We evaluate the integrals on the Riemann surface of genus N. If the Riemann surface is of genus N, then there are 2N + 1 or 2N + 2 branch points.
Case 1. The number of branch points is odd (2N + 1 branch points)
Figure 40 1 ≤ k ≤ N in Figure 40. Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +1) =
2N +1
Y
j=1
pz − zj.
(1) To evaluateR
akf (z)dz
Figure 41 Theoretical Evaluation
Along z2k
−→ z+ 2k+1, let d = |z2k+1− z2k|.
z = z2k+1+ rei(−π)= z2k+1− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2k,
arg(z − zj) = 0 =⇒pz − zj = q
(z2k+1− r) − zj For j = 2k + 1, 2k + 2, · · · , 2N + 1,
arg(z − zj) = −π =⇒pz − zj = −i q
zj − (z2k+1− r) Let
u(r) =
2k
Y
j=1
q
(z2k+1− r) − zj
! 2N +1 Y
j=2k+1
q
zj − (z2k+1− r)
! .
Then
Note that the value of the integral is a pure imaginary number.
Using Mathematica
(2) To evaluateR
bkf (z)dz
Figure 42 Before our compuation, we first discuss the integrals of the two kinds of path drawn in Figure 43 and Figure 44.
Class 1. Along a path that there is a cut on it
Figure 43 Since f (z)|II = −f (z)|I,
Z
zm−1
L99z− m
f (z) dz = − Z
zm−1
←−z− m
f (z) dz
= − Z
zm−1
−→z+ m
f (z) dz
So, we have
Z
zm−1 +
−→zm
f (z) dz + Z
zm−1
L99z− m
f (z) dz = 0. (49)
Class 2. Along a path that there is no cut on it
Figure 44 Since f (z)|II = −f (z)|I,
Z
zm−1
L99z− m
f (z) dz = − Z
zm−1
←−z− m
f (z) dz
= −
− Z
zm−1
−→z+ m
f (z) dz
= Z
+ f (z) dz
So, we have
Thus, we only need to evaluate the integralsR
zj
−→z+ j+1f (z) dz for j = 1, 2, · · · , 2k − 1 and add them together. That is,
Z
Thus,
Z
bk
f (z) dz = 2
k
X
m=1
Z
z2m−1−→z2m
f (z) dz (52)
= 2
k/2
X
m=1
(−1)N −m Z 0
d
u(r) dr (53)
Note that the value of the integral is a real number.
Using Mathematica For j = 1, 2, · · · , 2m − 1,
arg(z − zj) = 0 =⇒pz − zj = math pz − zj For j = 2m, 2m + 1, · · · , 2N + 1,
arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj Then,
Z
z2m−1
−→z+ 2m
f (z) dz = (−1)(2N +1)−2m+1
· math
Z z2m
z2m−1
f (z) dz
= math
Z z2m
z2m−1
f (z) dz
Thus,
Z
bk
f (z) dz = 2
k
X
m=1
Z
z2m−1−→z2m
f (z) dz (54)
= 2
k
X
m=1
math
Z z2m
z2m−1
f (z) dz
(55)
Case 2. The number of branch points is even (2N + 2 branch points)
Figure 46
1 ≤ k ≤ N in Figure 46. Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +2) =
2N +2
Y
j=1
pz − zj.
(1) To evaluateR
akf (z)dz
Figure 47 Theoretical Evaluation
Along z2k+1−→ z+ 2k+2, let d = |z2k+2− z2k+1|.
z = z2k+2+ rei(−π)= z2k+2− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2k + 1,
arg(z − zj) = 0 =⇒pz − zj = q
(z2k+2− r) − zj For j = 2k + 2, 2k + 3, · · · , 2N + 2,
arg(z − zj) = −π =⇒pz − zj = −i q
zj − (z2k+2− r) Let
u(r) =
2k+1
Y
j=1
q
(z2k+2− r) − zj
! 2N +2 Y
j=2k+2
q
zj− (z2k+2− r)
! . Then
Z
z2k+1
−→z+ 2k+2
f (z) dz = −(−i)(2N +2)−(2k+2)+1
Z 0 d
u(r) dr
= −(−i)2N −2k+1 Z 0
d
u(r) dr
= (−1)N −k· i Z 0
d
u(r) dr Thus,
Z
ak
f (z) dz = 2 Z
z2k+1
−→z+ 2k+2
f (z) dz (56)
= (−1)N −k· 2i Z 0
d
u(r) dr (57)
Note that the value of the integral is a pure imaginary number.
Using Mathematica For j = 1, 2, · · · , 2k + 1,
arg(z − zj) = 0 =⇒pz − zj = math pz − zj
For j = 2k + 2, 2k + 3, · · · , 2N + 2,
arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj
Then,
Z
z2k+1
−→z+ 2k+2
f (z) dz = (−1)(2N +2)−(2k+2)+1
· math
Z z2k+2
z2k+1
f (z) dz
!
= (−1) · math
Z z2k+2
z2k+1
f (z) dz
!
Thus,
Z
ak
f (z) dz = 2 Z
z2k+1
−→z+ 2k+2
f (z) dz (58)
= (−2) · math
Z z2k+2
z2k+1
f (z) dz
!
(59)
(2) To evaluateR
bkf (z)dz
Figure 48 Using similar arguments in class 1 and class 2 of Case 1, we also can see that
Z
bk
f (z) dz = 2 Z
z2−→z3
f (z) dz + Z
z4−→z5
f (z) dz + · · · + Z
z2k−→z2k+1
f (z) dz
!
= 2
k
X
m=1
Z
z2m−→z2m+1
f (z) dz (60)
Figure 49 Theoretical Evaluation
Along z2m−→z2m+1, let d = |z2m+1− z2m|.
z = z2m+1+ rei(−π)= z2m+1− r, r : d −→ 0 =⇒ dz = −dr For j = 1, 2, · · · , 2m,
arg(z − zj) = 0 =⇒pz − zj = q
(z2m− r) − zj
For j = 2m + 1, 2m + 2, · · · , 2N + 2,
arg(z − zj) = −π =⇒pz − zj = −i q
zj − (z2m− r) Let
u(r) =
2m
Y
j=1
q
(z2m− r) − zj
! 2N +2 Y
j=2m+1
q
zj− (z2m− r)
! . Then
Z
z2m−→z2m+1
f (z) dz = −(−i)(2N +2)−(2m+1)+1
Z 0 d
u(r) dr
= −(−i)2N −2m+2 Z 0
d
u(r) dr
= (−1)N −m Z 0
d
u(r) dr Thus,
Z
bk
f (z) dz = 2
k
X
m=1
Z
z2m−→z2m+1
f (z) dz (61)
= 2
k
X
m=1
(−1)N −m Z 0
d
u(r) dr (62)
Note that the value of the integral is a real number.
Using Mathematica For j = 1, 2, · · · , 2m,
arg(z − zj) = 0 =⇒pz − zj = math pz − zj For j = 2m + 1, 2m + 2, · · · , 2N + 2,
arg(z − zj) = −π =⇒pz − zj = (−1) · math pz − zj
Then, Z
z2m−→z2m+1
f (z) dz = (−1)(2N +2)−(2m+1)+1
· math
Z z2m+1
z2m
f (z) dz
= math
Z z2m+1
z2m
f (z) dz
Thus,
Z
bk
f (z) dz = 2
k
X
m=1
Z
z2m−→z2m+1
f (z) dz (63)
= 2
k
X
m=1
math
Z z2m+1
z2m
f (z) dz
(64)
Example 8. Let
f (z) =
2N +2
Y
j=1
pz − zj.
Suppose that Im(z2j−1) = Im(z2j), j = 1, 2, · · · , N + 1. The cuts are drawn in Figure 50.
Let Re(zj) = xj, j = 1, 2, · · · , 2N + 2 and Im(z2j−1) = Im(z2j) = yj, j = 1, 2, · · · , N + 1.
Evaluate R
akf (z)dz and R
bkf (z)dz.
Figure 51 Figure 52 Solution.
(1) To evaluateR
akf (z)dz (Figure 51) Theoretical Evaluation
Along z2k+1−→ z+ 2k+2, let d = |z2k+2− z2k+1|.
z = z2k+2+ rei(−π)= z2k+2− r, r : d −→ 0 =⇒ dz = −dr For j = 2k + 2,
arg(z − z2k+2) = −π =⇒√
z − z2k+2= −i√ r For other j,
arg(z − zj) 6= −π =⇒pz − zj = q
(z2k+2− r) − zj Let
u(r) =
2k+1
Y
j=1
q
(z2k+2− r) − zj
! √ r
2N +2
Y
j=2k+3
q
zj − (z2k+2− r)
! . Then
Z
z2k+1
−→z+ 2k+2
f (z) dz = −(−i) Z 0
d
u(r) dr = i Z 0
d
u(r) dr Thus,
Z
ak
f (z) dz = 2i Z 0
d
u(r) dr .
Using Mathematica For other j,
arg(z − zj) 6= −π =⇒pz − zj = math pz − zj
(2) To evaluateR
bkf (z)dz (Figure 52)
Theoretical Evaluation For other j,
arg(z − zj) 6= −π =⇒pz − zj = For other j,
arg(z − zj) 6= −π =⇒pz − zj = math pz − zj
Then,
4 Integrals for Vertical Cuts
4.1 Cut Structures for Vertical Cuts
We first define the branches for vertical cuts. Let f (z) = √
z and let z = reiθ, where θ = arg z. We define two single-valued branches of f as
f (z) =√
re12iθ, −3π
2 ≤ θ < π 2, and
f (z) =√
re12iθ, π
2 ≤ θ < 5π 2 . And we define sheet I and sheet II as
sheet I = {z ∈ C| −3π
2 ≤ arg z < π 2}, and
sheet II = {z ∈ C|π
2 ≤ arg z < 5π 2 }.
To Label the second quadrant with a + and label the first quadrant with a −.
Figure 53
Then we can use the same method used in section 2.1 to construct the Riemann surface for f. (see Figure 6)
4.2 The Problem in Using Mathematica
We use (I) to denote sheet I and (II) to denote sheet II. We can see z ∈ (I) =⇒ −3π
2 ≤ arg z < π
2 =⇒ −3π 4 ≤ 1
2arg z < π 4. f maps the points on sheet I into the region {z ∈ C| −3π4 ≤ arg z < π4}.
Figure 54 And,
z ∈ (II) =⇒ π
2 ≤ arg z < 5π
2 =⇒ π 4 ≤ 1
2arg z < 5π 4 . f maps the points on sheet II into the region {z ∈ C|π4 ≤ arg z < 5π4 }.
Figure 55
Let z ∈ Ic = [−3π2 , −π] ⊆ (I). For example, suppose that z = −1 + i ∈ (I). Then arg z = −5π4 and z = ei(−5π4 ).
arg z = −5π
4 ∈ Ic=⇒ arg √
z = −5π 8
=⇒ f (z) = √
−1 + i = (ei(−5π4 ))12 = ei(−5π8 ) But in Mathematica,
−1 + i = ei(3π4) =⇒√
−1 + i = ei(3π8 ) We find that ei(3π8) = (−1) · ei(−5π8 ).
Figure 56
Thus we have the result
z ∈ sheet I and − 3π
2 ≤ arg z ≤ −π =⇒√
z = (−1) · math √ z Let θ = arg z, and let
A = {z ∈ C| − π
2 < θ < π 4}, BT = {z ∈ C| − 3π
4 ≤ θ ≤ −π 2}, BM = {z ∈ C|π
4 ≤ θ ≤ π 2}.
BT
BM
A
Figure 57 Theoretically,
f (sheet I) = A ∪ BT. In Mathematica,
f (sheet I) = A ∪ BM.
4.3 Evaluating Integrals Using Mathematica
Example 9. Let f (z) =√
z and let γ be the positively oriented (counterclockwise oriented) circular path z = eiθ, −3π2 ≤ θ < π2. Evaluate the integral R
γf (z) dz.
Solution.
(1) Integral along the circular path
Figure 58
z ∈ γ =⇒ z = eiθ, −3π
2 ≤ θ < π 2
=⇒ dz = ieiθdθ.
Then,
Z
γ
f (z) dz = math (−1) · Z −π
−3π
2
f (eiθ)ieiθ dθ + Z π2
−π
f (eiθ)ieiθ dθ
!
= math (−1) · Z −π
−3π
2
√
eiθieiθ dθ + Z π2
−π
√
eiθieiθ dθ
!
= −0.942809 + 0.942809i . (2) Deformation of path
Figure 59 Theoretical Evaluation
Along i−→ 0(z ∈ γ+ ∗)
z = ri, r : 1 −→ 0 =⇒ dz = idr arg z = −3
2π =⇒ √
z =p|ri| ei(−3π4 )
Z
γ∗
f (z) dz = Z 0
1
p|ri| ei(−3π4 )i dr
= −0.471405 + 0.471405i .
To use the similar method of deriving Equation(18), we can know that Z
γ
f (z) dz = 2 Z
γ∗
f (z) dz
= −0.942809 + 0.942809i . Using Mathematica
3 √ √
Z
γ∗
f (z) dz = (−1) · math
Z 0 1
f (ri) i dr
= (−1) · math
Z 0 1
√ ri i dr
. Then,
Z
γ
f (z) dz = 2 Z
γ∗
f (z) dz
= −0.942809 + 0.942809i .
Example 10. Suppose that f (z) =p(z − i)(z − 2i)(z − 3i)(z − 4i) and γ is a positively oriented simple closed curve that encloses all cuts. Evaluate the integral R
γf (z)dz.
Figure 60 Solution.
Theoretical Evaluation (1)Along 4i−→ 3i+
z = ri, r : 4 −→ 3 =⇒ dz = idr For k = 1, 2, 3,
arg(z − ki) = −3
2π =⇒√
z − ki =p|ri − ki| ei(−3π4 ), and
arg(z − 4i) = −1
2π =⇒√
z − 4i = p|ri − 4i| ei(−π4) Then,
Z
4i−→3i+
f (z) dz = Z 3
4 4
Y
k=1
p|ri − ki|
!
ei(−3π4 )3
ei(−π4 )i dr
= −0.76002 .
(2)Along 2i−→ i+
arg(z − i) = −3
2π =⇒√
z − i = (−1) · math√
z − i
For k = 2, 3, 4,
arg(z − ki) = −1
2π =⇒ √
z − ki = math√
z − ki
Then,
Z
2i−→i+
f (z) dz = (−1) · math Z 1
2 4
Y
k=1
p|ri − ki|
! i dr
!
= 0.76002 . Thus,
Z
γ
f (z) dz = 2
Z
4i−→3i+
f (z) dz + Z
2i←−i+
f (z) dz
= 0 . Example 11. Suppose that
f (z) =p
(z − i)(z − 2i)(z − 3i))(z − 4i)(z − 5i)(z − 6i) .
Let a1, a2 be two a − cycles and let b1, b2 be two b − cycles drawing in Figure 61. Evaluate the four integrals R
akf (z)dz and R
bkf (z)dz, k = 1, 2 using the method of deformation of path.
Figure 61 Solution.
Figure 62 1. To evaluate R
a1f (z)dz Theoretical Evaluation Along 2i−→ i,+
z = ri, r : 2 −→ 1 =⇒ dz = idr arg(z − i) = −3
2π =⇒ √
z − i =p|ri − i| ei(−3π4), For k = 2, 3, 4, 5, 6,
arg(z − ki) = −1
2π =⇒√
z − ki =p|ri − ki| ei(−π4) Then,
Z
a1
f (z) dz = 2 Z
2i−→i+
f (z) dz
= 2 Z 1
2 6
Y
k=1
p|ri − ki|
!
ei(−3π4 )
ei(−π4 )5
i dr
= 2 Z 1
2 6
Y
k=1
p|ri − ki|
! i dr
= −6.08344i .
Using Mathematica Along 2i−→ i,+
z = ri, r : 2 −→ 1 =⇒ dz = idr arg(z − i) = −3
2π =⇒√
z − i = (−1) · math
√ z − i
,
For k = 2, 3, 4, 5, 6,
2. To evaluate R
a2f (z)dz
For k = 4, 5, 6,
arg(z − ki) = −1
2π =⇒ √
z − ki = math√
z − ki Then,
Z
a2
f (z) dz = 2 Z
4i−→3i+
f (z) dz
= 2 · (−1)3· math Z 3
4 6
Y
k=1
√
ri − ki
! i dr
!
= 2.88937i .
Figure 63 3. To evaluate R
b1f (z)dz Theoretical Evaluation (1)Along 5i −→ 4i
z = ri, r : 5 −→ 4 =⇒ dz = idr For k = 1, 2, 3, 4,
arg(z − ki) = −3
2π =⇒√
z − ki =p|ri − ki| ei(−3π4) For k = 5, 6,
arg(z − ki) = −1
2π =⇒√
z − ki =p|ri − ki| ei(−π4)
Then,
For k = 1, 2, 3, 4, 4. To evaluate R
b2f (z)dz : We have done in 3.
Z
f (z) dz = 2 Z
f (z) dz = 3.43541 .
Next, we discuss how to determine the region needed to change the sign of a given function f .
Example 12. Let f (z) = √
z − z1√
z − z2. Determine the region needed to change the sign in sheet I of the cut plane for f .
Solution.
Figure 64
We separate the cut plane to six region, R1, R2, R3, R4, R5, R6 (Figure 64). Let Ic = [−3π2 , −π]. We investigate the sign of each√
z − zk for all z ∈ sheet I.
If z ∈ R1,
arg(z − z1) ∈ Ic=⇒√
z − z1 = (−1) · math √
z − z1 arg(z − z2) ∈ Ic=⇒√
z − z2 = (−1) · math √
z − z2 . Then,
f (z) = (−1)2· math (f(z)) = math (f(z)) . If z ∈ R2,
arg(z − z1) ∈ Ic=⇒√
z − z1 = (−1) · math √
z − z1 arg(z − z2) /∈ Ic=⇒√
z − z2 = math √
z − z2 . Then,
f (z) = (−1) · math (f (z)) . If z ∈ R3,
arg(z − z1) /∈ Ic =⇒√
z − z1 = math √ z − z1
arg(z − z2) /∈ Ic =⇒√
z − z2 = math √
z − z2 .
Then,
f (z) = math (f (z)) . If z ∈ R4∪ R5∪ R6,
arg(z − z1) /∈ Ic =⇒√
z − z1 = math √
z − z1 arg(z − z2) /∈ Ic =⇒√
z − z2 = math √
z − z2 . Then,
f (z) = math (f (z)) .
Thus, the region needed to change the sign is R2. We call such region the sign-region of f .
Example 13. Let f (z) = √
z − z1√
z − z2√
z − z3. Determine the region needed to change the sign in sheet I of the cut plane for f .
Solution.
Figure 65
From example 11, we know that it does not to change sign on the right-half side of sheet I. So we only discuss the left-half side.
We separate the left-half side to four region, R1, R2, R3, R4 (Figure 65). Let Ic = [−3π2 , −π].
If z ∈ R1,
arg(z − z1) ∈ Ic=⇒√
z − z1 = (−1) · math √
z − z1 arg(z − z2) ∈ Ic=⇒√
z − z2 = (−1) · math √
z − z2 arg(z − z3) ∈ Ic=⇒√
z − z3 = (−1) · math √
z − z3 . Then,
f (z) = (−1)3· math (f(z)) = (−1) · math (f(z)) .
If z ∈ R2,
arg(z − z1) ∈ Ic=⇒√
z − z1 = (−1) · math √
z − z1 arg(z − z2) ∈ Ic=⇒√
z − z2 = (−1) · math √ z − z2
arg(z − z3) ∈ Ic=⇒√
z − z3 = math √
z − z3 . Then,
f (z) = (−1)2· math (f(z)) = math (f(z)) . If z ∈ R3,
arg(z − z1) ∈ Ic=⇒√
z − z1 = (−1) · math √
z − z1 arg(z − z2) ∈ Ic=⇒√
z − z2 = math √
z − z2 arg(z − z3) ∈ Ic=⇒√
z − z3 = math √
z − z3 . Then,
f (z) = (−1)math (f (z)) . If z ∈ R4,
arg(z − z1) ∈ Ic =⇒√
z − z1 = math √
z − z1 arg(z − z2) ∈ Ic =⇒√
z − z2 = math √
z − z2 arg(z − z3) ∈ Ic =⇒√
z − z3 = math √
z − z3 . Then,
f (z) = math (f (z)) . Thus, the sign-region are R1 and R3.
We generalize the result of example 11 and example 12 in next two examples.
Example 14. (There are odd branch points) Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +1) =
2N +1
Y
j=1
pz − zj,
where Re(zk), k = 1, 2, · · · , 2N + 1, are all the same (Figure 62). Determine the region needed to change the sign in sheet I of the cut plane for f .
Solution.
We separate the left-half side to the 2N + 1 region, R1, R2, · · · , R2N +1. Let Ic = [−3π2 , −π].
Figure 66 (1) z ∈ R2j−1, j = 1, 2, · · · , N + 1.
For k = 1, 2, · · · , 2j − 2,
arg(z − zk) ∈ Ic =⇒√
z − zk = (−1) · math √
z − zk For k = 2j − 1, 2j, · · · , 2N + 1,
arg(z − zk) /∈ Ic=⇒√
z − zk= math √ z − zk
Then,
f (z) = (−1)2j−2· math (f(z)) = math (f(z)) . (2) z ∈ R2j, j = 1, 2, · · · , N + 1.
For k = 1, 2, · · · , 2j − 1,
arg(z − zk) ∈ Ic =⇒√
z − zk = (−1) · math √
z − zk For k = 2j, 2j + 1, · · · , 2N + 2,
arg(z − zk) /∈ Ic=⇒√
z − zk= math √
z − zk Then,
f (z) = (−1)2j−1· math (f(z)) = (−1) · math (f(z)) .
Example 15. (There are even branch points) Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +2) =
2N +2
Y
j=1
pz − zj,
where Re(zj), j = 1, 2, · · · , 2N + 2, are all the same (Figure 67). Determine the region needed to change the sign in sheet I of the cut plane for f .
Solution.
We separate the left-half side to the 2N + 2 region, R1, R2, · · · , R2N +2. Let Ic = [−3π2 , −π].
Figure 67 (1) z ∈ R2j−1, j = 1, 2, · · · , N + 2.
For k = 1, 2, · · · , 2j − 2,
arg(z − zk) ∈ Ic =⇒√
z − zk = (−1) · math √ z − zk
For k = 2j − 1, 2j, · · · , 2N + 1,
arg(z − zk) /∈ Ic=⇒√
z − zk= math √
z − zk Then,
f (z) = (−1)2j−2· math (f(z)) = math (f(z)) . (2) z ∈ R2j, j = 1, 2, · · · , N + 1.
For k = 1, 2, · · · , 2j − 1,
arg(z − zk) ∈ Ic =⇒√
z − zk = (−1) · math √
z − zk
For k = 2j, 2j + 1, · · · , 2N + 2,
arg(z − zk) /∈ Ic=⇒√
z − zk= math √
z − zk Then,
f (z) = (−1)2j−1· math (f(z)) = (−1) · math (f(z)) . Thus, the sign-region are R2, R4, · · · , R2N +2.