The above statement and result all in horizontal cut, but the way is the same in any cut. Take w2 = a(z − z1)(z − z2)(z − z3), where z1, z2, z3 are distinct for example. We let f (z) =p(z − z1)p(z − z2)p(z − z3) and discuss f (z). Cause√
a does not influence the cuts. For f (z) the factor p(z − zk) change the sign when when arg (z − zk) changes by 2π. We cut complex plane from z1 to z2 and from z3 to ∞. Label left of cut with (+)edge and right of cut with (−)edge
Figure 31: The cut-plane and a, b cycle in sheets
Using similarly way to construct the Riemann Surface. Image this two sheets are made of rubber, and together the (+)edges of sheet-I with the (−)edges of sheet-II. We
get correspond Riemann Surface R1. The curve a,b correspond to the meridian curve a and latitude curve b on Riemann Surface R1, respectively.
Figure 32: Corresponding Riemann Surface
For arbitrary cut, if f(z) has 2N-1 or 2N roots, then 1. There are N cuts in complex plane.
2. It’s geometric graph has N-1 holes, that is construct corresponding Riemann Surface of genus N-1.
3. There are N-1 a-cycles and N-1 b-cycles.
2 The integrals of
f (z)1over a,b cycles for horizontal cut
We will use Mathematica help us to obtain the values of integrals of f (z)1 over a,b cycles. First, discuss the values in sheet-I, sheet-II and Mathematica for horizontal cuts.
f (z) = s n
Q
k=1
(z − zk), using polar form
n
Q
k=1
(z − zk) = reθi. Let θ1 denotes θ in sheet-I and θ2 denotes in sheet-II. So
θ2 = θ1+ 2π We have
f (z)|(II)=√ reθ22 i
=√
reθ1+2π2 i
=√
reθ12 ieπi
= −√
reθ12 i = −f (z)|(I) (4)
where f (z)|(I) denote the value of f (z) with z in sheet-I and f (z)|(II) means z in sheet-II.
Because the difference of argument between z in sheet-I and in sheet-II is 2π , that is the difference between f (z)|(I) and f (z)|(II) is π . So f (z)|(I) = −f (z)|(II).
Now discuss the difference in sheet-I of theory and in Mathematica. First,√
−1. From the definition of argument in sheet-I, √
−1 = −i, but we compute √
−1 in Mathematica obtain√
−1M ath.= i. Why? We found that θ ∈ (−π, π] of reiθ in Mathematica, actually.
Figure 33: Domain and range in Mathematica
For any other θ of reiθ which does not belong to (−π, π], Mathematica will conversion reiθ into reiθ∗, θ∗ ∈ (−π, π] where reiθ = reiθ∗.
Compare the value of f (z) with z in sheet-I and in Mathematica, we discover that Lemma 1. If
n
Q
k=1
(z − zk) = reiθ in sheet-I for horizontal cut
f (z)|(I) = f (z)|M athematica if θ ∈ (−π, π),
−f (z)|M athematica if θ = −π Proof.
Since−π does not in (−π, π], Mathematica will conversion re−πi into reπi and re−πi= reπi but f (z) will different.
In theory: − 1 = e−πi
√
→√
−1 = e−πi2 = −i.
In Mathematica: − 1 = e−πi M ath.= eπi
√
→√
−1 = eπi2 = i So f (z)M ath.= −f (z) if θ = −π in Mathematica.
In hole paper, f (z) M ath.= −f (z) denotes the polynomial f (z) in front of M ath.= is the value of f (z) in theory and the polynomial f(z) behind the M ath.= is the value of f (z) in Mathematica.
After we known the state above, we must modify the computation when we want to use Mathematica to calculate the value. Take example to explain: evaluate R
r 1
f (z)dz where f (z) =pz(z − 1)(z − 2), z ∈ R and r = r1S r2 where r1 = the path on a horizontal cut from 1 to 2 with (+)edge of sheet-I and r2 = the path on a horizontal cut from 2 to 1 with (−)edge of sheet-I.
Figure 34: cuts in complex plane of f (z) =pz(z − 1)(z − 2)
Analysis the integrals: f (z) =pz(z − 1)(z − 2) =√ z√
z − 1√ z − 2
1. If z ∈ r1.
Compare (1) and (2), we found there a difference of a minus sign with the value in sheet-I and in Mathematica.
2. z ∈ r2
Clearly, there is a mistake when θ = −π. When we use Mathematica to get the value of integration we want, we need modify some range where the value will wrong.
Determine the difference of sign(f) (same or negative) and then modify the computation of Mathematica to get right value. Because sometimes the form of integration is complex, if we could simplify the way about modify the difference of sigh(f), it will help us to get right value easier.
Example: Same f(z) as the example before, using lemma1 to modify.
1. If z ∈ r1, z : 1 → 2
Take another example to consider how to modify computation in Mathematica such that numeral result is right for horizontal cuts. And discuss the difference between the value in theory and in Mathematica.
Example 2 : Evaluate R 1
f (z)dz over a1, a2, a3, b1, b2 and b3 cycles. where
f (z) = p(z + 3)(z + 1)(z − 1)(z − 3)(z − 4)(z − 6)(z − 9) . We analysis the integral in
Figure 35: a-cycles and their equivalent path a∗
Mathematica and in theory to compare the result and using the result of angle to modify the computation to get value. Let z1 = 9, z2 = 6, z3 = 4, z4 = 3, z5 = 1, z6 = −1, z7 = −3
Solution:
1. Let a1 is a cycle center at 152 with radius 2 and enclosed the cut [6, 9]. So let z = 152 + 2eiθ, we have
Z
a1
1
f (z)dz = Z π
−π
2ieiθ
7
Q
k=1
q15
2 + 2eiθ− zk dθ
= 1.0842 × 10−19+ 0.0776642i
By Cauchy Theorem. Since ak cycle is simple connected, we can use some equivalent paths, say a∗k, to easily compute the integrals for ak cycle.
(1) If z ∈ a∗1 of theory in sheet-I where a∗1 = 6 → 9+ S 6← 9−
(a) 6→ 9 : the path along x-axis from 6 to 9 on (+)edge of sheet-I.+ z − 9 = −|z − 9| = |z − 9|e−πi then 1
√z − 9 = |z − 9|−12eπ2i = |z − 9|−12i z − zk = |z − zk| then 1
√z − zk = |z − zk|−12, k = 2, 3, 4, 5, 6, 7
Z
6→9+
1
f (z)dz = Z 9
6
i
7
Y
k=1
|z − zk|−12dz
(b) 6← 9 : the path along x-axis from 9 to 6 on sheet-I with (−)edge.− z − 9 = −|z − 9| = |z − 9|eπi then 1
√z − 9 = |z − 9|−12e−π2i = −i|z − 9|−12 z − zk = |z − zk| then 1
√z − zk
= |z − zk|−12, k = 2, 3, 4, 5, 6, 7
Z
(2) Analysis the integral over a∗1 in Mathematica
(a) 6→ 9 : the path along x-axis from 6 to 9 on sheet-I with (+)edge+ A difference of a minus sign with in sheet-I
(b) 6← 9 : the path along x-axis from 9 to 6 of sheet-I with (−)edge− as same as in sheet-I.
But in Mathematica
Z
a∗1
1
f (z)dz = 0 (3) Using the results before to modify
(a) 6→ 9: the path along x-axis from 6 to 9 on sheet-I with (+)edge+
(b) 6← 9: the path along x-axis from 9 to 6 on sheet-I with (−)edge− arg(z − z1) = π then √
z − z1 M ath.= √ z − z1 arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk, k = 2, . . . , 7 So f (z)M ath.= f (z)
We have
Z
a∗1
1
f (z)dz M ath.= −2 Z 9
6
1 f (z)dz
= 0.0776642i
2. Let a2is a cycle center at 72 with radius 1 and enclosed the cut [3, 4]. So let z = 72+eiθ, we have
Z
a2
1
f (z)dz = Z π
−π
ieiθ
7
Q
k=1
q7
2 + eiθ − zk dθ
= 0. − 0.200969i
Same as a1 , by Cauchy Theorem to compute equivalent path a∗2 where a∗2 = 3 →+ 4S 3← 4−
(1) Analysis the integral of a∗2 in sheet-I
(a) 3→ 4: the path along x-axis from 3 to 4 on sheet-I with (+)edge+ z − zk= −|z − zk| = |z − zk|e−πi
then 1
√z − zk = |z − zk|−12eπ2i = |z − zk|−12i, k = 1, 2, 3 z − zk= |z − zk| then 1
√z − zk = |z − zk|−12, k = 4, 5, 6, 7 Z
3→4+
1
f (z)dz = Z 4
3
i3
7
Y
k=1
|z − zk|−12dz
(b) 3← 4 : the path along x-axis from 4 to 3 on sheet-I with (−)edge− z − zk = −|z − zk| = |z − zk|eπi
then 1
√z − zk = |z − zk|−12e−π2i = −i|z − zk|−12, k = 1, 2, 3 z − zk = |z − zk| then 1
√z − zk
= |z − zk|−12, k = 4, 5, 6, 7
Z
(2) Analysis the integral of a∗2 in Mathematica
(a) 3→ 4 : the path along x-axis from 6 to 9 on sheet-I with (+)edge+ But by (a),(b) we obtain different value in Mathematica
Z
a∗2
1
f (z)dz = 0 (3) Using lemma 1 to modify
(a) 3→ 4 : the path along x-axis from 3 to 4 on sheet-I with (+)edge+
(b) 3← 4 : the path along x-axis from 4 to 3 on sheet-I with (−)edge− arg(z − zk) = π then √
z − zk M ath.= √
z − zk , k = 1, 2, 3 arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk , k = 4, 5, 6, 7 So f (z)M ath.= f (z)
We have
Z
a∗3
1
f (z)dz M ath.= −2 Z 4
3
1
f (z)dz = 0. − 0.200969i
3. a3 : Let a3 is a cycle center at 0 with radius 2 and enclosed the cut [−1, 1]. So let z = 2eiθ, we have
Z
a3
1
f (z)dz = Z π
−π
2ieiθ
7
Q
k=1
√2eiθ− zk dθ
= 3.46945 × 10−18+ 0.151409i
Same as a1 , by Cauchy Theorem to compute equivalent path a∗3 where a∗3 = −1→+ 1S 1← −1−
(1) Analysis of a∗3 in theory
(a) −1→ 1 : the path along x-axis from −1 to 1 on sheet-I with (+)edge+
z − zk = −|z − zk| = |z − zk|e−πi
then 1
√z − zk = |z − zk|−12eπ2i = i|z − zk|−12, k = 1, 2, 3, 4, 5 z − zk = |z − zk| then 1
√z − zk = |z − zk|−12, k = 6, 7
Z
−1→1+
1
f (z)dz = Z 1
−1
i5
7
Y
k=1
|z − zk|−12dz
(b) −1← 1 : the path along x-axis from 1 to -1 on (−)edge of sheet-I−
z − zk= −|z − zk| = |z − zk|eπi
then 1
√z − zk = |z − zk|−12e−πi2 = −i|z − zk|−12, k = 1, 2, 3, 4, 5 z − zk= |z − zk| then 1
√z − zk = |z − zk|−12, k = 6, 7
Z
(2) Consider a∗3 in Mathematica (a) −1→ 1 :+ But we obtain different value in Mathematica
Z
a∗3
1
f (z)dz = 0 (3) Using lemma 1 to modify
(a) −1→ 1 : the path along x-axis from −1 to 1 on sheet-I with (+)edge+ arg(z − zk) = −π then √
z − zk M ath.= −√
z − zk, k = 1, 2, 3, 4, 5 arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk, k = 6, 7 So f (z)M ath.= −f (z)
(b) 1← −1 : the path along x-axis from 1 to -1 on sheet-I with (−)edge− arg(z − zk) = π then √
z − zk M ath.
= √
z − zk, k = 1, 2, 3, 4, 5 arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk, k = 6, 7 So f (z)M ath.= f (z)
Z
a3
1
f (z)dz = Z
a∗3
1 f (z)dz
M ath.
= −2
Z 1
−1
1 f (z)dz
= 0.151409i
Figure 36: b-cycles
4. b3 : Let b3 is a cycle which center at −2 with radius 2. We could write down the parameter, let z = −2 + 2eiθ and θ ∈ [−π, 0)S[2π, 3π). Notice that f (z)|(II) =
−f (z)|(I), so we have Z
b3
1
f (z)dz = Z 0
−π
2ieiθ
7
Q
k=1
√−2 + 2eiθ− zk dθ −
Z π 0
2ieiθ
7
Q
k=1
√−2 + 2eiθ− zk dθ
= −0.0765026 + 6.93889 × 10−18i
Since bk cycle is simple connected, we can use some equivalent paths, say b∗k, such that bk ≈ b∗k to easily compute the integrals for bk cycle. Here b3 ≈ b∗3.
Figure 37: b3’s equivalent path b∗3
(1) Consider b∗3 of theory in sheet-I (a) −3 → −1
z + z3 = |z + z3| then 1
√z + z3 = |z + z3|−12 z − zk = −|z − zk| = |z − zk|e−πi
then 1
√z − zk = |z − zk|−12eπ2i = |z − zk|−12i, k = 1, 2, 3, 4, 5, 6
Z
−3→−1
1
f (z)dz = − Z −1
−3
i6
7
Y
k=1
|z − zk|−12dz
(b) −1 99K −3 : the path along x-axis from -1 to -3 of sheet-II. We known that f (z)|(I) = −f (z)|(II) , so consider −1 → −3
z + z3 = |z + z3| then 1
√z + z3 = |z + z3|−12 z − zk = −|z − zk| = |z − zk|e−πi
then 1
√z − zk = |z − zk|−12eπ2i = |z − zk|−12i, k = 1, 2, 3, 4, 5, 6
Z
−3L99−1
1
f (z)dz = − Z
−3←−1
1
f (z)dz = Z −3
−1
i6
7
Y
k=1
|z − zk|−12dz By (1), (2), we obtain
Z
b∗3
1
f (z)dz = 2 Z −3
−1
i6
7
Y
k=1
|z − zk|−12dz
= −2 Z −3
−1 7
Y
k=1
|z − zk|−12dz
= −0.0765026
(2) Consider b∗3 in Mathematica
(a) −3 → −1 (3) Using Lemma1 to modify
(a) −3 → −1 : the path along x-axis from -3 to -1 of sheet-I
Z
Figure 38: The equivalent path b∗2
(1) Consider b∗2 of theory in sheet-I (a) −1→ 1 :+
(b) −1L99 1 ≡ −1− ← 1 i.e. the path on horizontal cut from -1 to 1 on (−)edge+ in sheet-II equals the path on horizontal cut from −1 to 1 of (+)edge in sheet-I. So consider z ∈ −1← 1+
(2) Analysis integral over b∗2 in Mathematica
(a) −1→ 1+
But in Mathematica we obtain different value Z
b∗2
1
f (z)dz = 0 (3) Using Lemma1 to modify
(a) −1→ 1+ (−)edge in sheet-II is equal the path on horizontal cut from −1 to 1 of (+)edge in sheet-I. So consider z ∈ −1← 1+ By (1), (2), (3) and Cauchy Integral Theorem
Z
6. b1: Let b1is a cycle center at 3 with radius 3 . So we could write down the parameter, let z = 3 + 3eiθ and θ ∈ [−π, 0)S[2π, 3π). Notice that f (z)|(II) = −f (z)|(I), so we have
Z
b1
1
f (z)dz = Z 0
−π
3ieiθ
7
Q
k=1
√3 + 3eiθ− zk dθ −
Z π 0
3ieiθ
7
Q
k=1
√3 + 3eiθ− zk dθ
= 0.0565161
Consider equivalent path b∗1 = b∗2S b∗3S 3→ 4+ S 3L99 4− S 4 → 6 S 4 L99 6
Figure 39: The equivalent path b∗1
(1) Analysis the integration over b∗1 in sheet-I (a) 3→ 4:+
z − zk = −|z − zk| = |z − zk|e−πi then 1
√z − zk = |z − zk|−12eπ2i = |z − zk|−12i, k = 1, 2, 3 z − zk = |z − zk| then 1
√z − zk = |z − zk|−12, k = 4, 5, 6, 7
Z
3→4+
1
f (z)dz = Z 3
4
i3
7
Y
k=1
|z − zk|−12dz
(b) 3L99 4 ≡ 3− ← 4 that is the path on horizontal cut from 3 to 4 of (−)edge+ in sheet-II is equal to the path on horizontal cut from 3 to 4 of (+)edge in sheet-I. So we consider z ∈ −1← 1+
z − zk = −|z − zk| = |z − zk|e−πi then 1
√z − zk = |z − zk|−12eπ2i = |z − zk|−12i, k = 1, 2, 3 z − zk = |z − zk| then 1
√z − zk
= |z − zk|−12, k = 4, 5, 6, 7
Z
(2) Consider b∗1 in Mathematica (a) 3→ 4+
(b) 3L99 4−
(3) Using Lemma1 to modify (a) 3→ 4+ sheet-II is equal to the path on horizontal cut from 3 to 4 of (+)edge in
sheet-I. So consider z ∈ 3← 4.+ By (1), (2), (3) and Cauchy Integral Theorem
Z Discuss in general situation:
ComputeR 1
f (z)dz over a,b cycles for horizontal cut where f (z) = s m
Figure 40: a-cycles for 2N-1 points
Figure 41: a-cycles for 2N points
There are N cuts (N-1 holes), we give that aj is a cycle center at x with radius r enclosed [z2j, z2j−1] and doesn’t intersect with other cuts.
If z ∈ aj, let z = x + reiθ where θ ∈ [−π, π) Z
aj
1
f (z)dz = Z
aj
1 sm
Q
k=1
(z − zk)
dz (5)
= Z π
−π
rieiθ
m
Q
k=1
√x + reiθ− zk
dθ (6)
2. Consider R
a∗j 1
f (z)dz where a∗j is an equivalent path for aj and it’s from z2j to z2j−1 in (+)edge and then from z2j−1 to z2j in (−)edge.
Figure 42: a∗-cycles for 2N-1 points
Figure 43: a∗-cycles for 2N points
By Cauchy theorem, we can get that Z
aj
1 f (z)dz =
Z
a∗j
1
f (z)dz (7)
Similarly, we use some ideas of complex number to analysis the integrations, first.
(1) z2j → z+ 2j−1: That is consider the path from z2j to z2j−1 on (+)edge
(a) Analysis in theory
We can use this with Mathematica to get value, and we compare with the result below to know the difference.
(b) Analysis in Mathematica:
z − zk > 0 then arg(z − zk) = 0, k = 2j, 2j + 1, ..., m
We can find that the difference of value between theory in sheet-I and Mathe-matica. is a minus and it must be pure imaginary number.
(2) z2j ← z− 2j−1: That is consider the path from z2j−1 to z2j in (−) edge Same as above
(a) In theory (b) In Mathematica, same as above
But if we can modify the computation it will more quick and easier.
(3) Using Lemma 1 to modify the computation.
(a) z2j
3. b-cycles
Figure 44: bj-cycle for 2N-1 points
Give bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and intersect at the points on [z2j, z2j−1] and [z2N −1, z2N].
Figure 45: bj-cycle for 2N points
Give bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and intersect at the points on [z2j, z2j−1] and [z2N −1, ∞). If z ∈ bj, z = x + reiθ where θ ∈ [−π, 0)S[2π, 3π). From
f (z)|(II)= −f (z)|(I)
Z
bj
1
f (z)dz = Z 0
−π
rieiθ
m
Q
k=1
√x + reiθ− zk dθ +
Z 3π 2π
rieiθ
m
Q
k=1
√x + reiθ− zk
dθ (10)
= Z 0
−π
rieiθ
m
Q
k=1
√x + reiθ− zk dθ −
Z π 0
rieiθ
m
Q
k=1
√x + reiθ− zk
dθ (11)
4. The equivalent path b∗j :
Figure 46: b∗j-cycle for 2N-1 points
Figure 47: b∗j-cycle for 2N points From Cauchy Theorem, we have
Z
bj
1 f (z)dz =
Z
b∗j
1
f (z)dz (12)
where b∗j is a path from zm to z2j in sheet-I and then from z2j to zm in sheet-II (1) the path on the cut that is the path from z2s+2 to z2s+1, s = j, j + 1, ..., N − 2
on (+)edge in sheet-I and the path from z2s+1 to z2s+2, s = j, j + 1, · · · , N − 2 on (−)edge in sheet-II.
(a) In theory
(i) z2s+2 → z+ 2s+1:
z − zk> 0 then arg(z − zk) = 0 then arg(√
z − zk) = 0 then √
z − zk= |z − zk|12, k = 2s + 2, 2s + 3, · · · , m z − zk< 0 then arg(z − zk) = −π then arg(√
z − zk) = −π2 then √
z − zk= |z − zk|12e−π2i = −i|z − zk|12, k = 1, 2, · · · , 2s + 1 So we have
f (z) = i2s+1
m
Y
k=1
|z − zk|−12
(ii) z2s+2 L99 z− 2s+1 in (−) edge of sheet-II is as same as in (+)edge of sheet-I, so consider z2s+2 → z+ 2s+1.
z − zk> 0 then arg(z − zk) = 0 ⇒ arg(√
z − zk) = 0 then √
z − zk= |z − zk|12, k = 2s + 2, 2s + 1, ..., m z − zk< 0 then arg(z − zk) = −π then arg(√
z − zk) = −π2 then √
z − zk= |z − zk|12e−π2i = −i|z − zk|12, k = 1, 2, ..., 2s + 1 We found that same as above
f (z) = i2s+1
m
Y
k=1
|z − zk|−12
(b) In Mathematica:
(i) z2s+2 → z+ 2s+1
z − zk> 0 then arg(z − zk) = 0 then arg(√
z − zk) = 0 then √
z − zk= |z − zk|12 , k = 2s + 2, · · · , m z − zk< 0 then arg(z − zk) = π then arg(√
z − zk) = π2 then √
z − zk= |z − zk|12eπ2i = i|z − zk|12 , k = 1, 2, · · · , 2s + 1 We have
f (z) = (−i)2s+1
m
Y
k=1
|z − zk|−12
(ii) z2s+2
L99 z− 2s+1: z2s+2
→ z+ 2s+1.
z − zk> 0 then arg(z − zk) = 0 then arg(√
z − zk) = 0 then √
z − zk = |z − zk|12 , k = 2s + 2, 2s + 1, ..., m z − zk< 0 then arg(z − zk) = π then arg(√
z − zk) = π2 then √
z − zk = |z − zk|12eπ2i = i|z − zk|12, k = 1, 2, ..., 2s + 1 So we obtain same value of f (z) as (i), but different with in theory
f (z) = (−i)2s+1
m
Y
k=1
|z − zk|−12
(c) Using Lemma 1 to modify the computation.
(i) z ∈ z2s+2 → z+ 2s+1: arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk , k = 2s + 2, ..., m arg(z − zk) = −π then √
z − zkM ath.= −√
z − zk , k = 1, 2, ..., 2s + 1 We obtain
f (z)M ath.= (−1)2s+1f (z) = −f (z)
(ii) z2s+2 → z+ 2s+1: in (−) edge of sheet-II is as same as in (+)edge of sheet-I, z2s+2 → z+ 2s+1
arg(z − zk) = 0 then √
z − zk M ath.= √
z − zk , k = 2s + 2, ..., m arg(z − zk) = −π then √
z − zk M ath.
= −√
z − zk , k = 1, 2, ..., 2s + 1 We have
f (z)M ath.= (−1)2s+1f (z) = −f (z)
(2) In no cuts that is the path from z2s+1 to z2s, s = j, j + 1, ..., N − 2 in sheet-I
(ii) z2s+1 L99 z2s: So we have same value of f (z) as (i), but different with in theory
f (z) = (−i)2s
(c) Using Lemma 1 to modify the computation in Mathematica:
(i) If z ∈ z2s+1 → z2s:
3 The integrals of
f (z)1over a,b cycles for vertical cut
After knowing the integrals in horizontal cut, we will discuss the integrals for vertical cuts. In this case, we define that
z − zk = reiθ, θ ∈ [−3π2 ,π2) iff z in sheet-I
reiθ, θ ∈ [π2,5π2 ) iff z in sheet-II (15) the cut in each sheet has two edges, label the starting edge with ”+” and the terming edge with ”−” and zk is the end point of the vertical cut.
Analysis the value of f(z) in sheet-I and sheet-II of theory.
Example f (z) =√
z. If z = ri ⊂ sheet-I
z = |z|eiθ, θ ∈ [−3π 2 ,π
2) then √
z = |z|12eiθ2,θ
2 ∈ [−3π 4 ,π
4) (16)
If z = ri ⊂ sheet-II
z = |z|eiθ, θ ∈ [π 2,5π
2 ) then √
z = |z|12eiθ2,θ 2 ∈ [π
4,5π
4 ) (17)
Figure 48: Example of f (z) =√ z
If f (z) = s n
Q
k=1
(z − zk), then
n
Y
k=1
(z − zk) = reiθ1, θ1 ∈ [−3π 2 ,π
2) in sheet-I
n
Y
k=1
(z − zk) = reiθ2, θ2 ∈ [π 2,5π
2 ) in sheet-II From the idea of definition, reiθ1 = reiθ2 and θ2 = θ1+ 2π.
f (z)|(II)=√
reθ22 i =√
reθ1+2π2 i =√
reθ12ieπi = −f (z)|(I) (18)
Discuss the difference between the value in theory and in Mathematica and find out how to modify the computation.
Figure 49: The value in sheet-I and Mathematica of √ z
So we need to modify the computation in Mathematica s.t. the numerical result of Mathematica is identical to the numerical result of theory when θ ∈ [−3π2 , −π].
Lemma 2. When z in sheet-I for vertical cut whose one of the end points is zk
√z − zkM ath.= −√
z − zk if arg(z − zk) ∈ [−3π2 , −π],
√z − zk if arg(z − zk) ∈ (−π,π2)
Proof.
Let z in sheet-I and using polar form z − zk= reiθ. When θ ∈ (−π,π2), the argument in theory or Mathematica is the same. When θ ∈ [−3π2 , −π], Mathematica will conversion
θ into θ + 2π ∈ [π2, π] where θ + 2π ∈ [π2, π] and reθi= reθi+2πi, but
In theory: √
z − zk =√ reθ2i
In Mathematica: √
z − zk =√
reθ+2π2 i = −√ reθ2
So if θ ∈ [−3π2 , −π]
√z − zk M ath.= −√ z − zk
As same as horizontal cut. We first discuss the difference between the value in theory and the value in Mathematica. Compare their sign(f) is different or not? Using statement before about modify and get value, the result will be the same or not?
Example: The integrals of f (z)1 over a,b cycles for vertical cut where f (z) = p(z − i)(z − 2i)(z − 3i)(z − 4i)(z − 5i)(z − 6i). Let f(z) = Q6
k=1
√z − zk where zk= ki, k = 1, 2, 3, 4, 5, 6
Figure 50: a and its equivalent path a∗
1. Compute R
a∗1 1
f (z)dz where a∗1 is equivalent path for a1 and a∗1 = the path along vertical cut from i to 2i on (+)edge of sheet-I (called a∗11) and then back from 2i to i on (−)edge of sheet-I (a∗12)
(1) a∗11: Let z = ri, r : 2 → 1, dz = idr
(a) Analysis in theory:
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − i) = −32π then arg√
z − i = −34π arg(z − ki) = −π2 then arg√
z − ki = −π4, k = 2, . . . , 6
f (z) = (
6
Y
k=1
|z − ki|−12)e−34π(e−14π)5 = (
6
Y
k=1
|z − ki|−12)e−84π
=
6
Y
k=1
|z − ki|−12 = R
(Let
6
Q
k=1
|z − ki|−12 = R)
(b) Analysis in mathematica (no matter in which sheet):
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − i) = π2 then arg√
z − i = π4 arg(z − ki) = −π2 then arg√
z − ki = −π4, k=2,3,4,5,6
f (z) =
6
Y
k=1
|z − ki|−12e14π(e−π4)5 =
6
Y
k=1
|z − ki|−12e−44π
= −
6
Y
k=1
|z − ki|−12 = −R
Compare with (a) and (b), we find that when you want to get true value, the value which we get from Mathematica should multiply −1 that is sign(f (z)|(I))=-sign(f (z)|M ath.)
(c) Using the Lemma 2 to modify:
arg(z − i) = −32π than√
z − iM ath.= −√ z − i arg(z − ki) = −π2 than √
z − kiM ath.= √
z − ki, k=2, 3, 4, 5, 6
So f (z) M ath.= −f (z), the same result as above difference between theory and Mathematica, the difference is a minus.
(2) a∗12 : Let z = ri, r : 1 → 2 and then dz = idr (a) Analysis in theory:
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − i) = π2 than arg(√
(b) Analysis in Mathematica (no matter in which sheet):
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:
arg(z − i) = π2 than √ vertical cut from 4i to 3i on (+)edge of sheet-I (called a∗21) and then back from 3i to 4i on (−)edge of sheet-I (a∗22)
(1) a∗21 : Let z = ri, r : 4 → 3, dz = idr (a) Analysis in theory:
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = −32π than arg(√
z − ki) = −34π, k = 1, 2, 3 arg(z − ki) = −π2 than arg(√
z − ki) = −π4, k = 4, 5, 6
f (z) =
6
Y
k=1
|z − ki|−12(e−34π)3(e−π4)3 =
6
Y
k=1
|z − ki|−12e−124π
= −
6
Y
k=1
|z − ki|−12
(b) analysis in Mathematica (no matter in which sheet):
Since z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = π2 than arg(√
z − i) = π4, k=1,2,3 arg(z − ki) = −π2 than arg(√
z − ki) = −π4, k=4,5,6
f (z) =
6
Y
k=1
|z − ki|−12(eπ4)3(e−π4)3 =
6
Y
k=1
|z − ki|−12
Compare with (a) and (b) we find that when you want to obtain true value, the value which we have from Mathematica should multiply −1, sign(f (z)|(I)) = −sign(f (z)|M athewatica)
(c) Using Lemma 2 to modify:
arg(z − i) = −32π than√
z − iM ath.= −√
z − i , k = 1, 2, 3 arg(z − ki) = −π2 than √
z − kiM ath.= √
z − ki , k = 4, 5, 6
f (z) M ath.= −f (z)
same as the above result (2) a∗22 : Let z = ri, r : 1 → 2, dz = idr
(a) Analysis in theory:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = π2 than arg(√
z − ki) = π4, k=1,2,3
arg(z − ki) = −π2 than arg(√
(b) Analysis in Mathematica (no matter in which sheet):
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:
arg(z − i) = π2 then √
same as the above result
Z
Figure 51: b and b∗ (a) Analysis in theory:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = −32π then arg(√
z − ki) = −34π, k=1,2,3,4 arg(z − ki) = −π2 then arg(√
z − ki) = −π4, k=5,6
f (z) =
6
Y
k=1
|z − ki|−12(e−34π)4(e−π4)2 =
6
Y
k=1
|z − ki|−12e−144π
= i
6
Y
k=1
|z − ki|−12
(b) Analysis of Mathematica (no matter in which sheet):
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = π2 then arg(√
z − ki) = π4, k=1,2,3,4 arg(z − ki) = −π2 then arg(√
z − ki) = −π4, k=5,6
f (z) =
6
Y
k=1
|z − ki|−12(e−π4)2(eπ4)4 =
6
Y
k=1
|z − ki|−12e24π
= i
6
Y
k=1
|z − ki|−12
Compare with (a) and (b) we find the value is same.
(c) Using Lemma 2 to modify:
arg(z − i) = π2 then √
z − iM ath.= √
z − i, k=1, 2, 3, 4 arg(z − ki) = −π2 then √
z − kiM ath.= √
z − ki, k=5, 6
f (z) M ath.= f (z) same as the above result
(2) b∗22: We known that f (z)|(I) = −f (z)|(II) , so we can consider b∗∗22 is the path along vertical line from 4i to 5i on sheet-I.
Let z = ri, r : 4 → 5 then dz = idr (a) Analysis in theory:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = −32π then arg(√
z − ki) = −34π, k=1,2,3,4 arg(z − ki) = −12π then arg(√
z − ki) = −14π, k=5,6 f (z) =
6
Y
k=1
|z − ki|−12(e−34π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e−144π
= i
6
Y
k=1
|z − ki|−12 So
f (z)|b∗22 = −f (z)|b∗∗22 = −i
6
Y
k=1
|z − ki|−12 (b) Analysis in Mathematica (no matter in which sheet):
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = 12π then arg(√
z − ki) = 14π, k=1,2,3,4 arg(z − ki) = −12π then arg(√
z − ki) = −14π, k=5,6 f (z) =
6
Y
k=1
|z − ki|−12(e14π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e24π
= i
6
Y
k=1
|z − ki|−12
Compare with (a) and (b) we find when you want get true value, the value which we obtain from Mathematica should multiply −1
(c) Using Lemma 2 to modify: same as the above result
Z path along vertical cut from 3i to 4i on (−)edge of sheet-II, b∗13 = the path along vertical line from 3i to 2i on sheet-I, b∗14= the path along vertical line from 2i to 3i on sheet-II.
(b) In Mathematica (no matter in which sheet):
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = 12π then arg(√
z − ki) = 14π, k=1,2 z − i = −|z − ki|i arg(z − ki) = −12π then arg(√
z − ki) = −14π, k=3,4,5,6
f (z) =
6
Y
k=1
|z − ki|−12(e−14π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e−24π
= −i
6
Y
k=1
|z − ki|−12
Compare with (a) and (b) we find that the value is same (c) Using Lemma 2 to modify:
arg(z − ki) = 12π then √
z − kiM ath.= √
z − ki, k=1,2 arg(z − ki) = −12π then √
z − kiM ath.= √
z − ki, k=3,4,5,6
f (z) M ath.= f (z)
same as the above result
(3) b∗13: Let z = ri, r : 3 → 2, so dz = idr (a) In theory:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = −32π then arg(√
z − ki) = −34π, k=1,2 arg(z − ki) = −12π then arg(√
z − ki) = −14π, k=3,4,5,6
f (z) =
6
Y
k=1
|z − ki|−12(e−34π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e−104π
= −i
6
Y
k=1
|z − ki|−12
(b) In Mathematica:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = 12π then arg(√
z − ki) = 14π, k=1,2
arg(z − ki) = −12π then arg(√
z − ki) = −14π, k=3,4,5,6
f (z) =
6
Y
k=1
|z − ki|−12(e−14π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e−24π
= −i
6
Y
k=1
|z − ki|−12
Compare with (a) and (b) we find the value is same (c) Using Lemma 2 to modify:
arg(z − i) = 12π then√
z − iM ath.= √
z − i, k=1,2 arg(z − ki) = −12π then √
z − kiM ath.= √
z − ki, k=3,4,5,6
f (z) M ath.= f (z)
same as the above result
(4) b∗14 : We known that f (z)|(I) = −f (z)|(II) , so we can consider b∗∗14= the path along vertical line from 2i to 3i on sheet-I.
Let z = ri, r : 2 → 3, so dz = idr (a) Analysis in theory:
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = 12π then arg(√
z − ki) = 14π, k=1,2 arg(z − ki) = −12π then arg(√
z − ki) = −14π , k=3,4,5,6
f (z) =
6
Y
k=1
|z − ki|−12(e14π)4(e−14π)2 =
6
Y
k=1
|z − ki|−12e−104π
= −i
6
Y
k=1
|z − ki|−12
So
f (z)|b∗
14 = −f (z)|b∗∗
14 = i
6
Y
k=1
|z − ki|−12
(b) In Mathematica (no matter in which sheet):
Using polar form z − ki = |z − ki|earg(z−ki)i, consider arg(z − ki).
arg(z − ki) = 12π then arg(√
Compare with (a) and (b) we find when you want get true value, the value which we got from Mathematica should multiply -1
(c) Using Lemma 2 to modify:
arg(z − i) = 12π then√ same as the above result
By (1), (2), (3), (4) and Cauchy Integral Theorem Z
We find that when compute the integral by Mathematica, the difference of integral form between theory and Mathematica can help us know how to modify. We also can use angle to decide how to modify. It will the same so we only use angle to decide how to modify below.
In general situation: Compute R 1
f (z)dz over a∗, b∗ for vertical cut where f (z) = sm
Q
k=1
(z − zk) and zk = aki, ak ∈ R.
Figure 52: a-cycle and their equivalent path a∗
1. a-cycles: aj cycle is a cycle center at x with radius r enclosed [z2j, z2j−1] and doesn’t intersect with other cuts. R
aj
1
f (z)dz = R
a∗j 1
f (z)dz in sheet-I. The equivalent path a∗j =the path on a vertical cut from z2j to z2j−1 with (+)edge of sheet-I and then from z2j−1 to z2j with (−)edge of sheet-I.
Using lemma to modify:
(a) z ∈ z2j → z+ 2j−1: Let z = ri, r : Im(z2j) → Im(z2j−1), so dz = idr arg(z − zk) = −3π2 then √
z − zk M ath.
= −√
z − zk, k = 1, 2, 3, ...2j − 1 arg(z − zk) = −π2 then √
z − zkM ath.= √
z − zk, k = 2j, 2j + 1, ..., m
f (z) M ath.= (−1)2j−1f (z)M.= −f (z)
(b) z ∈ z2j−1 → z− 2j : Let z = ri, r = Im(z2j) → Im(z2j−1), so dz = idr arg(z − zk) = −π2 then √
z − zk M ath.
= √
z − zk,k=1,2,3,...2j-1 arg(z − zk) = −π2 then √
z − zk M ath.
= √
z − zk, k = 2j, 2j + 1, ..., m
f (z)M ath.= f (z)
By (a),(b) and Cauchy Theory:
Z
aj
1
f (z)dz = Z
a∗j
1 f (z)dz
M ath.
= 2
Z Im(z2j) Im(z2j−1)
m
Y
k=1
√ 1
z − zkidr (19)
2. b-cycles :
Figure 53: bj and b∗j of 2N − 1 points and 2N points
bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and intersect at the points on [z2j, z2j−1] and [z2N −1, z2N] or bj is a circle centered at x with radius r and enclosed the [z2N −1, z2j] and intersect at the points on [z2j, z2j−1] and [z2N −1, ∞) By Cauchy Theorem, we know that
Z
bj
1 f (z)dz =
Z
b∗j
1
f (z)dz (20)
Equivalent path b∗j is a path from zm to z2j in sheet-I and then from z2j to zm in sheet-II
(1) the path in cut i.e. the path from z2s+2 to z2s+1, s = j, j + 1, ..., N − 2 on (+)edge in sheet-I and path from z2s+1 to z2s+2, s = j, j + 1, · · · , N − 2 on (−)edge in sheet-II.
(a) z2s+2 → z+ 2s+1 arg(z − zk) = −π
2 then √ z − zk
M ath.
= √
z − zk, k = 2s + 2, ..., m arg(z − zk) = −3
2π then√ z − zk
M ath.
= −√
z − zk, k = 1, 2, ..., 2s + 1 So
f (z)M ath.= (−1)2s+1f (z)M ath.= −f (z) (b) if z2s+2 → z− 2s+1
arg(z − zk) = −π2 then √
z − zkM ath.= √
z − zk, k = 2s + 2, ..., m arg(z − zk) = 12π then√
z − zk M ath.= √
z − zk, k = 1, 2, ..., 2s + 1 So
f (z)M ath.= f (z)
(2) In no cuts that is the path from z2s+1 to z2s, s = j, j + 1, ..., N − 2 in sheet-I and the path from z2s to z2s+1, s = j, j + 1, ..., N − 2 in sheet-II.
(a) z ∈ z2s+1 → z2s arg(z − zk) = −π
2 then√
z − zk M ath.= √
z − zk, k = 2s + 1, ..., m arg(z − zk) = −3π
2 then √
z − zk M ath.= −√
z − zk, k = 1, 2, ..., 2s
So
f (z)M ath.= (−1)2sf (z)M ath.= f (z)
(b) If z2s+1 L99 z2s: Since f (z)|(II)= −f (z)|(I), we consider z2s+1 ← z2s arg(z − zk) = −π
2 then√
z − zk M ath.= √
z − zk, k = 2s + 1, ..., m arg(z − zk) = −3π
2 then √
z − zk M ath.= −√
z − zk, k = 1, 2, ..., 2s
So
z ∈ z2s+1 L99 z2s then f (z)M ath.= −(−1)2sf (z)M ath.= −f (z) From (1), (2) we have
Z
b∗j
1
f (z)dz = Z
b∗j
1 f (z)dz
M ath.
=
N −1
X
s=j
(2 Z z2s
z2s+1
1
f (z)dz) (21)
When we want to modify the computation of f (z) and f (z) has m roots. We needs to consider √
z − zk, k=1,...,m. There are m steps of modifying the computation, if m is large it will become troublesome. Here we provide a way to reduce the step. We can divided domain R into many areas to discuss the way to modify on vertical cuts We call it the area to modify. Take a example to explain f (z) =p(z − i)(z − 2i) =√
z − i√ z − 2i
Figure 54: 6 blocks of domain
Only discuss in sheet-I and divided it into six blocks (A),(B),(C),(D),(E) and (F) where
(A) = {z = x + yi : x < 0, 2 ≤ y}, (B) = {z = x + yi : x < 0, 1 ≤ y < 2}, (C) = {z = x + yi : x < 0, y < 1}, (D) = {z = x + yi : x > 0, 2 ≤ y}, (E) = {z = x + yi : x > 0, 1 ≤ y < 2}, (F) = {z = x + yi : x > 0, y < 1}
1. z ∈ (A)
arg(z − i), arg(z − 2i) ∈ (−3π2 , −π) (where we need to modify) So √
z − iM ath.= −√
z − i and √
z − 2iM ath.= −√ z − 2i f (z)M ath.= f (z)
2. z ∈ (B)
arg(z − i) ∈ (−3π2 , −π) then √
z − iM ath.= −√ z − i arg(z − 2i) ∈ (−π, −π2) then√
z − 2iM ath.= √ z − 2i
f (z)M ath.= −f (z)
3. z ∈ (C), (D), (E), (F )
arg(z − i), arg(z − 2i) ∈ (−π, π) So √
z − iM ath.= √
z − i and √
z − 2iM ath.= √ z − 2i f (z)M ath.= f (z)
4. z ∈ (+)edge of the cut [i, 2i].
arg(z − i) = −3π2 then √
z − iM ath.= −√ z − i arg(z − 2i) = −π2 then √
z − 2iM ath.= √ z − 2i f (z)M ath.= −f (z)
5. z ∈ (+)edge of the cut [i, 2i].
arg(z − i) = π2 then √
z − iM ath.= √ z − i arg(z − 2i) = −π2 then √
z − 2iM ath.= √ z − 2i
f (z)M ath.= f (z)
Conclusion:
f (z)M ath.= f (z) if z ∈ (B)S(+)edge of the cut [i,2i],
−f (z) otherwise. (22)
If f (z) = s n
Q
k=1
(z − zk) for vertical cut.
Figure 55: The areas with points 2N − 1 and 2N in vertical cuts
In each case we can find the z-region where f (z)M ath.= f (z) and where f (z)M ath.= −f (z).
Case 1. zk= aki, ak∈ R, k = 1, 2, ..., 2N − 1 = m
region − (1) = {(x, y) : x < 0, a1 ≤ y < a2}[
{z1z2 cut in (+)edge}
region − (2) = {(x, y) : x < 0, a2 ≤ y < a3}[
{z2z3 cut in (+) edge } region − (2j − 1) = {(x, y) : x < 0, a2j−1 ≤ y < a2j}[
{z2j−1z2j cut in (+) edge}
region − (2N − 1) = {(x, y) : x < 0, a2N −1 ≤ y}[
{z2N −1 → −∞ cut in (+) edge}
1. z ∈ region−(2j − 1)
arg(z − zk) ∈ [−3π2 , −π] then √
z − zkM ath.= −√
z − zk, k=1, 2, ..., 2j − 1 arg(z − zk) ∈ (−π, π) then √
z − zk M ath.= √
z − zk , k=2j − 1, 2j, ..., 2N − 1 So
f (z) M.= (−1)2j−1f (z) M ath.= −f (z)
2. z ∈ region−(2j)
arg(z − zk) ∈ [−3π2 , −π] √
z − zkM ath.= −√
z − zk, k = 1, 2, ..., 2j arg(z − zk) ∈ (−π, π) then √
z − zk M ath.= √
z − zk, k = 2j, 2j + 1, ..., 2N − 1 So
f (z)M.= (−1)2jf (z)M ath.= f (z)
Case 2. zk= aki, ak∈ R, k = 1, 2, ..., 2N = m
region − (1) = {(x, y) : x < 0, a1 ≤ y < a2}[
{z1z2 cut in (+)edge}
region − (2) = {(x, y) : x < 0, a2 ≤ y < a3}[
{z2z3 cut in (+) edge } region − (2j − 1) = {(x, y) : x < 0, a2j−1 ≤ y < a2j}[
{z2j−1z2j cut in (+) edge}
region − (2N − 1) = {(x, y) : x < 0, a2N −1 ≤ y}[
{z2N −1 → −∞ cut in (+) edge}
region − (2N ) = {(x, y) : x < 0, a2N −1< y}
1. z ∈ region−(2j − 1), j = 1, 2, ..., N arg(z − zk) ∈ [−3π2 , −π] then √
z − zkM ath.= −√
z − zk, k=1, 2, ..., 2j − 1 arg(z − zk) ∈ (−π, π) then √
z − zk M ath.= √
z − zk, k=2j-1,2j,...,2N-1 So
f (z)M ath.= (−1)2j−1f (z) = −f (z) 2. z ∈ region−(2j) , j = 1, 2, ..., N
arg(z − zk) ∈ [−3π2 , −π] then √
z − zkM ath.= −√
z − zk, k = 1, 2, ..., 2j arg(z − zk) ∈ (−π, π) then √
z − zk M ath.= √
z − zk, k = 2j, 2j + 1, ..., 2N − 1 So
f (z)M ath.= (−1)2jf (z) = f (z) Conclusion:
Case 1.
f (z)M ath.= −f (z) if z ∈ (2j − 1), j=1,2,...,N-1,
f (z) otherwise. (23)
Case 2.
f (z)M ath.= −f (z) if z ∈ (2j − 1), j=1,2,...,N,
f (z) otherwise. (24)
Take an example to show this way how to help us modify : ComputeR 1
f (z)dz over a1, a2, a3, b1, b2 and b3 cycles where f (z)=
p(z − 2 + 2i)(z − 2 − 2i)(z − 1 + 3i)(z − 1 − 3i)(z + i)(z − i)(z + 1 + 3i)(z + 1 − 3i) Let z1 = 2−2i, z2 = 2+2i, z3 = 1−3i, z4 = 1+3i, z5 = −i, z6 = i, z7 = −1−3i, z8 = −1+3i.
Using the way of blocks to modify the computation in Mathematica to get right value.
Figure 56: cut plane
Figure 57: a-cycles and their equivalent path
1. ComputeR
a∗1 1
f (z)dz where a∗1 is equivalent path for a1 and a∗1 = a∗11S a∗12
(1) z ∈ a∗11= the path for vertical cut from 2 + 2i to 2 − 2i on (+)edge in sheet-I by (1), (2) and Cauchy Integrate Theorem
Z
path on vertical cut from 1 + 2i to 1 − 2i on (+)edge in sheet-I, a∗24 = the path on vertical cut from 1 − 2i to 1 + 2i on (−)edge in sheet-I, a∗25 = the path on vertical cut from 1 − 2i to 1 − 3i on (+)edge in sheet-I and a∗26 = the path on vertical cut from 1 − 3i to 1 − 2i on (−)edge in sheet-I.
(1) z ∈ a∗21: Let z = 1 + ri, r : 3 → 2 and dz = idr
(1) z ∈ a∗21: Let z = 1 + ri, r : 3 → 2 and dz = idr