1 Discussion on multi-valued functions

Transcription

1 Week 3 notes, Math Discussion on multi-valued functions Log function : Note that if z is written in its polar representation: z = r e iθ, 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 or θ (θ, 2 π +θ ], (2) [θ, 2 π + θ ) where θ 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, for n Z. log 1 z log 2 z = 2 π i n (4) Proof. Let w 1 = log 1 z and w 2 = log 2 z. Then, exp(w 1 w 2 ) = e w 1 /e w 2 = 1. Since for ζ = x+iy, e ζ 1 = implies e x cosy 1 = and e x siny =, we deduce y = 2nπ, and x =, 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, 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 [, 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, is the same for log 1 and log 2 functions. 1

2 Definition 2 The particular choice n = in (1), with θ = π in (2) defines a particular logarithmic function, called the principal branch of the logarithm, with the notation log p z or Log z. Definition 3 The set of all z for which log is undefined are called it s branch points: ( 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 } is the branch cut. Definition 5 For specificied branch cut, i.e. θ, 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 at some z = z and obtain the value of θ at other values from continuity along a path connecting z 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, log p (z 1 z 2 ) log p z 1 + log p 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 e i(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 is a branch point of a function f(z) if for all sufficiently small ǫ >, f(z + ǫ e i(φ + 2π) ) f(z + ǫ e iφ ) when f(z +ǫe iφ ) changes continuously with φ. This means that if we circle around z = z in a sufficiently small circle, we do not to the same value of f. 2

3 Eg. log (z+1) at z = 1. If z+1 = ǫ e iφ, then log (z+1) = log ǫ + i φ + 2 i n π. When φ is replaced by φ + 2 π, we do not return to the same value. Definition 7 z = is a branch point of f(z) if f(1/w) has a branch point at w =. Eg. log z has a branch point at since log 1/w = log w + 2inπ has a branch point at w =. However, log ( z+1 z 1) has no branch points at z =. Recall, we mentioned for a specific branch and branch-cut choice, it is not always true log(z 1 z 2 ) = log(z 1 ) + log(z 2 ) (8) However, if we do not assign any explicit restriction on arg (z 1 z 2 ), but instead define it as: arg (z 1 z 2 ) = arg z 1 + arg z 2 (9) then (8) will hold. For instance, if we take arg ( 1) = π, and apply definition (9) arg ( 1) 2 = 2 arg( 1) = 2 π and hence log ( 1) 2 = 2 i π and not. In defining a composite function involving logarithm, say. log (z 2 1), it is convenient to place a 2 π interval restriction on arg (z 1) and arg (z +1), but not on arg (z 2 1); instead we require arg (z 2 1) = arg (z 1) + arg(z +1), (1) in accordance to (9) Thus, if z 1 = r 1 e iθ 1 and z 2 = r 2 e iθ 2, then, using (1), log (z 2 1) = log (r 1 r 2 ) + iarg(z 2 1) + 2inπ = log r 1 + log r 2 + i(θ 1 + θ 2 ) + 2inπ (11) If θ 1 θ 2 ( π,π], the branch cuts for log (z 2 1) will bethose shown in Fig.1. z = ±1 and z = are branch points as may be readily checked. There is of course a denumerably infinite set of branches of the function, corresponding to differing integral values of the integer n. Alternately, if we restricted θ 1 [, 2 π), θ 2 ( π, π], then the corresponding branch cuts are as given below in Fig.1. Once again we have infinitely many branches characterized by integer n in (11) onsider another example of of a composite function involving logarithm: ( ) z 1 log (12) z +1 3

4 Im z z -1 θ 2 θ 1 1 Re z Figure 1: Branch cuts for log (z 2 1) In this case, let z 1 = r 1 e i θ 1 and z+1 = r 2 e iθ 2. Then, if we put 2 π interval restriction on arg(z ±1), but avoid putting direct restrictions on arg ( ) z 1 z+1 itself, then if is possible to define ( ) z 1 arg = arg (z 1) arg (z +1) (13) z +1 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. 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 [,2π), then the branch cut situation is still the same as in Fig. 1. 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 [, 2π) and θ 2 ( π, π], then the corresponding branch cut situation is described by Fig. 1. 4

5 Im z -1 1 Figure 2: Another choice of cuts for log (z 2 1) Im z -1 1 Re z Figure 3: Branch cuts for log ( ) z 1 z+1 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. z k for nonintegral k Definition 8 For nonintegral k z k = exp[k log z] = exp[k (log r + iθ + 2 i n π)] (15) 5

6 Im z -1 1 Figure 4: Another possible cut for log ( ) z 1 z+1 where for the purposes of uniqueness on the plane for fixed n, θ [θ, 2 π +θ ) or (θ, 2 π +θ ] (16) Remark 5 Since k is not an integer, exp[2 i n k π] 1; hence z k can have more than one values for the same z, depending on n. A fixed choice of n in (15) and θ in (16) specifies as unique function z k on the plane. The value of n characterizes the branch, and the choice θ defines the branch cut. However, note that, unlike the case of log, z k need not have infinite set of distinct branches since differing values of n can give the same value of exp[2 i n k π]. In particular, if k = p/q, p and q being integers, then clearly exp[2 i n k π] can have q distinct values corresponding to the choice n =,1,2,..,(q 1). Note that n = q gives rise to the same value as n =, n = q+1 the same as n = 1 and so on. For instance z 1/2 has only two distinct branches corresponding to even and odd n in (15) z 1/2 = r 1/2 e iθ/2, (17) or z 1/2 = r 1/2 e iθ/2 (18) Remark 6 We note that for a particular choice of n (i.e. branch) and a specific 2π restriction on the arg ( ), generally, (z 1 z 2 ) k z k 1 z k 2 (19) On the other hand, if we only impose 2 π interval restriction on arg z 1 and arg z 2, and not on arg (z 1 z 2 ), then as before in the context of log function, it is possible to define arg (z 1 z 2 ) so that (z 1 z 2 ) k = z k 1 zk 2 (2) 6

7 Exercise for reader: Determine appropriate branch cuts and branches for (i) (z 2 1) 1/2 and (ii) ( z+1 1/2 z 1). Suppose we have a branch for which f(2) >. 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 introduce 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 e iθ/2, g 2 (z) = r 1/2 e iθ/2 (21) where r = z. There is a branch point of each of g 1 (z) and g 2 (z) at z = and z =. These are clearly branch points of f(z) = log(1+g(z)) as well. Im z 1 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 7

8 in Fig. 1. we define log(1 + w) = log 1+w + i arg(1 + w) + i 2 m π (22) Im w -1 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) = log m (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) = log m (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 8

9 Im z 1 Figure 7: Branch cut for f 2,m (z), as given by (24) interested to note that if we start from z = 1 where f(1) = log 2, i.e. choose f 1, (z) for this point evaluation, and follow the change of f(z) on the Riemann surface along the path shown in Fig. 1, then on crossing the cut in Fig. 1, f(z) ceases to be f 1, (z). Instead, it becomes f 2, (z) for which Fig. 1 describes the branch cuts and points. We have used dotted lines on the path in Fig. 1 to denote our journey to the other sheet for which Fig. 1 is applicable. 9

10 2 ontour Integration Remark 7 auchy s theorem and its extension are of course directly related to determination of closed contour integrals of analytic functions that contain only isolated singularities within the contour. We simply use the Residue theorem and compute the sum of the residues. However, it can also be used to compute: 1. ertain definite integrals 2. 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 contribution. Sometimes, change of variables relates the contour integration to a complex closed path integration. Eg. 1: ompute I = dx x Solution: onsider the closed path contribution from the contour shown in Fig. 2. i R +R Figure 8: losed contour for evaluation of Eg. 1 dz 1 + z 4 = R dx 1 + x 4 + π/2 i R e iθ dθ 1 + R 4 e 4iθ + R i dr 1 + r 4 (25) 1

11 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 i R e iθ dθ 1 + R 4 e 4iθ < π/2 R dθ R 4 1 as R Thus, from (25), as R dz = (1 i) I = 2π Residue at z = e i π/4 (26) 1 + z 4 since z = e iπ/4 1 is the only contour within. Residue at that point found to be. 4 e i 3 π/4 So, from (26), 1 I = 4(1 i) 2 π i e i 3 π/4 = π Remark 8 For integrands of the type, it is prudent, to choose closed path with a leg 1 + x p on the real axis, another one a straight line, at an angle 2π/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 e iθ ) as R for θ in [, π], then dz g(z) e iα z as R R where α > and R is a semi-circle in the upper-half plane of radius R about the origin. Proof. We note that on R, z = R e iθ = R cos θ + i R sin θ, dz = i R e iθ dθ. Further for any ǫ, there exists R (independent of θ) so that g(r e iθ < ǫ. Further, we note from the graph of sin θ that on the interval [, π/2], sin θ 2 θ Using these information, we π get for sufficiently large R, dz g(z) e iα z < R Hence the lemma follows. π < 2 ǫ R R dθ ǫ e α R sin θ = 2 ǫ R π/2 π/2 dθ exp[ α R 2 π θ] < ǫ π α α R sin θ dθ e [1 exp[ α R]] (27) 11

12 orollary 1 If g(r e iθ ) as R for θ in [ π, ], then dz g(z) e i α z as R R where α > and R is a semi-circle in the lower-half plane of radius R about the origin. The proof The proof is very similar to previous Lemma, and left to the reader. orollary 11 If g(r e iθ ) as R for θ in [ π/2, π/2], then dz g(z) e α z as R R where α > and R is a semi-circle in the righthalf plane of radius R about the origin. Proof is left to the reader. orollary 12 If g(r e iθ ) as R for θ in [π/2, 3π/2], then dz g(z) e α z as R R where α > and R is a semi-circle in the left-half plane of radius R about the origin. Proof is left to the reader. orollary 13 If g(r e iθ ) /R p 1, as R for θ in [, π/p], then dz g(z) e i α zp as R R where α > and R is an arc of a circle of radius R about the origin, starting on the positive real axis and subtending an angle π/p at the origin. Proof. This follows simply using from Jordan s Lemma, after we substitute z 1 = z p. Exercise 2: ompute I(a) = cos ax 1 + x 2 dx Solution: Note that, without loss of generality, a >. Further, I(a) = 1 2 cos ax 1 + x 2 dx = 1 2 Re I 1(a), (28) 12

13 R -R R Figure 9: losed contour for (29) where I 1 (a) = Now consider the integral on the closed path, shown in Fig. 2. The round trip contour integral can be broken up into parts: e i a z dz 1 + z 2 dz = R R e iax dx 1 + x 2 (29) e i a x 1 + x 2 dx + R e i a z dz (3) 1 + z2 In the limit R, from Jordan s Lemma, there is no contribution from R. On the other hand from (3), in this limit, we get I 1 (a). So I 1 (a) = 2πi (Residue at z = i) = 2 π i e a 2i = π e a Hence I(a) = π 2 e a. Remark 9 : Note if we were to consider the complex integral cos az dz instead of 1 + z 2 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 x p 1+x dx (31) 13

14 In this case, it is prudent to consider z p dz 1 + z where arg z is in [, 2π), (32) and is the contour shown in Fig. 2. R L 1 ε L 2 Figure 1: ontour for Exercise 7.2 The contour consists of straight segments L 1, L 2 and near circles of radius R and ǫ. Note that the cut from to along the positive real axis is being avoided. Now, as ǫ, z p dz ǫ 1 + z < Further, as R, z p dz R 1 + z < 2π 2π i ǫ i R 1 p dθ ei(1 p)θ ǫ1 p 2 π 1 + ǫ eiθ 1 ǫ 1 p dθ ei(1 p)θ Further, we note that on the straightline segment L 1, z p dz R L z = x p dx 1 + x while on L 2, z p dz L z = ǫ R R1 p 2 π 1 + R eiθ R 1 ǫ r p e i2πp dr 1 + r 14 = e 2πip L 1

15 Therefore, as ǫ and R, z p dz 1 + z = (1 e 2πip ) I = 2π i ( residue at z = e iπ) = 2 π i e iπp Therefore, I = 2 π i e iπp 1 e 2 π i p = π sin π p Remark 1 The method of calculating above also tells us that I exists in a limiting process involving R, with ǫ and R. ǫ x p dx 1+x Remark 11 It is to be noted from this exercise that if we have contour integration around a small circular arc of radius ǫ around a branch point point z, where the integrand is like (z z ) p for p < 1, then there is no contribution from such an integral as ǫ. This fact will simplify future calculations of such integral. Remark 12 The contour in Fig. 2 can be used to calculate by considering Exercise 7.3 ompute Solution: onsider dx Polynomial in x of degree 2 I(a) = log z dz Polynomial in z e i a x2 dx (9) dz e i a z2, (33) where is the contour shown in Fig. 2. Note from a corollary of Jordan s lemma, the contribution for R must go to zero as R. Also, the contribution from the straightline segment along the real axis as R isclearlyi. Also, onthestraightline, wherearg z = π/4,sinceiaz 2 = ia(re iπ/4 ) 2 = a r 2,, the contribution, as R is R dr e iπ/4 e a r2 e iπ/4 π 2 a (34) 15

16 R e i π/4 R R Figure 11: ontour for exercise 7.3 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, the =. ombining with (34), we therefore get I(a) to be negative of the answer in (34). So, I(a) = e iπ/4 π 2 a (35) 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 < to conclude no contribution from the circular arc. Remark 14 On taking the real and imaginary parts of the (??), it is possible to compute cos a x 2 dx and sin a x 2 dx. Exercise 7.4 ompute e ix I = x Solution: It is prudent to choose a contour as shown in Fig. 2. It is clear there are no singularities within the contour, So, = e iz z dz = = ( ǫ R + R ǫ ) e ix dx x dx (36) + R e iz z dz + ǫ e iz z dz (37) 16

17 R ε Figure 12: ontour for Exercise 7.4 From Jordan s lemma, there is no contribution from R as R. Further, as ǫ, since on ǫ, e i z 1, i ǫ e iθ dθ = π i (15) ǫ e iθ Therefore, from (37), in the limit R, ǫ, ǫ π ( ǫ R + R ǫ ) e ix dx x But the integral in (38) in this limit is precisely I. π i (38) Remark 15 By taking the imaginary part of (38), and using the even nature of the integrand, it is clear that sin x = π x 2 Note in this case, the principal value integral reduces to a regular integral, since the integrand has a finite limit as x. Remark 16 On taking the real part of (38), it follows that Exercise 7.5 For π < α < π, I(α) = cos x x 17 = e αx cosh π x dx (39)

18 Solution: It is to be noted that if the integrand in (39) is denoted by f(x), then it has the property (using representation of cosh interms of exp) that f(x + i) = e iα f(x) (4) From property (4), it is prudent to consider e αz dz (41) cosh π z where is a rectangular contour, shown in Fig. 2. -R+i i/2 -R R R+i Figure 13: ontour for Exercise 7.5 Since the integrand f(±r + iy) = e ±αr e iαy e πr e iπy + e πr e iπy is bounded in absolute value by e α R, which tends to zero as R. In this limit, e πr 1 there is no contribution from the vertical segments of (since the integration is over a fixed range in y, independent of R). Thus, in this limit, using (4), we get = (1 + e i α ) I (42) Now consider the residues within the contour. This is where cosh π z =. From the exponential representation, it is clear that this happens where e 2πz = 1, i.e. z = i (1/2 + n) for integer n. The only one inside the contour is at i/2, where the residue is e iα/2 lim (z i/2) f(z) = z i/2 π sinh (i π/2) = eiα/2 (43) i π So, from residue theorem and (42), I = (1 + e iα ) 1 2πi 18 e iα/2 i π = 1 cos (πα/2) (44)

19 Remark 17 The result (44) is independent of the sign of α. Hence it follows that and cosh αx cosh π x dx = 1 cos (α/2) cosh αx cosh π x dx = 1 2 cos (α/2) 19