Consider the line with the slope m = tan α, 0 < α ≤ π, and for any cuts on this line, we define that
z = {
reiθ, θ∈ [α − 2π, α) iff z is in the sheet-I, reiθ, θ∈ [α, α + 2π) iff z is in the sheet-II.
To begin with, we consider the values in sheet-I and sheet-II.
If we have f (z) =
√∑n
k=1
(z− zk), and by using the polar form,
∑n k=1
(z− zk) = reiθ1 = reiθ2,
where θ1 and θ2 is in sheet-I and sheet-II respectively, so that θ2 = θ1+ 2π.
Therefore,
f (z)|(II) =√ reiθ22
=√
reiθ1+2π2
=√
reiθ12 eiπ
=−√
reiθ12 =−f(z)|(I),
where f (z)|(I) and f (z)|(II) denotes the value of f (z) with z in sheet-I and sheet-II re-spectively.
Notice that the difference of argument between z in these two sheets is 2π, so the difference of argument between f (z)|(I) and f (z)|(II) is π, and this implies f (z)|(II) =
−f(z)|(I).
If we could find the equivalent straight paths of a, b-cycles, it would be easier to calculate the integrals over these cycles.
First, consider an a-cycle is shown in Fig. 2.24.
Fig. 2.24 Cycle a crossover a cut
We consider these paths are shown in Fig. 2.25.
Fig. 2.25 Cycle a = a1∪ a2 crossover a cut and some auxiliary paths
Notice that, by the Cauchy’s integral theorem,
Again, by the Cauchy integral theorem, { ∫
Next, consider a b-cycle is shown in Fig. 2.27.
Fig. 2.27 Cycle b crossover two cuts
We consider these paths are shown in Fig. 2.28.
Fig. 2.28 Cycle b = b1∪ b2 crossover two cuts and some auxiliary paths
Since
And, by the Cauchy integral theorem, { ∫
Proposition 2.1. If f (z) =
√∑n
k=1
(z− zk), where zk are distinct, and the paths r1 and r2 are shown in Fig. 2.30.
Fig. 2.30 Paths r1 and r2 for Pro. 2.1
f (z)dz, and this implies
∫
Proposition 2.2. If f (z) =
√∑n
k=1
(z− zk), where zk are distinct, and the paths r1 and r2 are shown in Fig. 2.31.
Fig. 2.31 Paths r1 and r2 for Pro. 2.2 Prove that
∫
r1
f (z)dz =
∫
r2
f (z)dz.
Proof.
We know that f (z)|(II) =−f(z)|(I), so
∫
r2
f (z)dz =−
∫
r2′
f (z)dz, where r′2 is a straight path from zs+1 to zs in the sheet-I
=
∫
r1
f (z)dz.
Sometimes the integrals over some complicated paths are difficult to compute, so we want to use a computational software program, MATHEMATICA, to obtain the correct value of the integrals over cycles.
Now, we have known the difference between the value in sheet-I and sheet-II of theory, so we just discuss the difference between the value in the sheet-I of theory and in MATHEMATICA, then we can modify the computation in MATHEMATICA such that the numerical result we modified equals the numerical result of theory.
Notice that θ ∈ (−π, π] of reiθ in MATHEMATICA, so if ϕ /∈ (−π, π] of reiϕ, MATHEMATICA will change reiϕ into reiθ∗, such that reiϕ = reiθ∗ and θ∗ ∈ (−π, π].
Lemma 2.1. If z is in the sheet-I for a cut whose one of the end points is zk. If this cut on the line with the slope m = tan α, 0 < α≤ π,
√z− zk|(I) =
{ √z− zk|M AT HEM AT ICA if arg(z− zk)∈ (−π, α),
−√
z− zk|M AT HEM AT ICA if arg(z− zk)∈ [α − 2π, −π].
Proof.
Let z be in the sheet-I, and using polar form, z− zk = reiθ for some θ ∈ [α − 2π, α).
Notice that MATHEMATICA will change reiθ into reiθ∗ such that reiθ = reiθ∗, and θ∗ ∈ (−π, π].
Case 1: arg(z− zk) = θ∈ (−π, α) Notice that θ ∈ (−π, α) ⊆ (−π, π].
Then θ∗ = θ, and this implies √
z− zk|(I) =√
z− zk|M AT HEM AT ICA. Case 2: arg(z− zk) = θ∈ [α − 2π, −π]
Notice that θ ∈ [α − 2π, −π] " (−π, π], so θ∗ = θ + 2π∈ [α, π] ⊂ (−π, π].
Then,
√z− zk|(I) =√
reiθ2, and
√z− zk|M AT HEM AT ICA =√ reiθ∗2
=√ reiθ+2π2
=−√ reiθ2, and this implies √
z− zk|(I) =−√
z− zk|M AT HEM AT ICA.
Definition 2.1.
Define that {
f (z)M AT H.= f (z) means f (z)|(I) = f (z)|M AT HEM AT ICA, f (z)M AT H.= −f(z) means f(z)|(I) =−f(z)|M AT HEM AT ICA.
Next, we will discuss some situations of the domain such that the value in MATH-EMATICA must be modified.
Case 1:
If z is in the sheet-I, and we consider a cut on the line l1(z) = 0 with the slope m = tan α, 0 < α < π, from z1 to∞, where z1 = x1+ iy1 ∈ C (Fig. 2.32).
Fig. 2.32 Cut with end point z1
1. z ∈ + edge of this cut:
arg(z− z1) = α− 2π implies √ z− z1
M AT H.
= −√
z− z1. 2. z ∈ − edge of this cut:
arg(z− z1) = α implies √ z− z1
M AT H.
= √
z− z1. 3. z is not on this cut:
Fig. 2.33 Modified-Value Domain for the cut with end point z1
Case 2:
We consider a cut on the line l1(z) = 0 with the slope m = tan α, 0 < α < π.
Assume without loss of generality that end points of the cut are z1 and z2, ∀zk = xk+ iyk∈ C (Fig. 2.34).
Fig. 2.34 Cut with end points z1 and z2
Now, we want to find the domain where z belongs, such that
√z− z1
√z− z2
M AT H.
= −√ z− z1
√z− z2.
1. z ∈ + edge of this cut:
By the Lemma 2.1, we know that
√z− zk
M AT H.
= −√
z− zk if arg(z− zk)∈ [α − 2π, −π], and in this sense, consider
Fig. 2.35 S+, S−, T+, and T− Then, √
z− z1
√z− z2
M AT H.
= −√
z− z1
√z− z2 if and only if z ∈ S+ ∩ T− or z ∈ S−∩ T+.
Notice that S+∩ T− = ∅, so √ z− z1
√z− z2
M AT H.
= −√
z− z1
√z− z2 if and only if z ∈ S−∩ T+.
Therefore, if z ∈ S−∩ T+ ={z|z ∈ C, l1(z) < 0, y1 ≤ lm(z) < y2}, shown in Fig.
2.36,
√z− z1
√z− z2
M AT H.
= −√ z− z1
√z− z2.
Fig. 2.36 S−∩ T+
Case 3:
We consider a horizontal cut from z1 to ∞.
Since the cut is horizontal, its slope m = tan α = 0, and this implies α = π (Fig.
2.37).
Fig. 2.37 Cut with end point z1
1. z ∈ + edge of this cut:
arg(z− z1) = −π implies √ z− z1
M AT H.
= −√
z− z1. 2. z ∈ − edge of this cut:
arg(z− z1) = π implies √ z− z1
M AT H.
= √
z− z1. 3. z is not in this cut:
Since z is not in this cut, arg(z− z1)∈ (−π, π).
By the Lemma 2.1, we know that
√z− z1
M AT H.
= √
z− z1.
Case 4:
We consider a horizontal cut with end points are z1 and z2.
Since the cut is horizontal, its slope m = tan α = 0, and this implies α = π (Fig.
3. z is not on this cut:
Suppose that √ z− z1
√z− z2
M AT H.
= −√ z− z1
√z− z2. By the Lemma 2.1, we know that
√z− zk
M AT H.
= −√
z− zk if arg(z− zk) =−π, and in this sense, √
z− z1
√z− z2
M AT H.
= −√
z− z1
√z− z2 if and only if only one of arg(z− z1) and arg(z− z2) is −π.
Then, z is on the cut, but it is a contradiction.
So, for a horizontal cut with end points are z1 and z2,
√z− z1
√z− z2
M AT H.
= −√ z− z1
√z− z2 if z ∈ + edge of this cut.
For these different cases, we know that where z belongs for different cuts such that the value in MATHEMATICA must be modified.
Now, the followings are some examples.
First, we consider f (z) =
√
∑7 k=1
(z− zk), where
z1 =−1 − i, z2 =−1 + 3i, z3 = 0,
z4 = 2, z5 = 3− 2i, z6 = 9
2 + i, and z7 = 6− i.
And, we shown the branch cuts in Fg. 2.39.
Fig. 2.39 Branch points and the branch cuts of f (z)
Example 2.6. Evaluate ∫
c1f (z)dz, c1 is shown in Fig. 2.40.
Solution.
Fig. 2.40 Cycle c1 and the equivalent paths c11 and c12
We have f (z) =
By the discussion in this subsection, we have
√z− z1
By the Proposition 2.1,
∫
Example 2.7. Evaluate ∫
c2f (z)dz, c2 is shown in Fig. 2.41.
Solution.
Fig. 2.41 Cycle c2 and the equivalent paths c21, c22, c23, and c24
We consider the equivalent paths, say c∗2 = c21∪ c22∪ c23∪ c24, shown in Fig. 2.41, such that c2 ≈ c∗2, where
c21 is the straight path on slant cut from z6 to 7
2− i on the + edge of sheet-I, c22 is the straight path on slant cut from 7
2 − i to z6 on the − edge of sheet-I, c23 is the straight path on slant cut from 7
2 − i to z5 on the + edge of sheet-I, and c24 is the straight path on slant cut from z5 to 7
2− i on the − edge of sheet-I.
1. If z ∈ c21:
By the discussion in this subsection, we have
By the Proposition 2.1,
∫
By the discussion in this subsection, we have
√z− z1
4. If z ∈ c24:
By the Proposition 2.1,
∫
Example 2.8. Evaluate ∫
d1f (z)dz, d1 is shown in Fig. 2.42.
Solution.
We consider the equivalent paths, say d∗1 = d11∪ d12∪ d13∪ d14, shown in Fig. 2.42, such that d1 ≈ d∗1, where
d11 is the straight path from z5 to 5
2 − i in the sheet-I, d12 is the straight path from 5
2− i to z5 in the sheet-II, d13 is the straight path from 5
2− i to z4 in the sheet-I, and
By the discussion in this subsection, we have
√z− z1
By the Proposition 2.2,
∫
3. If z ∈ d13:
By the discussion in this subsection, we have
√z− z1
By the Proposition 2.2,
∫
Example 2.9. Evaluate ∫
d2f (z)dz, d2 is shown in Fig. 2.43.
Solution.
Fig. 2.43 Cycle d2 and the equivalent paths d∗1, c12, d21, d22, d23, d24, and d25
We consider the equivalent paths, say d∗2 = d∗1 ∪ c12∪ d21∪ d22 ∪ d23∪ d24∪ d25, shown in Fig. 2.43, such that d2 ≈ d∗2, where
d21 is the straight path on horizontal cut from z3 to z4 on the + edge of sheet-II, d22 is the straight path from z3 to−1
3 + i in the sheet-I, d23 is the straight path from −1
3 + i to z3 in the sheet-II, d24 is the straight path from −1
3 + i to z2 in the sheet-I, and d25 is the straight path from z2 to−1
3 + i in the sheet-II.
To start with, we know that d21 is also the straight path on horizontal cut from z3 to z4 on the− edge of sheet-I, that is, d21=−c12, and this implies
∫
c12
f (z)dz +
∫
d21
f (z)dz = 0
1. If z ∈ d22:
By the discussion in this subsection, we have
√z− z1
By the Proposition 2.2,
∫
By the discussion in this subsection, we have
By the Proposition 2.2,
∫