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Week 3 notes, Math 7651
1 Discussion on multi-valued functions
Log function : Note that if z is written in its polar representation: z = r eiθ^ , where r = |z| and θ = arg z, then
log z ≡ log r + i θ + 2inπ (1)
for n ∈ Z is consistent with exp [log z] = z. This is easily shown through substitution. However, (1) is not unambiguous on the plane. To make (1) a single-valued function (for a particular n) on the complex plane, one needs to restrict
θ ∈ (θ 0 , 2 π + θ 0 ] , (2)
or [θ 0 , 2 π + θ 0 ) where θ 0 is some constant (3)
Remark 1 One might wonder if (1) provides the most general representation of the inverse of exp function. The following lemma answers this in the positive.
Lemma 1 If log 1 and log 2 are two inverse functions of exp, then for any particular z,
log 1 z − log 2 z = 2 π i n (4)
for n ∈ Z.
Proof. Let w 1 = log 1 z and w 2 = log 2 z. Then, exp(w 1 − w 2 ) = ew^1 /ew^2 = 1. Since for ζ = x + iy, eζ^ − 1 = 0 implies ex^ cos y − 1 = 0 and ex^ sin y = 0, we deduce y = 2nπ, and x = 0, implying w 1 − w 2 = ζ = 2 π i n
Remark 2 : It is possible that log 1 z = log 2 z for some set of z ∈ C, not for others. For instance, if we define
log 1 z = log r + i θ, with θ in (−π, π] , (5) log 2 z = log r + i θ, with θ in [0, 2 π) , (6)
then clearly log 1 z = log 2 z for z in the first and second quadrant but not equal in the third and fourth quadrant. In general, n ∈ Z appearing in (4) varies with choice of z unless the choice of branch cut, i.e. choice of θ 0 , is the same for log 1 and log 2 functions.
Definition 2 The particular choice n = 0 in (1), with θ 0 = −π in (2) defines a particular logarithmic function, called the principal branch of the logarithm, with the notation logp z or Log z.
Definition 3 The set of all z for which log is undefined are called it’s branch points: ( 0 and ∞ in this case).
Definition 4 The set of points across which log function undergoes a discontinuity for a particular restriction of the function in a plane is called a branch cut, i.e. {z | arg z = θ 0 } is the branch cut.
Definition 5 For specificied branch cut, i.e. θ 0 , the value of n is referred to as the branch of the logarithm.
Remark 3 Branch cuts (and therefore discontinuities) of log function cannot be avoided if log z is to be uniquely defined in the complex plane. However, a branch cut is quite arbitrary. It can be completely avoided if we equate the domain through the polar pair (r, θ), with no restriction on θ. In that case, one can define uniquely
log z = ln r + iθ , with θ ∈ (−∞, ∞) (7)
In this representation, specification of z is not enough; we need to specify θ = θ 0 at some z = z 0 and obtain the value of θ at other values from continuity along a path connecting z 0 to z. The identification of the complex domain, based on value of θ, is referred to a Riemann surface. Note that while log z is not continuous (across cuts), when restricted on the plane; it is indeed continuous on the Riemann surface.
Remark 4 Note that, for a general z 1 , z 2 ∈ C, logp (z 1 z 2 ) 6 = logp z 1 + logp z 2 , though this is true on the Riemann surface, since on a Riemann surface, with appropriate choice of the argument,
log (z 1 z 2 ) = log
[
r 1 r 2 ei(θ^1 +θ^2 )
]
= log r 1 + log r 2 + i θ 1 + iθ 2 = log z 1 + log z 2
Definition 6 A finite point z 0 is a branch point of a function f (z) if for all sufficiently small ǫ > 0 , f (z 0 + ǫ ei(φ^ + 2π)) 6 = f (z 0 + ǫ eiφ)
when f (z 0 + ǫeiφ) changes continuously with φ. This means that if we circle around z = z 0 in a sufficiently small circle, we do not to the same value of f.
-1^1
z
Im z
Re z
Figure 1: Branch cuts for log (z^2 − 1)
In this case, let z −1 = r 1 ei θ^1 and z +1 = r 2 eiθ^2. Then, if we put 2 π interval restriction on arg(z ± 1), but avoid putting direct restrictions on arg
( (^) z− 1 z+
itself, then if is possible to define
arg
z − 1 z + 1
= arg (z − 1) − arg (z + 1) (13)
Then
log
z − 1 z + 1
= log r 1 − log r 2 + i(θ 1 − θ 2 ) + 2 i n π (14)
If θ 1 , θ 2 ∈ (−π, π], then the corresponding branch cut is shown in Fig. 1. Note that there is no cut on the real axis, left of z = − 1. The reason in this case is that the two cuts, for log (z − 1) and log (z + 1) have cancelled each other out. To see this, note that if we approach from the top a point on the real axis to the left of -1, θ 1 , θ 2 = π; approaching from below, one gets θ 1 , θ 2 = − π. In either case, (θ 1 − θ 2 ) appearing in (14) is equal to 0. So, for any fixed branch n, the function as defined in (14) has no discontinuity on the real axis to the left of −1 and hence no cut over there. Note if we instead restrict θ 1 , θ 2 ∈ [0, 2 π), then the branch cut situation is still the same as in Fig.
- In this case, the branch cut cancellation is on the positive real axis to the right of the branch point z = 1. On the other hand if we restrict θ 1 ∈ [0, 2 π) and θ 2 ∈ (− π, π], then the corresponding branch cut situation is described by Fig. 1.
Im z
-1^1
Figure 2: Another choice of cuts for log (z^2 − 1)
-1 (^1) Re z
Im z
Figure 3: Branch cuts for log
(z− 1 z+
It is to be noted that the branch cut situation described in Fig. 1 and Fig. 1 are not the only ones possible. One can have other 2 π interval restrictions on θ 1 and θ 2 that will give rise to other kinds of branch cuts for the function in (5). In practice, specific choices are made according to the requirement that the function be continuous in some part of the complex plane.
zk^ for nonintegral k
Definition 8 For nonintegral k
zk^ = exp[k log z] = exp[k (log r + iθ + 2 i n π)] (15)
Exercise for reader: Determine appropriate branch cuts and branches for (i) (z^2 − 1)^1 /^2
and (ii)
( (^) z+ z− 1
. Suppose we have a branch for which f (2) > 0. Determine what is the value of f (−2) consistent with a choice of cut such that f (z) changes continously from z = 2 to z = −2 on a path connecting the two points. Remark: Sometimes in determining branch cuts and branches of a composite function, involving logarithm and nonintegral powers in a complicated manner, it is useful to intro- duce suitable intermediate transformation w = g(z) and determine the branch cut in the w variable. The pre-image of that cut in the z-variable gives the cut location in the z-plane. The example below illustrates this point. Example: Describe the branch cuts, branch points and branches for the function
f (z) = log (1 + z^1 /^2 )
Solution: Let w = g(z) = z^1 /^2. Then f (z) = log [1 + g(z)]. If we restrict arg z = θ to (−π, π] (corresponding cut shown in Fig. 1), then there are two distinct possibilities for g(z): g 1 (z) = r^1 /^2 eiθ/^2 , g 2 (z) = − r^1 /^2 eiθ/^2 (21)
where r = |z|. There is a branch point of each of g 1 (z) and g 2 (z) at z = 0 and z = ∞. These are clearly branch points of f (z) = log(1 + g(z)) as well.
Im z
Figure 5: Branch cuts for z^1 /^2 and f 1 ,m(z). A cut-crossing path also shown
We also note that w = −1 is not in the range of g 1 , but it is of g 2 (since g 2 (1) = −1). Now, consider log (1 + w). We choose the restriction arg (1 + w) ∈ (−π, π], with the cut
in Fig. 1. we define
log(1 + w) = log |1 + w| + i arg(1 + w) + i 2 m π (22)
Im w
Figure 6: Branch cut in the w plane for log (1 + w)
Since no point on the chosen cut in the w plane is in the range of g 1 , it is clear that
f 1 ,m(z) = logm (1 + g 1 (z)) (23)
will have no branch points and cuts in the z plane corresponding to the branch point and cut in the w-plane. So, the only branch point and branch cut for the function f 1 ,m(z) are those shown in Fig. 1– only the ones corresponding to g 1 (z). However, the pre-image in the z-plane of the chosen branch cut in the w-plane under g 2 is the positive real z axis, to the right of the branch point z = 1 (corresponds to the cut in the w-plane in Fig. 1). Thus, the function
f 2 ,m(z) = logm (1 + g 2 (z)) (24)
will have branch cuts shown in Fig. 1. The function f (z) has two set of infinite number of branches given by (23) and (24), corresponding to integer m. The cuts are different for (23) and (24). The reader might be
2 Contour Integration
Remark 7 Cauchy’s theorem and its extension are of course directly related to determi- nation of closed contour integrals of analytic functions that contain only isolated singular- ities within the contour. We simply use the Residue theorem and compute the sum of the residues. However, it can also be used to compute:
- Certain definite integrals
- Principal value integrals
The main objective in such exercises is to relate the desired integral to some closed contour in the complex plane, either by appending additional open contours to the original open contour so as to make the union closed. This is helpful if the answer on the additional open contours are known independently or can be related to the original open contour contribu- tion. Sometimes, change of variables relates the contour integration to a complex closed path integration.
Eg. 1: Compute I =
0
dx x^4 + 1 Solution: Consider the closed path contribution from the contour shown in Fig. 2.
0 +R
i R
Figure 8: Closed contour C for evaluation of Eg. 1
C
dz 1 + z^4
∫ R
0
dx 1 + x^4
∫ (^) π/ 2
0
i R eiθ^ dθ 1 + R^4 e^4 iθ^
R
i dr 1 + r^4
It is to the noted that the first and third integral on the right side of (25) combine to give us, in the limit R → ∞, (1 − i) I. Now, as far as the second integral, we note that
∫ (^) π/ 2
0
i R eiθ^ dθ 1 + R^4 e^4 iθ^
∫ (^) π/ 2
0
R dθ R^4 − 1
→ 0 as R → ∞
Thus, from (25), as R → ∞
∮
C
dz 1 + z^4
= (1 − i) I = 2 π Residue at z = ei π/^4 (26)
since z = eiπ/^4 is the only contour within C. Residue at that point found to be (^4) ei 13 π/ 4. So, from (26),
I =
4(1 − i)
2 π i e−i^3 π/^4 =
π 2
Remark 8 For integrands of the type (^) 1 +^1 xp , it is prudent, to choose closed path with a leg on the real axis, another one a straight line, at an angle 2 π/p with respect to the real axis while the third is a circular arc of radius R, which is found not to contibute as R → ∞.
Lemma 9 Jordan’s Lemma If |g(R eiθ)| → 0 as R → ∞ for θ in [0, π], then
∫
CR
dz g(z) eiα z^ → 0 as R → ∞
where α > 0 and CR is a semi-circle in the upper-half plane of radius R about the origin.
Proof. We note that on CR, z = R eiθ^ = R cos θ + i R sin θ, dz = i R eiθ^ dθ. Further for any ǫ, there exists R (independent of θ) so that |g(R eiθ^ | < ǫ. Further, we note from the graph of sin θ that on the interval [0, π/2], sin θ ≥ (^2) π θ Using these information, we get for sufficiently large R,
CR
dz g(z) eiα z^ | <
∫ (^) π
0
R dθ ǫ e−α R sin θ^ = 2 ǫ R
∫ (^) π/ 2
0
dθ e−α R sin θ
< 2 ǫ R
∫ (^) π/ 2
0
dθ exp[−α R
π
θ] <
ǫ π α
[1 − exp[−α R]] (27)
Hence the lemma follows.
-R R
C R
Figure 9: Closed contour C for (29)
where
I 1 (a) =
−∞
eiax^ dx 1 + x^2
Now consider the integral on the closed path C, shown in Fig. 2. The round trip contour integral can be broken up into parts: ∮
C
ei a z^ dz 1 + z^2
dz =
∫ R
−R
ei a x 1 + x^2
dx +
CR
ei a z 1 + z^2
dz (30)
In the limit R → ∞, from Jordan’s Lemma, there is no contribution from
CR. On the other hand from (30), in this limit, we get I 1 (a). So
I 1 (a) = 2 πi (Residue at z = i) = 2 π i
e−a 2 i
= π e−a
Hence I(a) = π 2 e−a.
Remark 9 : Note if we were to consider the complex integral
C
cos az 1 + z^2 dz^ instead of the chosen integrand in (6), we could not chose a convenient contour, since Jordan type lemma is invalid for cos; it does not go to zero in the upper-half or lower-half plane at large distances.
Exercise 7.2 Evaluate (^) ∫ (^) ∞
0
x−p 1 + x
dx (31)
In this case, it is prudent to consider ∮
C
z−p^ dz 1 + z
where arg z is in [0, 2 π) , (32)
and C is the contour shown in Fig. 2.
L 1
L 2
C
R
C ε
Figure 10: Contour C for Exercise 7.
The contour C consists of straight segments L 1 , L 2 and near circles of radius R and ǫ. Note that the cut from 0 to ∞ along the positive real axis is being avoided. Now, as ǫ → 0,
|
Cǫ
z−p^ dz 1 + z
2 π
i ǫ^1 −p^
dθ ei(1−p)θ 1 + ǫ eiθ^
| ≤ 2 π
ǫ^1 −p | 1 − ǫ|
Further, as R → ∞,
CR
z−p^ dz 1 + z
∫ (^2) π
0
i R^1 −p^
dθ ei(1−p)θ 1 + R eiθ^
| ≤ 2 π
R^1 −p R − 1
Further, we note that on the straightline segment L 1 , ∫
L 1
z−p^ dz 1 + z
∫ R
ǫ
x−p^ dx 1 + x
while on L 2 , (^) ∫
L 2
z−p^ dz 1 + z
∫ (^) ǫ
R
r−p^ e−i^2 πp^ dr 1 + r
= − e−^2 πip
L 1
C
R
0 R
R e
i π/
Figure 11: Contour for exercise 7.
from well known calculus result about area under a Gaussian. Since (33) involves an integral of an analytic function without any singularity on and within C, the
C = 0. Combining with (34), we therefore get I(a) to be negative of the answer in (34). So,
I(a) = eiπ/^4
π 2
a
Remark 13 If a in exercise 7.3 were negative, we would have chosen a contour to return to the origin at an angle −π/ 4 , so as to allow use of Jordan’s lemma for a < 0 to conclude no contribution from the circular arc.
Remark 14∫ On taking the real and imaginary parts of the (??), it is possible to compute ∞ 0 cos a x
(^2) dx and ∫^ ∞ 0 sin a x
(^2) dx.
Exercise 7.4 Compute
I = −
−∞
eix x
dx (36)
Solution: It is prudent to choose a contour as shown in Fig. 2. It is clear there are no singularities within the contour, So,
C
eiz z
dz = =
(∫ (^) −ǫ
−R
∫ R
ǫ
eix^ dx x
CR
eiz z
dz +
Cǫ
eiz z
dz (37)
0
C R
Cε
Figure 12: Contour C for Exercise 7.
From Jordan’s lemma, there is no contribution from CR as R → ∞. Further, as ǫ → 0, since on Cǫ, ei z^ → 1, (^) ∫
Cǫ
π
i ǫ eiθdθ ǫ eiθ^
= −π i (15)
Therefore, from (37), in the limit R → ∞, ǫ → 0,
(∫ (^) −ǫ
−R
∫ R
ǫ
eix^ dx x
→ π i (38)
But the integral in (38) in this limit is precisely I.
Remark 15 By taking the imaginary part of (38), and using the even nature of the inte- grand, it is clear that (^) ∫ ∞
0
sin x x
π 2
Note in this case, the principal value integral reduces to a regular integral, since the inte- grand has a finite limit as x → 0.
Remark 16 On taking the real part of (38), it follows that
−∞
cos x x
Exercise 7.5 For −π < α < π,
I(α) =
−∞
eαx cosh π x
dx (39)
Remark 17 The result (44) is independent of the sign of α. Hence it follows that
∫ (^) ∞
−∞
cosh αx cosh π x
dx =
cos (α/2)
and (^) ∫ (^) ∞
0
cosh αx cosh π x
dx =
2 cos (α/2)
