eRdzzf (as R∞).
HGF (radius )
iez , iaaa ez
idedz i
0)1(
)(2
0
)1(12
02
deiidee
edzzf aiai
i
iaa
HGF
(as 0).
Then we have
)(Re2)1( 12
21 iai ezsiIeII
or
)sin(1
)(2
1
)(Re2221
a
a
e
eai
e
ezsiI
ai
ia
ai
i
where
iaai
ez
a
ez
ai aeaeaz
z
zz
dz
dezs ii
)1(1
22 |]
)1()1[(
)!12(
1)(Re
_____________________________________________________________________ 9.21.2 Application II Evaluate
03)( xax
dxI (a>0).
zazzf
3)(
1)(
f(z) has a pole at z = -a. We note that z = 0 is a branch point and the positive x axis is a cut line.
x
y
B
D
H
F
A
EGC
)(Re2)()()()( i
HGFEFABCDEHA
aezsidzzfdzzfdzzfdzzf
HA
0ixez , 2/12/02/12/1 xexz ia
0idxedz
03
.
1)(
1)( dx
xaxdzzfI
HA
EF
ixez 2 , xexz i 2/12/1
dxdxedz i 2
1
02
0
3.
2 )1()(
1)( Idx
x
xdx
xaxdzzfI
a
EF
ABCD (radius R)
iz Re , 2/2/12/1 ieRz
iddz iRe
0)2(Re)(Re
1)( 2/13
2
03
2/12
02/2/13
RdR
Rid
eRadzzf i
iiABCDE
as R∞. HGF (radius )
iez , 2/2/12/1 iez
idedz i
0)(
1)(
2
0
2/3
2/12
02/2/13
dea
iide
eaedzzf ii
iiHGF
as 0. Then we have
2/52/5121 4
3)
8
3(2)(Re22
aa
iiaezsiIII i
or
2/51 8
3
aI
((Mathematica))
Residue 1
z12
1
z a3, z, a Exp Simplify, a 0 &
3
8 a52
Series 1
z12
1
z a3, z, a Exp , 2 Simplify, a 0 & Normal
5
16 a72
a a z3
2 a32 a z2
3
8 a52 a z 35 a z
128 a92
63 a z2
256 a112
Note that the residue is equal to the coefficient of 1/(z + a). 9.22 Kronig-Kramers relation
We consider the motion of a particle (mass m and charge q) in the presence of electric field.
FqExxxm )( 200
]Re[ tieXx
, ]Re[ tieFF
.
where is the angular frequency, x is the total displacement and F is the applied electric field.
qEFXim )( 200
2
Eim
EqqX )(
1
022
0
2
Here () is called response function and is defined by
022
0
2 1)(
im
q
We need not to assume this form of (), but we make use of three properties of the response function viewed as a function of the complex variables. Now we regard () as a function of complex variable .
f0 = q2/m, and
022
0
0)(
i
f
The pole of ():
0200
2 i ,
2
4 20
200
i
.
Case (1): 04 20
20 .
x
y
O
Two poles
Case (2): 04 20
20 .
2
4 20
200
ii
.
x
y
O
Two poles
Thus we have the following features. (1) The poles of () are all below the real axis. (2) The integral of ()/ vanishes when taken around an infinite semicircle in the
upper half of the complex plane. It suffices that ()/0 uniformly as || ∞.
0)(
lim||
.
(3) )('')(')( i ,
20
22220
2200
)(
)()(
if
where )(' is even and )(" is odd with respect to real . Now we calculate
C
dzz
z
)(
.
We note that Integrant has a simple pole at the real axis.
x
y
C
C1
O z=w
GR
-¶ ¶
0)(Re)()()(
zsidx
x
xPdz
z
zdz
z
z
RC
Since
R
dzz
z
)(
,
)()(Re)(
izsidx
x
xP
,
or
dx
x
xP
i
)(1
)( .
Now we have
])(")('
[1
)(")(')(
dxx
xixP
i
i
,
])(")("
[1
)("1)('
0
0
dxx
xdx
x
xP
dxx
xP
.
Since , )(")(" xx (odd function), we have
022
00
)("2
])(")("
[1
)('
dxx
xxP
dxx
xdx
x
xP
.
Similarly
022
00
)('2
])(")("
[1
)("
dxx
xP
dxx
xdx
x
xP
.
________________________________________________________________________ APPENDIX A.1 Mathematica Residue[expr, {x, x0}];
to find the residue of expression when x equals x0.
The residue is defined as the coefficient of 1/(z-z0) in the Laurent expansion of expr.
NResidue[expr, {x, x0}];
to find numerically the residue of expression when x equals x0. Series[f, {x, x0, n}];
to generate a power series expansion for f about the point x=x0 to order (x - x0)n.
_____________________________________________________________________ A.2 Principal part of an integral
We consider a real function f(x) that blows up at x = a. The principal part of the integral is defined as
])()([lim)(0
a
a
dxxfdxxfdxxfP
((Example))
0)2
1
2
1(lim]
11[lim
12203303
dx
xdx
xdx
xP