Case 1. The number of branch points is odd (2N + 1 branch points) Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +1) =
2N +1
Y
j=1
pz − zj.
Aussume that Im(zj) > Im(zj+1) for j = 1, 2, · · · , 2N . Let yj to denote the imaginary part of zj for all j. That is, yj = Im(zj).
Figure 68 Figure 69 Figure 70
(1) To evaluateR
Then,
(2) To evaluateR
bkf (z)dz (Figure 70)
To use the similar method of deriving Equation (49) in case 1 of (2) in section 3.4, we obtain
Thus,
Case 2. The number of branch points is even (2N + 2 branch points) Let
Figure 71 Figure 72 Figure 73 (1) To evaluateR
akf (z)dz (Figure 72) Theoretical Evaluation
Along z2k+1−→ z+ 2k+2,
z = ri, r : y2k+1−→ y2k+2=⇒ dz = idr For j = 1, 2, · · · , 2k + 1,
arg(z − zj) = −1
2π =⇒pz − zj = q
|ri − zj| ei(−π4) For j = 2k + 2, 2k + 3, · · · , 2N + 2,
arg(z − zj) = −3
2π =⇒pz − zj = q
|ri − zj| ei(−3π4)
Then,
Note that this value is a pure imaginary number.
Using Mathematica
(2) To evaluateR
bkf (z)dz (Figure 73)
Z
For j = 1, 2, · · · , 2m,
arg(z − zj) = −1
2π =⇒pz − zj = math
pri − zj For j = 2m + 1, 2m + 2, · · · , 2N + 2,
arg(z − zj) = −3
2π =⇒pz − zj = (−1) · math
pri − zj Then,
Z
bk
f (z) dz = 2 Z
z2m−→z2m+1
f (z) dz
= 2 · (−1)(2N +2)−2m· math
Z y2m+1
y2m
2N +2
Y
j=1
pri − zj
! i dr
!
= 2 · math
Z y2m+1
y2m
2N +2
Y
j=1
pri − zj
! i dr
! .
Next, we investigate the sign-regions for other complicated examples.
Example 16. Suppose that
f (z) =
7
Y
j=1
pz − zj
and the cutted plane (sheet I) is drawn below. To determine the sign-regions of f .
Figure 74
Solution.
First we draw a horizontal line from each end point of every cut to the direction of minus real asix, L1, L2, L3, L4, L5, L6. Then these horizontal lines and all cuts separate the complex plane to several regions, R1, · · · , R18. Let Ic= [−3π2 , −π].
z ∈ R1
=⇒ arg(z − zk) ∈ Ic for all k
=⇒ f (z) = (−1)7· math (f(z)) = (−1) · math (f(z)) .
z ∈ R8
=⇒
( arg(z − zk) ∈ Ic if k = 2, 3, 5, 6, 7, arg(z − zk) /∈ Ic otherwise.
=⇒ f (z) = (−1)5· math (f(z)) = (−1) · math (f(z)) .
z ∈ R15
=⇒
( arg(z − zk) ∈ Ic if k = 5, 7, arg(z − zk) /∈ Ic otherwise.
=⇒ f (z) = (−1)2· math (f(z)) = math (f(z)) .
To discuss the other regions using the above method, you can find that the sign-regions are the gray regions drawn in Figure 74.
We also give a simple method of finding the sign-regions. For z ∈ Rk, e,g, z ∈ R10, we imgine that there is a coordinate with orgin z. If there are odd branch points in the forth quadrant, then f (z) = (−1) · math (f (z)). If there are even branch points in the forth quadrant, then f (z) = math (f (z)). Finally the sign-regions is shown below.
Example 17. Suppose that f (z) =p
[z − (1 + 2i)][z − (1 − 2i)][z − (2 + i)][z − (2 − i)].
Evaluate the integral R
bf (z)dz.
Figure 76 Solution.
We evaluate this integral along two different, but equivalent paths, respectively.
(1) Along the paths in Figure 77
Figure 77 Since the path b∗1 lies in a sign-region, we obtain
Z
b∗1
f (z) dz + Z
b∗∗1
f (z) dz = 2 Z
b∗1
f (z) dz
= 2 · (−1) · math
Z
1−i−→1−2i+
f (z) dz
Since b∗3 and b∗∗3 are two paths in sheet II and f (z)|II = (−1) · f (z)|I, Furthermore, b∗3 lies in a sign-region, so
Z
(2) Along the paths in Figure 78
Figure 78 Z
1+2i−→1+i+
f (z) dz = (−1) · math
Z 1 2
f (1 + ri)i dr
and
Z
1+2iL991+i−
f (z) dz = (−1) Z
1+2i←−1+i+
f (z) dz
= (−1) · math
Z 2 1
f (1 + ri)i dr
= (−1) Z
1+2i−→1+i+
f (z) dz Thus,
Z
1+2i−→1+i+
f (z) dz + Z
1+2iL991+i−
f (z) dz = 0 . Similarly,
Z
1+i−→1−i+
f (z) dz + Z
1−iL991+i−
f (z) dz = 0 Z
1−i−→1−2i+
f (z) dz + Z
1−2iL991−i−
f (z) dz = 0
So, we have
Z
1+2i−→1−2i+
f (z) dz + Z
1−2iL991+2i−
f (z) dz = 0 and
Z
2+i−→2−i+
f (z) dz + Z
2−iL992+i−
f (z) dz = 0
It remains to evaluate the integrals along the slant pathsR
1−2i−→2+if (z) dz+R
2+iL991−2if (z) dz.
Z
1−2i−→2+i
f (z) dz + Z
2+iL991−2i
f (z) dz
= 2 Z
1−2i−→2+i
f (z) dz
= 2 Z
√ 10 3
0
f (1 − 2i + rei tan−13)ei tan−13dr + 2 · (−1) Z
√ 10
√ 10 3
f (1 − 2i + rei tan−13)ei tan−13dr
= −0.13095 − 11.5969i . Thus,
Z
b
f (z) dz = 2 Z
1−2i−→2+i
f (z) dz = −0.13095 − 11.5969i .
Example 18. Suppose that
f (z) =
N +1
Y
j=1
pz − zjpz − zj.
satisfying that Re(zj) < Re(zj+1) and Im(zj) = Im(zj+1) for all j and suppose that N + 1 is odd. The cut plane is drawn in Figure 79. Evaluate R
akf (z)dz and R
bkf (z)dz.
Figure 79 Solution.
1. EvaluateR
akf (z)dz
Figure 80 Let zj = xj + iyj for all j.
Z
zk+1
−→z+ k+1
f (z) dz = math
Z −yk+1 yk+1
f (xk+1+ ri)i dr
!
Z
zk+1←−z− k+1
f (z) dz = (−1) · math
Z yk+1
−yk+1
f (xk+1+ ri)i dr
!
= math
Z −yk+1 yk+1
f (xk+1+ ri)i dr
!
= Z
zk+1
−→z+ k+1
f (z) dz Thus,
Z
ak
f (z) dz = 2 Z
zk+1
−→z+ k+1
f (z) dz
= 2 · math
Z −yk+1 yk+1
f (xk+1+ ri)i dr
! .
2. EvaluateR
bkf (z)dz
Figure 81 For j = 1, 2, · · · , k,
Z
zk+1
−→z+ k+1
f (z) dz + Z
zk+1
L99z− k+1
f (z) dz = 0
So,
Z
bk
f (z)dz = 2
k
X
j=1
Z
zj−→zj+1
f (z) dz
Let zj = xj + iyj for all j and let
d = |zj+1− zj+1|
|zj+1− zj| = 2yj
xj+1− xj
For z ∈ zj −→ zj+1, let z = zj+ rei tan−1d, r : 0 −→ |zj+1− zj|. Then Z
zj−→zj+1
f (z) dz = (−1)j· math
Z |zj+1−zj| 0
f (zj + rei tan−1d)ei tan−1ddr
!
Therefore, Z
bk
f (z)dz = math 2
k
X
j=1
(−1)j
Z |zj+1−zj| 0
f (zj+ rei tan−1d)ei tan−1ddr
! .
Example 19. Suppose that f (z) =p
[z − (−1 + i)][z − 0][z − 2i][z − (1 + 3i)][z − (1 + 5i)][z − (2 + 2i)][z − (2 − 2i)].
The cut plane is drawn in Figure 82. Evaluate R
bkf (z)dz for k = 1, 2, 3.
Figure 82 Solution.
Figure 83 1. EvaluateR
b1f (z)dz
= 0.0529343 − 18.4855i . 2. EvaluateR
b2f (z)dz
Thus,
Z
b2
f (z)dz = 2 Z
−1+i−→2i
f (z) dz + 2 Z
0−→1+5i
f (z) dz
= (0.0529343 − 18.4855i) + (−67.2252 + 88.1591i)
= −67.1723 + 69.6736i . 3. EvaluateR
b3f (z)dz Z
b3
f (z)dz = 2 Z
−1+i−→2i
f (z) dz + 2 Z
0−→1+5i
f (z) dz + 2 Z
1+3i−→2+2i
f (z) dz
Along 1 + 3i −→ 2 + 2i, let z = (1 + 3i) + rei(−π4), r : 0 −→√ 2.
Z
1+3i−→2+2i
f (z) dz = 2 · math Z
√ 2
0
f ((1 + 3i) + rei(−π4))ei(−π4)dr
!
= 9.0209 + 17.2364i . Thus,
Z
b3
f (z)dz = 2 Z
−1+i−→2i
f (z) dz + 2 Z
0−→1+5i
f (z) dz + 2 Z
1+3i−→2+2i
f (z) dz
= (0.0529343 − 18.4855i) + (−67.2252 + 88.1591i) + (9.0209 + 17.2364i)
= −58.1514 + 86.91i .
5 Integrals for Slant Cuts
5.1 Cut Structures for Slant Cuts
Let f (z) =√
z and let z = reiθ where θ = arg z. Define two single-valued branches of f as
f (z) =√
re12iθ, −7π
4 ≤ θ < π 4, and
f (z) =√
re12iθ, π
4 ≤ θ < 9π 4 . Define sheet I and sheet II as
sheet I = {z ∈ C| −7π
4 ≤ arg z < π 4}, and
sheet II = {z ∈ C|π
4 ≤ arg z < 9π 4 }.
To Label a + and label a − as in Figure 84.
Figure 84
Then we can use the same method used in section 2.1 to build the Riemann surface for f. (see Figure 6)
More generally, suppose that 0 ≤ α ≤ π. we can define two single-valued branches of f as
f (z) =√
re12iθ, α − 2π ≤ θ < α, and
f (z) =√
re12iθ, α ≤ θ < α + 2π.
Define sheet I and sheet II as
sheet I = {z ∈ C|α − 2π ≤ arg z < α},
and
sheet II = {z ∈ C|α ≤ arg z < α + 2π}.
To Label a + and label a − as in Figure 85.
p a q= -2
a q =
a
p a q= +2
a q =
Figure 85
5.2 The Problem in Using Mathematica
Let (I) to denote sheet I and let (II) to denote sheet II. Then z ∈ (I) =⇒ −7π
4 ≤ arg z < π
4 =⇒ −7π
8 ≤ arg z < π 8. f maps the points on sheet I into the region {z ∈ C| −7π8 ≤ 12arg z < π8}.
Figure 86 And,
z ∈ (II) =⇒ π
4 ≤ arg z < 9π
4 =⇒ π 8 ≤ 1
2arg z < 9π 8 . f maps the points on sheet II into the region {z ∈ C|π8 ≤ arg z < 9π8 }.
Figure 87
Let z ∈ Ic = [−7π4 , −π] ⊆ (I). For example, suppose that z = 12 +
√3
2 i ∈ (I). Then arg z = −5π3 and z = ei(−5π3 ).
arg z = −5π
3 ∈ Ic=⇒ arg √
z = −5π 6
=⇒ f (z) = s
1 2 +
√3
2 i = (ei(−5π3 ))12 = ei(−5π6 ) But in Mathematica,
1 2+
√3
2 i = ei(π3)=⇒
s 1 2+
√3
2 i = ei(π6) Note that ei(π6) = (−1) · ei(−5π6 ).
Figure 88 Thus we have the result
z ∈ sheet I and − 7π
4 ≤ 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}.
Figure 89 Theoretically,
f (sheet I) = A ∪ BT. In Mathematica,
f (sheet I) = A ∪ BM.
5.3 Evaluating Integrals Using Mathematica
Example 20. Let f (z) =√
z and let γ be the positively oriented (counterclockwise ori-ented) circular path z = eiθ, −7π4 ≤ θ < π4. Evaluate the integralR
γf (z) dz.
Figure 90 Solution.
(1) Integral along the circular path
z ∈ γ =⇒ z = eiθ, −7π
4 ≤ θ < π
4 =⇒ dz = ieiθdθ.
Then, (2) Deformation of path
Theoretical Evaluation
To use the similar method of deriving Equation(18), we can know that Z
Example 21. Suppose that
f (z) =p
z − (1 + i)p
z − (2 + 2i).
Let γ be the positively oriented circular path γ : z = 3
2 +3
2i + eiθ, −7π
4 ≤ θ < π 4. Evaluate the integral R
γf (z) dz.
Figure 91 Solution.
Similar to the method of finding sign-regions for vertical cuts, the sign-region is shown in Figure 91.
(1) Integral along the circular path
z ∈ γ =⇒ z = 3 2+ 3
2i + eiθ, −7π
4 ≤ θ < π
4 =⇒ dz = ieiθdθ.
Z
γ
f (z) dz = math
Z −7π6
−7π4
f (3 2+3
2i + eiθ)ieiθ dθ
!
+ (−1) · math
Z −5π
6
−7π
6
f (3 2 +3
2i + eiθ)ieiθ dθ
!
+ math Z π4
−5π
6
f (3 2 +3
2i + eiθ)ieiθ dθ
!
= 1.5708 . (2) Deformation of path
Theoretical Evaluation Along 2 + 2i−→ 1 + i (z ∈ γ+ ∗)
z = 1 + i + rei(−7π4), r :√
2 −→ 0 =⇒ dz = ei(−7π4 )dr
Then,
z − (1 + i) = |1 + i + rei(−7π4 )− (1 + i)|ei(−7π4 )
=⇒p
z − (1 + i) = q
|1 + i + rei(−7π4)− (1 + i)| ei(−7π8 )
z − (2 + 2i) = |1 + i + rei(−7π4 )− (2 + 2i)|ei(−3π4)
=⇒p
z − (2 + 2i) = q
|1 + i + rei(−3π4)− (2 + 2i)| ei(−3π8 )
Z
γ
f (z) dz = 2 Z
γ∗
f (z) dz
= 2 Z 0
√ 2
q
|1 + i + rei(−7π4)− (1 + i)|
q
|1 + i + rei(−3π4)− (2 + 2i)| ei(−7π8 )ei(−3π8 )ei(−7π4 )dr
= 1.5708 . Using Mathematica
Along 2 + 2i−→ 1 + i (z ∈ γ+ ∗)
z = 1 + i + rei(−7π4), r :√
2 −→ 0 =⇒ dz = ei(−7π4 )dr
arg (z − (1 + i)) = −7
4π =⇒p
z − (1 + i) = (−1) · mathp
z − (1 + i) arg (z − (2 + 2i)) = −3
4π =⇒ p
z − (2 + 2i) = mathp
z − (2 + 2i)
Z
γ
f (z) dz = 2 Z
γ∗
f (z) dz
= 2 · (−1) · math
Z 0
√ 2
q
1 + i + rei(−7π4 )− (1 + i) q
1 + i + rei(−3π4 )− (2 + 2i) ei(−7π4)dr
= 1.5708 .
Example 22. Suppose that f (z) =p
z − (1 + i)p
z − (2 + 2i)p
z − (3 + 3i)p
z − (4 + 4i)p
z − (5 + 5i)
=
5
Y
k=1
pz − (k + ki).
Let a1, a2 be two a − cycles and let b1, b2 be two b − cycles drawing in Figure 92. Evaluate the four integrals R
akf (z)dz and R
bkf (z)dz, k = 1, 2 using the method of deformation of path.
Figure 92 Solution.
Figure 93 1. EvaluateR
a1f (z) dz Theoretical Evaluation Along 2 + 2i−→ 1 + i,+
z = 1 + i + rei(−7π4), r :√
2 −→ 0 =⇒ dz = ei(−7π4 )dr For k = 1,
arg(z − (1 + i)) = −7
4π =⇒ p
z − (1 + i) = q
|1 + i + rei(−7π4 )− (1 + i)| ei(−7π8). For k = 2, 3, 4, 5,
arg(z − (k + ki)) = −3
4π =⇒ p
z − (k + ki) = q
|1 + i + rei(−7π4)− (k + ki)| ei(−3π8 )
Then,
2. EvaluateR
a2f (z) dz
Then, Z
a2
f (z) dz = 2 Z
4+4i−→3+3i+
f (z) dz
= 2 Z 0
√ 2
5
Y
k=1
q
|3 + 3i + rei(−7π4)− (k + ki)|
!
ei(−7π8 )3
ei(−3π8 )2
ei(−7π4) dr
= 5.69991 − 2.36098i . Using Mathematica
Along 4 + 4i−→ 3 + 3i,+
z = 3 + 3i + rei(−7π4 ), r : √
2 −→ 0 =⇒ dz = ei(−7π4)dr For k = 1, 2, 3,
arg(z−(k+ki)) = −7
4π =⇒p
z − (k + ki) = (−1)·math
q
3 + 3i + rei(−7π4)− (k + ki)
For k = 4, 5,
arg(z − (k + ki)) = −3
4π =⇒p
z − (k + ki) = math
q
3 + 3i + rei(−7π4 )− (k + ki)
Then, Z
a2
f (z) dz = 2 Z
4+4i−→3+3i+
f (z) dz
= 2 · (−1)2· math Z 0
√2 5
Y
k=1
q
|3 + 3i + rei(−7π4 )− (k + ki)|
!
ei(−7π4) dr
!
= 5.69991 − 2.36098i .
3. EvaluateR
Thus,
Thus,
Z
b1
f (z) dz = 2 Z
5+5i−→4+4i
f (z) dz + 2 Z
3+3i−→2+2i
f (z) dz
= −4.99015 − 12.0473i . 4. EvaluateR
b2f (z) dz : We have done in 3.
Z
b2
f (z) dz = 2 Z
5+5i−→4+4i
f (z) dz = −3.67557 − 8.8736i . Example 23. Let
f (z) =p
(z − z1)(z − z2) · · · (z − z2N +1) =
2N +1
Y
j=1
pz − zj
and
g(z) =p
(z − z1)(z − z2) · · · (z − z2N +1) =
2N +2
Y
j=1
pz − zj
where zj is of the form zj = rjei(α−2π). That is, these zj are lies on a slant cut of angle α. The cuts of f and g are drwan in Figure 95 and Figure 96, respectively.
Figure 95 Figure 96
Let Ic= [α − 2π, −π]. To Apply the similar method using in example 13 and example 14 for vertical cuts, we are able to easily determine the sign-regions for f and g. The sign-regions for f and for g are all R2, R4, · · · , R2N +2 which are shown in Figure 95 and Figure 96, respectively.