5 Complex-Variable Theory

5.1 Analytic functions A complex-valued function f(z) of a complex variable z is di↵erentiable at z with derivative f 0(z)ifthelimit

f(z0) f(z) f 0(z)= lim (5.1) z z z z 0! 0 exists as z0 approaches z from any direction in the complex plane. The limit must exist no matter how or from what direction z0 approaches z. If the function f(z) is di↵erentiable in a small disk around a point z0, then f(z) is said to be analytic (or equivalently holomorphic) at z0 (and at all points inside the disk).

Example 5.1 (Polynomials). If f(z)=zn for some integer n, then for tiny dz and z = z + dz,thedi↵erencef(z ) f(z)is 0 0 n n n 1 f(z0) f(z)=(z + dz) z nz dz (5.2) ⇡ and so the limit

n 1 f(z0) f(z) nz dz n 1 lim =lim = nz (5.3) z z z z dz 0 dz 0! 0 ! n 1 exists and is nz independently of how z0 approaches z. Thus the function zn is analytic at z for all z with derivative

n dz n 1 = nz . (5.4) dz 186 Complex-Variable Theory A function that is analytic everywhere is entire. All polynomials

N n P (z)= cn z (5.5) n=0 X are entire.

Example 5.2 (A function that’s not analytic). To see what can go wrong when a function is not analytic, consider the function f(x, y)=x2 +y2 = zz¯ for z = x + iy. If we compute its derivative at (x, y)=(1, 0) by setting x =1+✏ and y = 0, then the limit is f(1 + ✏, 0) f(1, 0) (1 + ✏)2 1 lim =lim = 2 (5.6) ✏ 0 ✏ ✏ 0 ✏ ! ! while if we instead set x = 1 and y = ✏,thenthelimitis f(1,✏) f(1, 0) 1+✏2 1 lim =lim = i lim ✏ =0. (5.7) ✏ 0 i✏ ✏ 0 i✏ ✏ 0 ! ! ! So the derivative depends upon the direction through which z 1. !

5.2 Cauchy-Riemann conditions How do we know whether a complex function f(x, y)=u(x, y)+iv(x, y) of two real variables x and y is analytic? We apply the criterion (5.1) of analyticity and require that the change df in the function f(x, y) be proportional to the change dz = dx + idy in the complex variable z = x + iy @u @v @u @v + i dx + + i dy = f 0(z)(dx + idy). (5.8) @x @x @y @y ✓ ◆ ✓ ◆ Setting ﬁrst dy and then dx equal to zero, we have @u @v 1 @u @v + i = f 0(z)= + i . (5.9) @x @x i @y @y ✓ ◆ ✓ ◆ This complex equation implies the two real equations @u @v @v @u = and = (5.10) @x @y @x @y which are the Cauchy-Riemann conditions. In a notation in which partial derivatives are labeled by subscripts, the Cauchy-Riemann conditions are u = v and v = u . x y x y 5.3 Cauchy’s Integral Theorem 187 5.3 Cauchy’s Integral Theorem The Cauchy-Riemann conditions imply that the integral of a function along a closed contour (one that ends where it starts) vanishes if the function is analytic on the contour and everywhere inside it. To keep the notation simple, let’s consider a rectangle R of length ` and height h with one corner at the origin of the z plane. The integral along the four sides of the rectangle is

f(z) dz = (u(x, y)+iv(x, y)) (dx + idy) R R I I ` h = [u(x, 0) + iv(x, 0)] dx + [u(`, y)+iv(`, y)] idy (5.11) 0 0 Z 0 Z 0 + [u(x, h)+iv(x, h)] dx + [u(0,y)+iv(0,y)] idy. Z` Zh

The real part of this contour integral is

` h Re f(z) dz = [u(x, 0) u(x, h)] dx [v(`, y) v(0,y)] dy. R 0 0 ✓I ◆ Z Z (5.12) The di↵erences u(x, h) u(x, 0) and v(`, y) v(0,y) are deﬁnite integrals of the y derivative uy(x, y) and of the x derivative vx(x, y)

h u(x, h) u(x, 0) = u (x, y) dy y 0 (5.13) Z ` v(`, y) v(0,y)= v (x, y) dx. x Z0

The real part (5.12) of the contour integral therefore vanishes due to the second v = u of the Cauchy-Riemann conditions (??) x y

` h h ` Re f(z) dz = u (x, y) dy dx v (x, y) dx dy y x ✓IR ◆ Z0 Z0 Z0 Z0 ` h = [u (x, y)+v (x, y)] dy dx =0. (5.14) y x Z0 Z0

Similarly, the ﬁrst ux = vy of the Cauchy-Riemann conditions (??)implies 188 Complex-Variable Theory that the imaginary part of these integrals vanishes

` h Im f(z) dz = [v(x, 0) v(x, h)] dx + [u(`, y) u(0,y)] dy ✓IR ◆ Z0 Z0 ` h h ` = vy(x, y) dy dx + ux(x, y) dx dy 0 0 0 0 `Z hZ Z Z = [ v (x, y)+u (x, y)] dy dx =0. (5.15) y x Z0 Z0 A similar argument shows that the contour integral along the four sides of any rectangle vanishes as long as the function f(z) is analytic on and within the rectangle

f(z) dz = 0 (5.16) IR whether or not the rectangle has one corner at the origin z = 0. Suppose a function f(z) is analytic along a closed contour C and also at every point of the surface S that the contour encloses. We can tile the surface S with a suitable collection of contiguous rectangles some of which might be as small as pixels. The integral of f(z) along the perimeter of each rectangle will vanish (5.16) because each rectangle lies entirely within the region in which f(z) is analytic. Now consider two adjacent rectangles like the two squares in Fig. 5.1. The sum of the two contour integrals around the two adjacent squares is equal to the contour integral around the perimeter of the two squares because the up integral along the right side (dots) of the left square cancels the down integral along the left side of the right square. Thus the sum of the contour integrals around the perimeters of all the rectangles that tile the surface S amounts to just the integral along the outer contour C that encloses the surface S. The integral around each rectangle vanishes. Thus the integral of f(z) along the contour C must also vanish because it is the sum of these vanishing integrals around the rectangles that tile the surface S.ThisisCauchy’s integral theorem: The integral of a function f(z) along a closed contour vanishes

f(z) dz = 0 (5.17) IC if the function f(z) is analytic on the contour and at every point inside it. A region in the complex plane is simply connected if we can shrink every loop in the region to a point while keeping the loop in the region. A slice of American cheese is simply connected, a slice of Swiss cheese is not. A dime is simply connected, a washer isn’t. The surface of a sphere is 5.3 Cauchy’s Integral Theorem 189 simply connected, the surface of a bagel isn’t. So another version of Cauchy’s integral theorem is that the integral of a function f(z) along a closed contour vanishes if the contour lies within a simply connected region in which f(z) is analytic (Augustin-Louis Cauchy, 1789–1857).

Example 5.3 (Polynomials). Since dzn+1 =(n + 1) zn dz, the integral of the entire function zn along any contour C that ends and starts at the same point z must vanish for any integer n 0 0 1 1 zn dz = dzn+1 = zn+1 zn+1 =0. (5.18) n +1 n +1 0 0 IC IC 2 Thus the integral of any polynomial P (z)=c0 + c1z + c2z + ... along any closed contour C also vanishes m n P (z) dz = cnz dz =0. (5.19) C C n=0 I I X

Example 5.4 (Tiny circular contour). If f(z) is analytic at z0,thenthe deﬁnition (5.1) of the derivative f (z) shows that f(z) f(z )+f (z )(z z ) 0 ⇡ 0 0 0 0 near z to ﬁrst order in z z . The points of a small circle of radius ✏ and 0 0 center z are z = z + ✏ei✓.Sincez z = ✏ei✓ and dz = i✏ei✓d✓, the closed 0 0 0 contour integral around the circle

2⇡ i✓ f(z) dz = f(z )+f 0(z )(z z ) i✏e d✓ 0 0 0 I Z0 (5.20) # ⇥ 2⇡ ⇤ 2⇡ i✓ i✓ i✓ = f(z0) i✏e d✓ + f 0(z0) ✏e i✏e d✓ Z0 Z0 vanishes because the ✓-integrals are zero. Thus the contour integral of an analytic function f(z) around a tiny circle that lies within the region in which f(z) is analytic.

Example 5.5 (Tiny square contour). The analyticity of f(z) at z = z0 lets us expand f(z) near z as f(z) f(z )+f (z )(z z ). A tiny square 0 ⇡ 0 0 0 0 contour consists of four complex segments dz = ✏, dz = i✏, dz = ✏, and 1 2 3 dz = i✏. The integral of the constant f(z ) around the square vanishes 4 0 f(z ) dz = f(z ) dz = f(z )[✏ + i✏+( ✏)+( i✏)] = 0. (5.21) 0 0 0 I⇤ I⇤ The integral of the second term f (z )(z z ) is the sum of four integrals 0 0 0 along the four sides of the tiny square. If we set z0 =0tosimplifythe 190 Complex-Variable Theory notation, then the integral from left to right along the bottom of the square is ✏/2 ✏/2 i✏2 I1 = f 0(0) xdx+ if 0(0) y dx = f 0(0). (5.22) ✏/2 ✏/2 2 Z Z The integral up the right side of the square is ✏/2 ✏ i✏2 I2 = f 0(0) + iy idy = f 0(0). (5.23) ✏/2 2 2 Z ⇣ ⌘ The integral from right to left along the top side of the square is

✏/2 2 i✏ I = f 0(0) (x + iy) dx = f 0(z ). (5.24) 3 2 0 Z✏/2 Finally, the integral down the left side is

✏/2 2 i✏ I = f 0(0) (x + iy) idy = f 0(0). (5.25) 4 2 Z✏/2 These integrals cancel in pairs. Thus the contour integral of an analytic function f(z) around a tiny square of side ✏ is zero to order ✏2 as long as the square lies inside the region in which f(z) is analytic. Suppose a function f(z) is analytic in a simply connected region R and that C and C0 are two contours that lie inside R and that both run from z1 to z2. The di↵erence of the two contour integrals is an integral along a closed contour C00 that runs from z1 to z2 and back to z1 and that vanishes by Cauchy’s theorem

z2 z2 z2 z1 f(z) dz f(z) dz = f(z) dz + f(z) dz = f(z) dz =0. Zz1C Zz1C0 Zz1C Zz2C0 IC00 (5.26) It follows that any two contour integrals that lie within a simply connected region in which f(z) is analytic are equal if they start at the same point z1 and end at the same point z2. Thus we may continuously deform the contour of an integral of an analytic function f(z) from C to C0 without changing the value of the contour integral as long as as long as these contours and all the intermediate contours lie entirely within the region R and have the same ﬁxed end points z1 and z2 as in Fig. 5.2

z2 z2 f(z) dz = f(z) dz. (5.27) Zz1C Zz1C0 So a contour integral depends upon its end points and upon the function 5.3 Cauchy’s Integral Theorem 191

The Cauchy Integral Theorem

<< ∨ ∧ ∨ >> ∧

Figure 5.1 The sum of two contour integrals around two adjacent squares is equal to the contour integral around the perimeter of the two squares because the up integral along the right side of the left square cancels (dots) the down integral along the left side of the right square. A contour integral around a big square is equal to the sum of the contour integrals around the smaller interior squares that tile the big square. f(z) but not upon the actual contour as long as the deformations of the contour do not push it outside of the region R in which f(z) is analytic. If the end points z and z are the same, then the contour is closed, and 1 2 C we write the integral as

z1 f(z) dz f(z) dz (5.28) z1 ⌘ I C IC with a little circle to denote that the contour is a closed loop. The value of that integral is independent of the contour as long as our deformations of the contour keep it within the domain of analyticity of the function and as long as the contour starts and ends at z1 = z2. Now suppose that the function f(z) is analytic along the contour and at all points within it. Then 192 Complex-Variable Theory

Four Equal Contour Integrals

∧∧∧ ∧

region of analyticity

Figure 5.2 As long as the four contours are within the domain of analyticity of f(z) and have the same endpoints, the four contour integrals of that function are all equal. we can shrink the contour, staying within the domain of analyticity of the function, until the area enclosed is zero and the contour is of zero length—all this without changing the value of the integral. But the value of the integral along such a null contour of zero length is zero. Thus the value of the original contour integral also must be zero

z1 f(z) dz =0. (5.29) Iz1C And so we again arrive at Cauchy’s integral theorem: The contour integral of a function f(z) around a closed contour C lying entirely within the domain R of analyticity of the function vanishes

f(z) dz = 0 (5.30) IC as long as the function f(z) is analytic at all points within the contour. 5.4 Cauchy’s Integral Formula 193 Example 5.6 (A pole). The function f(z)=1/(z z ) is analytic in a 0 region that is not simply connected because its derivative 1 1 1 1 f 0(z)= lim = (5.31) dz 0 z + dz z z z dz (z z )2 ! ✓ 0 0 ◆ 0 exists in the whole complex plane except for the point z = z0.

5.4 Cauchy’s Integral Formula

Suppose that f(z) is analytic in a simply connected region R and that z0 a point inside this region. We ﬁrst will integrate the function f(z)/(z z0) along a tiny closed counterclockwise contour around the point z0.The i✓ contour is a circle of radius ✏ with center at z0 with points z = z0 + ✏e for 0 ✓ 2⇡, and dz = i✏ei✓d✓.Sincez z = ✏ei✓, the contour integral in   0 the limit ✏ 0is ! f(z) 2⇡ [f(z )+f (z )(z z )] dz = 0 0 0 0 i✏ei✓d✓ z z z z I✏ 0 Z0 0 2⇡ i✓ f(z0)+f 0(z0) ✏e i✓ = i✓ i✏e d✓ (5.32) 0 ⇥ ✏e ⇤ Z 2⇡ i✓ = f(z0)+f 0(z0) ✏e id✓ =2⇡i f(z0) 0 Z h i since the ✓-integral involving f 0(z0) vanishes. Thus f(z0) is the integral 1 f(z) f(z )= dz (5.33) 0 2⇡i z z I✏ 0 which is a miniature version of Cauchy’s integral formula. Now consider the counterclockwise contour in Fig. 5.3 which is a big C0 counterclockwise circle, a small clockwise circle, and two parallel straight lines, all within a simply connected region in which f(z) is analytic. The R function f(z)/(z z ) is analytic everywhere in except at the point z . 0 R 0 We can withdraw the contour to the left of the point z and shrink it C0 0 to a point without having the contour cross z .Duringthisprocess,the C0 0 integral of f(z)/(z z ) does not change. Its ﬁnal value is zero. So its initial 0 value also is zero 1 f(z) 0= dz. (5.34) 2⇡i z z0 IC0 We let the two straight-line segments approach each other so that they cancel. What remains of contour is a big counterclockwise contour C0 C 194 Complex-Variable Theory

′ The Contour C Around z 0

> z 0 < <

C ′

Figure 5.3 The full contour is the sum of a big counterclockwise contour and a small clockwise contour, both around z0, and two straight lines which cancel.

around z0 and a tiny clockwise circle of radius ✏ around z0.Thetinyclockwise circle integral is the negative of the counterclockwise integral (5.33), so we have

1 f(z) 1 f(z) 1 f(z) 0= dz = dz dz. (5.35) 2⇡i z z0 2⇡i z z0 2⇡i ✏ z z0 IC0 IC I

Using the miniature result (5.33), we ﬁnd

1 f(z) f(z0)= dz (5.36) 2⇡i z z0 IC which is Cauchy’s integral formula. 5.4 Cauchy’s Integral Formula 195

We can use this formula to compute the ﬁrst derivative f 0(z) of f(z) f(z + dz) f(z) f 0(z)= dz 1 1 1 1 = dz0 f(z0) 2⇡i dz z z dz z z I ✓ 0 0 ◆ 1 f(z0) = dz0 . (5.37) 2⇡i (z z dz)(z z) I 0 0 So in the limit dz 0, we get ! 1 f(z0) f 0(z)= dz0 . (5.38) 2⇡i (z z)2 I 0 The second derivative f (2)(z) of f(z)thenis

(2) 2 f(z0) f (z)= dz0 . (5.39) 2⇡i (z z)3 I 0 And its nth derivative f (n)(z)is

(n) n! f(z0) f (z)= dz0 . (5.40) 2⇡i (z z)n+1 I 0 In these formulas, the contour runs counterclockwise about the point z and lies within the simply connected domain in which f(z) is analytic. R Thus a function f(z) that is analytic in a region is infinitely di↵eren- R tiable there. Example 5.7 (Schlaefli’s Formula for the Legendre Polynomials). Rodrigues showed (section 8.2) that the Legendre polynomial Pn(x)isthenth derivative 1 d n P (x)= (x2 1)n. (5.41) n 2n n! dx ✓ ◆ Schlaefli used this expression and Cauchy’s integral formula (5.40) to represent Pn(z) as the contour integral (exercise 5.8) 2 n 1 (z0 1) P (z)= dz0 (5.42) n 2n 2⇡i (z z)n+1 I 0 in which the contour encircles the complex point z counterclockwise. This formula tells us that at z = 1 the Legendre polynomial is 2 n n 1 (z0 1) 1 (z0 + 1) P (1) = dz0 = dz0 = 1 (5.43) n 2n 2⇡i (z 1)n+1 2n 2⇡i (z 1) I 0 I 0 in which we applied Cauchy’s integral formula (5.36) to f(z)=(z + 1)n. 196 Complex-Variable Theory Example 5.8 (Bessel Functions of the First Kind). The counterclockwise integral around the unit circle z = ei✓ of the ratio zm/zn in which both m and n are integers is

m 2⇡ 2⇡ 1 z 1 i✓ i(m n)✓ 1 i(m+1 n)✓ dz = ie d✓e = d✓e . (5.44) 2⇡i zn 2⇡i 2⇡ I Z0 Z0 If m +1 n = 0, this integral vanishes because exp 2⇡i(m +1 n)=1 6 2⇡ 2⇡ i(m+1 n)✓ 1 i(m+1 n)✓ 1 e d✓e = =0. (5.45) 2⇡ 0 2⇡ i(m +1 n) Z " #0 If m +1 n = 0, the exponential is unity exp i(m +1 n)✓ = 1, and the integral is 2⇡/2⇡ = 1. Thus the original integral is the Kronecker delta

1 zm dz = . (5.46) 2⇡i zn m+1,n I The generating function (9.5) for Bessel functions Jm of the ﬁrst kind is

t(z 1/z)/2 1 m e = z Jm(t). (5.47) m= X1 Applying our integral formula (5.46) to it, we ﬁnd

m 1 t(z 1/z)/2 1 1 1 z dz e = dz J (t) 2⇡i zn+1 2⇡i zn+1 m m= I I X1 (5.48) 1 = m+1,n+1 Jm(t)=Jn(t). m= X1 Thus letting z = ei✓,wehave

2⇡ i✓ i✓ 1 e e J (t)= d✓ exp t in✓ (5.49) n 2⇡ 2 Z0 " # or more simply

2⇡ ⇡ 1 i(t sin ✓ n✓) 1 J (t)= d✓e = d✓ cos(t sin ✓ n✓). (5.50) n 2⇡ ⇡ Z0 Z0 (exercise 5.3). 5.5 Harmonic Functions 197 5.5 Harmonic Functions The Cauchy-Riemann conditions (5.10) u = v and u = v (5.51) x y y x tell us something about the laplacian of the real part u of an analytic function f = u + iv. First, the second x-derivative u is u = v = v = u . xx xx yx xy yy So the real part u of an analytic function f is a harmonic function

uxx + uyy = 0 (5.52) that is, one with a vanishing laplacian. Similarly v = u = v ,sothe xx yx yy imaginary part of an analytic function also is a harmonic function

vxx + vyy =0. (5.53) A harmonic function h(x, y) can have saddle points, but not local minima or maxima because at a local minimum both hxx > 0 and hyy > 0, while at a local maximum both hxx < 0 and hyy < 0. So in its domain of analyticity, the real and imaginary parts of an analytic function f have neither minima nor maxima. For static fields, the electrostatic potential (x, y, z) is a harmonic function of the three spatial variables x, y, and z in regions that are free of charge because the electric field is E = , and its divergence vanishes E =0 r r· where the charge density is zero. Thus the laplacian of the electrostatic potential (x, y, z) vanishes = + + = 0 (5.54) r·r xx yy zz and (x, y, z) is harmonic where there is no charge. The location of each positive charge is a local maximum of the electrostatic potential (x, y, z) and the location of each negative charge is a local minimum of (x, y, z). But in the absence of charges, the electrostatic potential has neither local maxima nor local minima. Thus one cannot trap charged particles with an electrostatic potential, a result known as Earnshaw’s theorem. The Cauchy-Riemann conditions imply that the real and imaginary parts of an analytic function are harmonic functions with two-dimensional gradients that are mutually perpendicular (u ,u ) (v ,v )=v v v v =0. (5.55) x y · x y y x x y And we know that the electrostatic potential is a harmonic function. Thus the real part u(x, y) (or the imaginary part v(x, y)) of any analytic function f(z)=u(x, y)+iv(x, y) describes the electrostatic potential (x, y) for some 198 Complex-Variable Theory electrostatic problem that does not involve the third spatial coordinate z. The surfaces of constant u(x, y) are the equipotential surfaces, and since the two gradients are orthogonal, the surfaces of constant v(x, y) are the electric field lines. Example 5.9 (Two-dimensional Potentials). The function f(z)=u + iv = Ez= Ex+ iEy (5.56) can represent a potential V (x, y, z)=Ex for which the electric-field lines E = E xˆ are lines of constant y. It also can represent a potential V (x, y, z)= Eyin which E points in the negative y-direction, which is to say along lines of constant x. Another simple example is the function f(z)=u + iv = z2 = x2 y2 +2ixy (5.57) for which u = x2 y2 and v =2xy. This function gives us a potential V (x, y, z) whose equipotentials are the hyperbolas u = x2 y2 = c2 and whose electric-field lines are the perpendicular hyperbolas v =2xy = d2. Equivalently, we may take these last hyperbolas 2xy = d2 to be the equipotentials and the other ones x2 y2 = c2 to be the lines of the electric field. For a third example, we write the variable z as z = rei✓ =exp(lnr + i✓) and use the function f(z)=u(x, y)+iv(x, y)= ln z = (ln r + i✓) (5.58) 2⇡✏0 2⇡✏0 which describes the potential V (x, y, z)= (/2⇡✏ )ln x2 + y2 due to a 0 line of charge per unit length = q/L. The electric-field lines are the lines p of constant v (x, y, 0) E = 2 2 (5.59) 2⇡✏0 x + y or equivalently of constant ✓.

5.6 Taylor Series for Analytic Functions Let’s consider the contour integral of the function f(z )/(z z) along a 0 0 circle inside a simply connected region in which f(z) is analytic. For C R any point z inside the circle, Cauchy’s integral formula (5.36) tells us that

1 f(z0) f(z)= dz0. (5.60) 2⇡i z0 z IC 5.6 Taylor Series for Analytic Functions 199

Taylor-Series Contour Around z 0

z ′

Figure 5.4 Contour of integral for the Taylor series (5.65).

We add and subtract the center z from the denominator z z 0 0 1 f(z0) f(z)= dz0 (5.61) 2⇡i z0 z0 (z z0) IC and then factor the denominator

1 f(z0) f(z)= dz0. (5.62) 2⇡i z z0 (z z ) 1 IC 0 0 z z0 0 ⇣ ⌘ From Fig. 5.4, we see that the modulus of the ratio (z z )/(z z )isless 0 0 0 than unity, and so the power series

1 n z z 1 z z 1 0 = 0 (5.63) z z z z 0 0 n=0 0 0 ✓ ◆ X ✓ ◆ by (4.32–4.35) converges absolutely and uniformly on the circle. We therefore 200 Complex-Variable Theory are allowed to integrate the series n 1 f(z0) 1 z z0 f(z)= dz0 (5.64) 2⇡i z z z z 0 0 n=0 0 0 IC X ✓ ◆ term by term

1 1 f(z ) dz f(z)= (z z )n 0 0 . (5.65) 0 2⇡i (z z )n+1 n=0 0 0 X IC By Eq.(5.40), the integral is just the nth derivative f (n)(z)dividedbyn- factorial. Thus the function f(z) possesses the Taylor series

1 (z z )n f(z)= 0 f (n)(z ) (5.66) n! 0 n=0 X which converges as long as the point z is inside a circle centered at z0 that lies within a simply connected region in which f(z) is analytic. R

5.7 Cauchy’s Inequality Suppose a function f(z) is analytic in a region that includes the disk z R | | and that f(z) is bounded by f(z) M on the circle z = R which is the | | | | perimeter of the disk. Then by using Cauchy’s integral formula (5.40), we may bound the nth derivative f (n)(0) of f(z) at z =0by n! f(z) dz f (n)(0) | || | | |2⇡ z n+1 I | | n!M 2⇡ Rd✓ n!M = (5.67)  2⇡ Rn+1 Rn Z0 which is Cauchy’s inequality. This inequality bounds the terms of the Taylor series (5.66)

1 (z z )n 1 (z z )n | 0 f (n)(z ) M | 0 (5.68) n! | 0 | Rn n=0 n=0 X X showing that it converges (4.35) absolutely and uniformly for z z

5.8 Liouville’s Theorem Suppose now that f(z) is analytic everywhere (entire) and bounded by f(z) M for all z R . (5.69) | | | | 0 5.9 The Fundamental Theorem of Algebra 201 Then by applying Cauchy’s inequality (5.67) at successively larger values of R,wehave n!M f (n)(0) lim = 0 (5.70) | |R Rn !1 which shows that every derivative of f(z) vanishes f (n)(0) = 0 for n 1 (5.71) at z = 0. But then the Taylor series (4.78) about z = 0 for the function f(z) consists of only a single term, and f(z) is a constant

1 zn f(z)= f (n)(0) = f (0)(0) = f(0). (5.72) n! n=0 X So every bounded entire function is a constant (Joseph Liouville, 1809–1882).

5.9 The Fundamental Theorem of Algebra Gauss applied Liouville’s theorem to the function 1 1 f(z)= = 2 N (5.73) PN (z) c0 + c1z + c2z + ...+ cN z which is the inverse of an arbitrary polynomial of order N. Suppose that the polynomial PN (z) had no zero, that is, no root anywhere in the complex plane. Then f(z) would be analytic everywhere. Moreover, for suciently large z , the polynomial P (z) is approximately P (z) c zN , and so | | N N ⇡ N f(z) would be bounded by something like 1 f(z) M for all z R0. (5.74) | | c RN ⌘ | | | N | 0 So if PN (z) had no root, then the function f(z) would be a bounded entire function and so would be a constant by Liouville’s theorem (5.72). But of course, f(z)=1/PN (z) is not a constant unless N = 0. Thus any polynomial PN (z) that is not a constant must have a root, a pole of f(z), so that f(z) is not entire. This is the only exit from the contradiction. If the root of PN (z) is at z = z1,thenPN (z)=(z z1) PN 1(z), in which PN 1(z) is a polynomial of order N 1, and we may repeat the argument for its reciprocal f1(z)=1/PN 1(z). In this way, one arrives at the fundamental N theorem of algebra: Every polynomial PN (z)=c0 + c1z + ...+ cN z has N roots somewhere in the complex plane P (z)=c (z z )(z z ) ...(z z ). (5.75) N N 1 2 N 202 Complex-Variable Theory 5.10 Laurent Series Consider a function f(z) that is analytic in an annulus that contains an outer circle of radius R and an inner circle of radius R as in Fig. 5.5. C1 1 C2 2 We integrate f(z) along a contour within the annulus that encircles the C12 point z in a counterclockwise fashion by following counterclockwise and C1 clockwise and a line joining them in both directions. By Cauchy’s integral C2 formula (5.36), this contour integral yields f(z)

1 f(z0) f(z)= dz0. (5.76) 2⇡i 12 z0 z IC The integrations in opposite directions along the line joining and can- C1 C2 cel, and we are left with a counterclockwise integral around the outer circle C1 and a clockwise one around or minus a counterclockwise integral around C2 C2 1 f(z0) 1 f(z00) f(z)= dz0 dz00. (5.77) 2⇡i 1 z0 z 2⇡i 2 z00 z IC IC

Now from the ﬁgure (5.5), the center z0 of the two concentric circles is closer to the points z on the inner circle than it is to z and also closer 00 C2 to z than to the points z on 0 C1 z z z z 00 0 < 1 and 0 < 1. (5.78) z z z z 0 0 0 We add and subtract z0 from each of the denominators (as in (5.61)) and absorb the minus sign before the second integral into its denominator

1 f(z0) 1 f(z00) f(z)= dz0 + dz00. 2⇡i z z (z z ) 2⇡i z z (z z ) I 1 0 0 0 I 2 0 00 0 C C (5.79) After factoring the two denominators

1 f(z0) f(z)= dz0 2⇡i 1 (z0 z0)[1 (z z0)/(z0 z0)] IC 1 f(z00) + dz00 (5.80) 2⇡i 2 (z z0)[1 (z00 z0)/(z z0)] IC we expand them, as in the series (5.64), in power series that converge abso- 5.10 Laurent Series 203

Two Contours Around z 0

> z 0 <

z ′′ ′ C 2 z

C < 1 >

Figure 5.5 The contour consisting of two concentric circles with center at z0 encircles the point z in a counterclockwise sense. The asterisks are poles or other singularities of the function f(z). lutely and uniformly on the two contours

1 n 1 f(z0) f(z)= (z z ) dz0 0 2⇡i (z z )n+1 n=0 1 0 0 X IC 1 1 1 m + (z00 z ) f(z00) dz00. (5.81) (z z )m+1 2⇡i 0 m=0 0 2 X IC

Since the functions being integrated are analytic between the two circles, we may move the contours, without changing the values of the integrals, to a common counterclockwise contour about any circle of radius R R R C 2   1 between the two circles and .Wethensetm = n 1, or n = m 1, C1 C2 204 Complex-Variable Theory so as to combine the two sums into one sum on n from to 1 1 1 n 1 f(z0) f(z)= (z z ) dz0. (5.82) 0 2⇡i (z z )n+1 n= 0 0 X1 IC This Laurent series often is written as

1 f(z)= a (z )(z z )n (5.83) n 0 0 n= X1 with 1 f(z) an(z0)= n+1 dz (5.84) 2⇡i (z z0) IC (Pierre Laurent, 1813–1854). The coecient a 1(z0) is called the residue of the function f(z) at z0.Its signiﬁcance will be discussed in section 5.13. Most functions have Laurent series that start at some least integer ` 1 f(z)= a (z )(z z )n (5.85) n 0 0 n= ` X rather than at . For such functions, we can pick o↵the coecients a 1 n one by one without doing the integrals (5.84). The ﬁrst one a ` is the limit ` a `(z0)= lim(z z0) f(z). (5.86) z z0 ! The second Laurent coecient a `+1(z0) is given by the recipe ` 1 ` a `+1(z0)= lim(z z0) f(z) (z z0) a `(z0) . (5.87) z z0 ! h i The third coecient requires a recipe with two subtractions, and so forth.

5.11 Singularities A function f(z) that is analytic for all z is entire. Entire functions have no singularities except possibly as z , which some call the point at | |!1 inﬁnity. A function f(z) has an isolated singularity at z0 if it is analytic in a small disk about z0 but not analytic that point. A function f(z) has a pole of order n>0 at a point z if (z z )n f(z)is 0 0 analytic at z but (z z )n 1 f(z) has an isolated singularity at z . A pole 0 0 0 of order n = 1 is called a simple pole. Poles are isolated singularities. A function is meromorphic if it is analytic for all z except for poles. 5.11 Singularities 205 Example 5.10 (Poles). The function n 1 f(z)= (5.88) (z j)j jY=1 has a pole of order j at z = j for j = 1, 2, . . . , n. It is meromorphic.

An essential singularity is a pole of inﬁnite order. If a function f(z) has an essential singularity at z0, then its Laurent series (5.82) really runs from n = and not from n = L as in (5.85). Essential singularities are spooky: 1 if a function f(z) has an essential singularity at w,theninsideeverydisk around w, f(z) takes on every complex number, with at most one exception, an inﬁnite number of times (Emile´ Picard, 1856–1941). Example 5.11 (Essential singularity). The function f(z)=exp(1/z) has an essential singularity at z = 0 because its Laurent series (5.82)

0 1 1 1 1 f(z)=e1/z = = zn (5.89) m! zm n ! m=0 n= X X1 | | runs from n = . Near z = 0, f(z)=exp(1/z) takes on every complex 1 number except 0 an inﬁnite number of times. Example 5.12 (Meromorphic function with two poles). The function f(z)= 1/z(z + 1) has poles at z = 0 and at z = 1 but otherwise is analytic; it is meromorphic. We may expand it in a Laurent series (5.83–5.84)

1 1 f(z)= = a zn (5.90) z(z + 1) n n= X1 about z = 0 for z < 1. The coecient a is the integral | | n 1 dz a = (5.91) n 2⇡i z n+2 (z + 1) IC in which the contour is a counterclockwise circle of radius r<1. Since C z < 1, we may expand 1/(1 + z) as the power series | | 1 1 = ( z)m. (5.92) 1+z m=0 X Doing the integrals, we ﬁnd

1 1 m dz 1 m m n 1 a = ( z) = ( 1) r (5.93) n 2⇡i zn+2 m,n+1 m=0 m=0 X IC X 206 Complex-Variable Theory for n 1 and zero otherwise. Thus the Laurent series about z = 0 for f(z) is 1 1 f(z)= = ( 1)n+1 zn. (5.94) z(z + 1) n= 1 X It starts at n = 1, not at n = , because f(z) is meromorphic with only 1 a simple pole at z = 0.

Example 5.13 (Argument principle). Consider the counterclockwise integral 1 g (z) f(z) 0 dz (5.95) 2⇡i g(z) IC along a contour that lies inside a simply connected region R in which f(z) C is analytic and g(z) meromorphic. If the function g(z) has a zero or a pole of order n at z = w R and no other singularity in R so that 2 g(z)=a (w)(z w)n, (5.96) n then the ratio g0/g is g (z) n(z w)n 1 n 0 = = (5.97) g(z) (z w)n z w and the integral is 1 g (z) 1 n f(z) 0 dz = f(z) dz = nf(w). (5.98) 2⇡i g(z) 2⇡i z w IC IC Any function g(z) meromorphic in R will have a Laurent series

1 g(z)= a (w)(z w)k (5.99) k kX=n about each point w R. One may show (exercise 5.18) that as z w the 2 ! ratio g0/g again approaches (5.97). It follows that the integral (5.95) is a sum of n f(w ) at the zeros and poles of g(z) that lie within the contour ` ` C 1 g0(z) 1 n` f(z) dz = f(z) = n` f(w`) (5.100) 2⇡i g(z) 2⇡i z w` IC X` IC X` in which n is the multiplicity of the `th zero or pole. | `| 5.12 Analytic Continuation 207 5.12 Analytic Continuation We saw in Sec. 5.6 that a function f(z) that is analytic within a circle of radius R about a point z0 possesses a Taylor series (5.66)

1 (z z )n f(z)= 0 f (n)(z ) (5.101) n! 0 n=0 X that converges for all z inside the disk z z

1 f(z)= zn (5.102) n=0 X converges and deﬁnes an analytic function for z < 1. But for such z,we | | may sum the series to 1 f(z)= . (5.103) 1 z By summing the series (5.102), we have analytically continued the function f(z) to the whole complex plane apart from its simple pole at z = 1.

Example 5.15 (The Gamma Function). Euler’s form of the gamma function is the integral

1 t z 1 (z)= e t dt =(z 1)! (5.104) Z0 which makes (z) analytic in the right half-plane Re z>0. But by successively using the relation (z + 1) = z (z), we may extend (z)intotheleft half-plane 1 1 1 1 1 1 (z)= (z + 1) = (z + 2) = (z + 3). (5.105) z z z +1 z z +1 z +2 The last expression deﬁnes (z) as a function that is analytic for Re z> 3 208 Complex-Variable Theory apart from simple poles at z = 0, 1, and 2. Proceeding in this way, we may analytically continue the gamma function to the whole complex plane apart from the negative integers and zero. The analytically continued gamma function is represented by Weierstrass’s formula

1 1 z 1 z z/n (z)= e 1+ e . (5.106) z n "n=1 # Y ⇣ ⌘

Example 5.16 (Riemann’s zeta function). Ser found an analytic continuation n 1 1 1 1 1 n ( 1)k ⇣(z)= = (5.107) nz z 1 n +1 k (k + 1)z 1 n=1 n=0 X X Xk=0 ✓ ◆ of Riemann’s zeta function (4.105) to the whole complex plane except for the point z = 1 (Joseph Ser, 1875–1954).

Example 5.17 (Dimensional Regularization). The loop diagrams of quantum ﬁeld theory involve badly divergent integrals like a d4q q2 I(4) = (5.108) (2⇡)4 2 2 b Z (q +↵ ) where often a = 0 and b = 2 and ↵2 > 0. Gerardus ’t Hooft (1946–) and Martinus J. G. Veltman (1931–) promoted the number of spacetime dimensions from 4 to a complex number d. The resulting integral has the value (Srednicki, 2007, p. 102)

a ddq q2 (b a d/2) (a + d/2) 1 I(d)= = (2⇡)d 2 2 b (4⇡)d/2 (b)(d/2) (↵2)b a d/2 Z (q +↵ ) (5.109) and so deﬁnes a function of the complex variable d that is analytic everywhere except for simple poles at d = 2(n a + b)wheren =0, 1, 2,..., . 1 At these poles, the formula

n ( 1)n 1 1 ( n + z)= + + O(z) (5.110) n! z k ! Xk=1 where =0.5772... is the Euler-Mascheroni constant (4.8) can be useful. 5.13 Calculus of residues 209 5.13 Calculus of residues A contour integral of an analytic function f(z) does not change unless the end points move or the contour crosses a singularity or leaves the region of analyticity (section 5.3). Let us consider the integral of a function f(z) along a counterclockwise contour that encircles n poles at z for k =1,...,n in C k a simply connected region in which f(z) is meromorphic. We may shrink R the area within the contour without changing the value of the integral until C the area is inﬁnitesimal and the contour is a sum of n tiny counterclockwise circles around the n poles Ck n f(z) dz = f(z) dz. (5.111) k IC Xk=1 IC These tiny counterclockwise integrals around the poles at zi are 2⇡i times the residues a 1(zi) deﬁned by Laurent’s formula (5.84) with n = 1. So the whole counterclockwise integral is 2⇡i times the sum of the residues of the function f(z) of the enclosed poles

n n f(z) dz =2⇡i a 1(zk)=2⇡i Res(f,zk) (5.112) IC Xk=1 Xk=1 a result is known as the residue theorem. In general, one must do each tiny counterclockwise integral about each pole zi, but simple poles are an important special case. If w is a simple pole of the function f(z), then near it f(z) is given by its Laurent series (5.83) as

a 1(w) 1 n f(z)= + a (w)(z w) . (5.113) z w n n=0 X In this case, its residue is by (5.86) with ` = 1 a 1(w)= lim(z w) f(z) (5.114) z w ! which usually is easier to do than the integral (5.84) 1 a 1(w)= f(z)dz. (5.115) 2⇡i IC Example 5.18 (A function with simple poles). The integral of the function

1 z f(z)= (5.116) z n s n=1 X 210 Complex-Variable Theory along a circle of radius 2 with center at z = 0 is just the sum of its residues

1 1 s 1 1 f(z) dz = lim (z n )f(z)= = ⇣(s) (5.117) 2⇡i z n s ns n=1 n=1 I X ! X which is the zeta function (4.105). Example 5.19 (Cauchy’s Integral Formula). Suppose the function f(z)is analytic within a region and that is a counterclockwise contour that R C encircles a point w . Then the counterclockwise contour encircles the 2R C simple pole at w of the function f(z)/(z w), which is its only singularity in . By applying the residue theorem and formula (5.114) for the residue R a 1(w) of the function f(z)/(z w), we ﬁnd f(z) f(z) dz =2⇡i a 1(w)=2⇡i lim (z w) =2⇡i f(w). (5.118) z w z w z w IC ! So Cauchy’s integral formula (5.36) is an example of the calculus of residues.

Example 5.20 (A meromorphic function). By the residue theorem (5.112), the integral of the function 1 1 f(z)= (5.119) z 1 (z 2)2 along the circle =4ei✓ for 0 ✓ 2⇡ is the sum of the residues at z =1 C   and z =2

f(z) dz =2⇡i [a 1(1) + a 1(2)] . (5.120) IC The function f(z) has a simple pole at z = 1, and so we may use the formula (5.114) to evaluate the residue a 1(1) as 1 a 1(1) = lim (z 1) f(z)=lim = 1 (5.121) z 1 z 1 (z 2)2 ! ! instead of using Cauchy’s integral formula (5.36) to do the integral of f(z) along a tiny circle about z = 1, which gives the same result 1 dz 1 1 a 1(1) = = =1. (5.122) 2⇡i z 1 (z 2)2 (1 2)2 I The residue a 1(2) is the integral of f(z) along a tiny circle about z = 2, which we do by using Cauchy’s integral formula (5.38) 1 dz 1 d 1 1 a 1(2) = = = = 1 2⇡i (z 2)2 z 1 dz z 1 (2 1)2 I z=2 (5.123) 5.14 Ghost Contours 211 getting the same answer as if we had used the recipe (5.86) for a 2 2 1 a 2(2) = lim (z 2) = 1 (5.124) z 2 (z 1)(z 2)2 ! and (5.87) for a 1 1 a 2(2) a 1(2) = lim (z 2) = 1. (5.125) z 2 (z 1)(z 2)2 (z 2)2 !  The sum of these two residues is zero, and so the integral (5.120) vanishes. Another way of evaluating this integral is to deform it, not into two tiny circles about the two poles, but rather into a huge circle z = Rei✓ and to notice that as R the modulus of this integral vanishes !1 2⇡ f(z) dz 0. (5.126) ⇡ R2 ! I This contour is an example of a ghost contour.

5.14 Ghost Contours Often one needs to do an integral that is not a closed counterclockwise contour. Integrals along the real axis occur frequently. One sometimes can convert a line integral into a closed contour by adding a contour along which the integral vanishes, a ghost contour. We have just seen an example (5.126) of a ghost contour, and we shall see more of them in what follows. Example 5.21 (Best Case). Consider the integral 1 I = 1 dx. (5.127) b (x i)(x 2i)(x 3i) Z1 We could do the integral by adding a contour Rei✓ from ✓ =0to✓ = ⇡. In the limit R , the integral of 1/[(z i)(z 2i)(z 3i)] along this !1 contour vanishes; it is a ghost contour. The original integral I and the ghost contour encircle the three poles, and so we could compute I by evaluating the residues at those poles. But we also could add a ghost contour around the lower half plane. This contour and the real line encircle no poles. So we get I = 0 without doing any work at all. Example 5.22 (Fourier Transform of a Gaussian). During our computation of the Fourier transform of a gaussian (3.16–3.19), we promised to justify the shift in the variable of integration from x to x+ik/2m2 in this chapter. So let us consider the contour integral of the entire function f(z)=exp( m2z2) 212 Complex-Variable Theory over a rectangular closed contour along the real axis from R to R and then from z = R to z = R + ic and then from there to z = R + ic and then to z = R.Sincef(z) is analytic within the contour, the integral is zero R R+ic R+ic R m2z2 m2z2 m2z2 m2z2 m2z2 dz e = dz e + dz e + dz e + dz e =0 R R R+ic R+ic I Z Z Z Z for all ﬁnite positive values of R and so also in the limit R .Thetwo !1 contours in the imaginary direction are of length c and are damped by the factor exp( m2R2), and so they vanish in the limit R . They are ghost !1 contours. It follows then from this last equation in the limit R that !1 1 m2(x+ic)2 1 m2x2 p⇡ dx e = dx e = (5.128) m Z1 Z1 which is the promised result. It implies (exercise 5.21) that

1 m2(x+z)2 1 m2x2 p⇡ dx e = dx e = (5.129) m Z1 Z1 for m>0 and arbitrary complex z.

Example 5.23 (A cosine integral). To compute the integral cos x I = 1 dx, q > 0, (5.130) c q2 + x2 Z0 we use the evenness of the integrand to extend the integration 1 cos x I = 1 dx, (5.131) c 2 q2 + x2 Z1 write the cosine as [exp(ix)+exp( ix)]/2, and factor the denominators 1 eix 1 e ix I = 1 dx + 1 dx. (5.132) c 4 (x iq)(x + iq) 4 (x iq)(x + iq) Z1 Z1 We promote x to a complex variable z and add the contours z = Rei✓ and i✓ z = Re as ✓ goes from 0 to ⇡ respectively to the ﬁrst and second integrals. The term exp(iz)dz/(q2 + z2)=exp(iR cos ✓ R sin ✓)iRei✓d✓/(q2 + R2e2i✓) vanishes in the limit R , so the ﬁrst contour is a counterclockwise ghost !1 contour. A similar argument applies to the second (clockwise) contour, and we have 1 eiz 1 e iz I = dz + dz. (5.133) c 4 (z iq)(z + iq) 4 (z iq)(z + iq) I I 5.14 Ghost Contours 213

Third-Harmonic Microscopy

−L unseenvisible unseen L

Figure 5.6 In the limit in which the distance L is much larger than the wavelength , the integral (5.135) is non-zero when an edge (solid line) lies where the beam is focused but not when a feature (...) lies where the beam is not focused. Only features within the focused region are visible.

The first integral picks up the pole at iq and the second the pole at iq i⇡ e q e q ⇡e q I = + = . (5.134) c 2 2iq 2iq 2q ✓ ◆ q So the value of the integral is ⇡e /2q. Example 5.24 (Third-Harmonic Microscopy). An ultra-short laser pulse intensely focused in a medium generates a third-harmonic electric field E3 in the forward direction proportional to the integral (Boyd, 2000) dz E (3) E3 1 ei kz (5.135) 3 / 0 (1 + 2iz/b)2 Z1 2 2 along the axis of the beam as in Fig. 5.6. Here b =2⇡t0n/ = kt0 in which n = n(!) is the index of refraction of the medium, is the wavelength of 214 Complex-Variable Theory the laser light in the medium, and t0 is the transverse or waist radius of the gaussian beam, defined by E(r)=E exp( r2/t2). 0 When the dispersion is normal, that is when dn(!)/d!> 0, the shift in the wave vector k =3![n(!) n(3!)]/c is negative. Since k<0, the exponential is damped when z = x + iy is in the lower half plane (LHP) i kz i k (x+iy) i kx ky e = e = e e . (5.136) So as we did in example 5.23, we will add a contour around the lower half plane (z = Rei✓, ⇡ ✓ 2⇡, and dz = iRei✓d✓) because in the limit   R , the integral along it vanishes; it is a ghost contour. !1 The function f(z)=exp(i kz)/(1+2iz/b)2 has a double pole at z = ib/2 which is in the UHP since the length b>0, but no singularity in the LHP y<0. So the integral of f(z) along the closed contour from z = R to z = R and then along the ghost contour vanishes. But since the integral along the ghost contour vanishes, so does the integral from R to R.Thuswhenthe dispersion is normal, the third-harmonic signal vanishes, E3 = 0, as long as the medium with constant (3)(z)e↵ectively extends from to so that 1 1 its edges are in the unfocused region like the dotted lines of Fig. 5.6. But an edge with k>0 in the focused region like the solid line of the figure does make a third-harmonic signal E3. Third-harmonic microscopy lets us see features instead of background. Example 5.25 (Green and Bessel). Let us evaluate the Fourier transform eikx I(x)= 1 dk (5.137) k2 + m2 Z1 of the function 1/(k2 + m2). If x>0, then the exponential deceases with Im k in the upper half plane. So as in example 5.23, the semicircular contour k = Rei✓ for 0 ✓ ⇡ on which dk = iRei✓d✓ is a ghost contour. So if   x>0, then we can add this contour to the integral I(x) without changing it. Thus I(x) is equal to the closed contour integral along the real axis and the semicircular ghost contour eikx eikx I(x)= dk = dk . (5.138) k2 + m2 (k + im)(k im) I I This closed contour encircles the simple pole at k = im and no other singularity, and so we may shrink the contour into a tiny circle around the pole. ikx mx Along that tiny circle, the function e /(k + im)issimplye /2im, and so e mx dk e mx ⇡e mx I(x)= =2⇡i = for x>0. (5.139) 2im k im 2im m I 5.14 Ghost Contours 215 Similarly if x<0, we can add the semicircular ghost contour k = Rei✓, ⇡ ✓ 2⇡, dk = iRei✓d✓ with k running around the perimeter of the lower   half plane. So if x<0, then we can write the integral I(x) as a shrunken closed contour that runs clockwise around the pole at k = im emx dk emx ⇡emx I(x)= = 2⇡i = for x<0. (5.140) 2im k + im 2im m I We combine the two cases (5.139) and (5.140) into the result

ikx 1 e ⇡ m x dk = e | |. (5.141) k2 + m2 m Z1 We can use this formula to develop an expression for the Green’s function of the laplacian in cylindical coordinates. Setting x = 0 and r = x = 0 | | ⇢2 + z2 in the Coulomb Green’s function (3.113), we have

3 p 1 1 d k 1 ik x G(r)= = = e · . (5.142) 4⇡r 2 2 (2⇡)3 k2 4⇡ ⇢ + z Z The integral over the z-componentp of k is (5.141) with m2 = k2 + k2 k2 x y ⌘ ikzz 1 e ⇡ k z dkz 2 2 = e | |. (5.143) kz + k k Z1 So with k x + k y k⇢ cos , the Green’s function is x y ⌘ 2⇡ 1 1 ⇡dk ik⇢ cos k z = de e | |. (5.144) 2 2 (2⇡)3 4⇡ ⇢ + z Z0 Z0

The integralp is a representation (5.50 & 9.7) of the Bessel function J0(k⇢)

2⇡ d J (k⇢)= eik⇢ cos . (5.145) 0 2⇡ Z0 Thus we arrive at Bessel’s formula for the Coulomb Green’s function

1 1 dk k z = J0(k⇢) e | | (5.146) 2 2 4⇡ 4⇡ ⇢ + z Z0 in cylindical coordinatesp (Schwinger et al., 1998, p. 166). Example 5.26 (Yukawa and Green). We saw in example 3.12 that the Green’s function for Yukawa’s di↵erential operator (3.126) is

3 ik x d k e · GY (x)= . (5.147) (2⇡)3 k2 + m2 Z 216 Complex-Variable Theory Letting k x = kr cos ✓ in which r = x ,weﬁnd · | | 2 1 ikr cos ✓ 1 k dk e 1 1 dk k ikr ikr GY (r)= 2 2 2 d cos ✓ = 2 2 2 e e 0 (2⇡) 1k + m ir 0 (2⇡) k + m Z Z Z 1 dk k 1 dk k⇣ ⌘ = 1 eikr = 1 eikr. ir (2⇡)2 k2 + m2 ir (2⇡)2 (k im)(k + im) Z1 Z1 We add a ghost contour that loops over the upper-half plane and get

mr 2⇡i im mr e G (r)= e = (5.148) Y (2⇡)2ir 2im 4⇡r which Yukawa proposed as the potential between two hadrons due to the exchange of a particle of mass m, the pion. Because the mass of the pion is 15 140 MeV, the range of the Yukawa potential is ~/mc =1.4 10 m. ⇥ Example 5.27 (Green’s function for the laplacian in n dimensions). The Green’s function for the laplacian G(x)=(n)(x)is 4 n 1 ik x d k G(x)= e · (5.149) k2 (2⇡)n Z in n dimensions. We use the formula

1 1 k2 = e d (5.150) k2 Z0 to write it as a gaussian integral n k2+ik x d k G(x)= e · d . (5.151) (2⇡)n Z We now complete the square in the exponent

k2 + ik x = (k ix/2)2 x2/4, (5.152) · use our gaussian formula (5.128), and with ↵ = x2/4 write the Green’s function as n n 1 d k x2/4 (k ix/2)2 1 d k x2/4 k2 G(x)= d e e = d e e (2⇡)n (2⇡)n Z0 Z Z0 Z 2 1 n/2 1 x2/4 d (x ) 1 ↵ n/2 2 = e = e ↵ d↵ (4⇡)n/2 4⇡n/2 Z0 Z0 (n/2 1) = . (5.153) n/2 2 (n/2 1) 4⇡ (x ) 1 Our formula (4.67) for (n + 2 ) says that (1/2) = p⇡, and so this formula 5.14 Ghost Contours 217 (5.153) for n = 3 gives G(x)=1/4⇡ x which is (3.113); since (1) = 1, it | | also gives for n =4 1 G(x)= . (5.154) 4⇡2x2

Example 5.28 (The Yukawa Green’s Function in n Dimensions). The Yukawa Green’s function which satisfies ( + m2)G(x)=(n)(x)inn 4 dimensions is the integral (5.149) with k2 replaced by k2 + m2 n 1 ik x d k G(x)= e · . (5.155) k2 + m2 (2⇡)n Z Using the integral formula (5.150), we write it as a gaussian integral n (k2+m2)+ik x dd k G(x)= e · . (5.156) (2⇡)n Z Completing the square as in (5.152), we have n n x2/4 (k ix/2)2 m2 dd k x2/4 (k2+m2) dd k G(x)= e e = e e (2⇡)n (2⇡)n Z Z 1 x2/4 m2 d = e . (5.157) (4⇡)n/2 Z0 We can relate this to a Bessel function by setting =(x /2m)exp( y) | | (n/2 1) 1 2m 1 mx cosh y+(n/2 1)y G(x)= e dy (4⇡)n/2 x ✓ ◆ Z1 (n/2 1) 2 2m 1 mx cosh y = e cosh(n/2 1)ydy (4⇡)n/2 x ✓ ◆ Z0 (n/2 1) 2 2m = Kn/2 1(mx) (5.158) (4⇡)n/2 x ✓ ◆ where x = x = px2 and K is a modified Bessel function of the second kind | | (9.110). If n = 3, this is (exercise 5.28) the Yukawa potential (5.148). Example 5.29 (A Fourier Transform). As another example, let’s consider the integral eikx J(x)= 1 dk. (5.159) (k2 + m2)2 Z1 We may add ghost contours as in the preceding example, but now the integrand has double poles at k = im, and so we must use Cauchy’s integral ± 218 Complex-Variable Theory formula (5.40) for the case of n = 1, which is Eq.(5.38). For x>0, we add a ghost contour in the UHP and find eikx d eikx J(x)= dk =2⇡i (k + im)2(k im)2 dk (k + im)2 I k=im ⇡ 1 mx = x + e . (5.160) 2m2 m ✓ ◆ If x<0, then we add a ghost contour in the LHP and find eikx d eikx J(x)= dk = 2⇡i (k + im)2(k im)2 dk (k im)2 I k= im ⇡ 1 mx = x + e . (5.161) 2m2 m ✓ ◆ Putting the two together, we get ikx 1 e ⇡ 1 m x J(x)= dk = x + e | |. (5.162) (k2 + m2)2 2m2 | | m Z1 ✓ ◆ as the Fourier transform of 1/(k2 + m2)2. Example 5.30 (Integral of a Complex Gaussian). As another example of the use of ghost contours, let us use one to do the integral

2 I = 1 ewx dx (5.163) Z1 in which the real part of the nonzero complex number w = u + iv = ⇢ei is negative or zero ⇡ 3⇡ u 0 . (5.164)  () 2   2 We ﬁrst write the integral I as twice that along half the x-axis

2 I =2 1ewx dx. (5.165) Z0 If we promote x to a complex variable z = rei✓,thenwz2 will be negative if +2✓ = ⇡, that is, if ✓ =(⇡ ) /2 where in view of (5.164) ✓ lies in the interval ⇡/4 ✓ ⇡/4.   The closed pie-shaped contour of Fig. 5.7 (down the real axis from z =0 to z = R, along the arc z = R exp(i✓0) as ✓0 goes from 0 to ✓, and then down the line z = r exp(i✓) from z = R exp(i✓)toz = 0) encloses no singularities of the function f(z)=exp(wz2). Hence the integral of exp(wz2) along that contour vanishes. 5.14 Ghost Contours 219

Pie-Shaped Contour

z = Reiθ

θ > z =0 z = R

Figure 5.7 The integral of the entire function exp(wz2) along the pie- shaped closed contour vanishes by Cauchy’s theorem.

To show that the arc is a ghost contour, we bound it by

✓ ✓ (u+iv)R2e2i✓0 2 2 e Rd✓0 exp uR cos 2✓0 vR sin 2✓0 Rd✓0  Z0 Z0 ✓ ⇥ ⇤ vR2 sin 2✓ e 0 Rd✓0. (5.166)  Z0 Here v sin 2✓ 0, and so if v is positive, then so is ✓ .Then0 ✓ ⇡/4, 0 0  0  and so sin(2✓ ) 4✓ /⇡.Thussinceu<0, we have the upper bound 0 0 2 ✓ ✓ 4vR ✓0/⇡ (u+iv)R2e2i✓0 4vR2✓ /⇡ ⇡(e 1) e Rd✓0 e 0 Rd✓0 = (5.167)  4vR Z0 Z0 which vanishes in the limit R .(Ifv is negative, then so is ✓ ,the !1 0 pie-shaped contour is in the fourth quadrant, sin(2✓ ) 4✓ /⇡, and the in- 0  0 equality (5.167) holds with absolute-value signs around the second integral.) 220 Complex-Variable Theory Since by Cauchy’s integral theorem (5.17) the integral along the pie- shaped contour of Fig. 5.7 vanishes, it follows that

0 1 2 I + ewz dz =0. (5.168) 2 i✓ ZRe But the choice ✓ =(⇡ ) /2 implies that on the line z = r exp(i✓)the quantity wz2 is negative, wz2 = ⇢r2.Thuswithdz =exp(i✓)dr,wehave Rei✓ R wz2 i✓ ⇢r2 I =2 e dz =2e e dr (5.169) Z0 Z0 so that as R !1 i✓ 1 ⇢r2 i✓ ⇡ ⇡ I =2e e dr = e = . (5.170) ⇢ ⇢e 2i✓ Z0 r r Finally from ✓ =(⇡ ) /2 and w = ⇢ exp(i), we ﬁnd that for Re w 0  2 ⇡ 1 ewx dx = (5.171) w Z1 r as long as w = 0. Shifting x by a complex number b,westillhave 6 1 w(x b)2 ⇡ e dx = (5.172) w Z1 r as long as Re w<0. If w = ia = 0, and b is real, then 6

1 ia(x b)2 i⇡ e dx = . (5.173) a Z1 r The simpler integral (5.129) applies when m>0 and z is an arbitrary complex number

1 m2(x+z)2 p⇡ e dx = . (5.174) m Z1 These last two formulas are used in chapter 19 on path integrals.

Let us try to express the line integral of a not necessarily analytic function f(x, y)=u(x, y)+iv(x, y) along a closed counterclockwise contour as an C integral over the surface enclosed by the contour. The contour integral is

(u + iv)(dx + idy)= (udx vdy)+i (vdx+ udy). (5.175) IC IC IC Now since the contour is counterclockwise, the di↵erential dx is negative at C 5.14 Ghost Contours 221 the top of the curve with coordinates (x, y+(x)) and positive at the bottom (x, y (x)). So the ﬁrst line integral is the surface integral

udx= [u(x, y (x)) u(x, y+(x))] dx IC Z y+(x) = uy(x, y)dy dx " y (x) # Z Z = u dxdy = u da (5.176) y | | y Z Z in which da = dxdy is a positive element of area. Similarly, we ﬁnd | | i vdx= i v dxdy = i v da. (5.177) y | | y IC Z Z The dy integrals are then:

vdy= v dxdy = v da (5.178) x | | x IC Z Z i udy = i u dxdy = i u da. (5.179) x | | x IC Z Z Combining (5.175–5.179), we ﬁnd

(u + iv)(dx + idy)= (u + v ) da + i ( v + u ) da. (5.180) y x y x IC Z Z This formula holds whether or not the function f(x, y) is analytic. But if f(x, y) is analytic on and within the contour , then it satisﬁes the Cauchy- C Riemann conditions (5.10) within the contour, and so both surface integrals vanish. The contour integral then is zero, which is Cauchy’s integral theorem (5.30). The contour integral of the function f(x, y)=u(x, y)+iv(x, y)di↵ers from zero (its value if f(x, y) is analytic in z = x + iy)) by the surface integrals of u + v and u v y x x y 2 2 2 2 f(z)dz = (u + iv)(dx + idy) = (u + v )da + (u v )da y x x y I I Z Z C C (5.181) which vanish when f = u + iv satisﬁes the Cauchy-Riemann conditions (5.10).

Example 5.31 (The integral of a nonanalytic function). The integral formula (5.180) can help us evaluate contour integrals of functions that are not 222 Complex-Variable Theory analytic. The function 1 1 f(x, y)= (5.182) x + iy + i✏ 1+x2 + y2 is the product of an analytic function 1/(z +i✏), where ✏ is tiny and positive, and a nonanalytic real one r(x, y)=1/(1+z⇤z).Thei✏ pushes the pole in u + iv =1/(z + i✏) into the lower half plane. The real and imaginary parts of f are x 1 U(x, y)=u(x, y) r(x, y)= (5.183) x2 +(y + ✏)2 1+x2 + y2 and y ✏ 1 V (x, y)=v(x, y) r(x, y)= . (5.184) x2 +(y + ✏)2 1+x2 + y2 We will use (5.180) to compute the contour integral I of f along the real axis from to and then along the ghost contour z = x + iy = Rei✓ for 1 1 0 ✓ ⇡ and R around the upper half plane   !1 1 1 I = f(x, y) dz = dx dy [ Uy Vx + i ( Vy + Ux)] . (5.185) 0 I Z1 Z Since u and v satisfy the Cauchy-Riemann conditions (5.10), the terms in the area integral simplify to U V = ur vr and V +U = vr +ur . y x y x y x y x So the integral I is

1 1 I = dx dy [ ury vrx + i( vry + urx)] (5.186) 0 Z1 Z or explicitly

2 2 1 1 2✏x 2i(x + y + ✏y) I = dx dy 2 . (5.187) 0 [x2 +(y + ✏)2](1+x2 + y2) Z1 Z We let ✏ 0 and ﬁnd ! 1 1 1 I = 2i dx dy 2 . (5.188) 0 (1 + x2 + y2) Z1 Z Changing variables to ⇢2 = x2 + y2,wehave ⇢ d 1 I = 4⇡i 1d⇢ =2⇡i 1d⇢ = 2⇡i. (5.189) (1 + ⇢2)2 d⇢ 1+⇢2 Z0 Z0 which is simpler than evaluating the integral (5.185) directly. 5.15 Logarithms and Cuts 223 5.15 Logarithms and Cuts

By definition, a function f is single valued; it maps every number z in its domain into a unique image f(z). A function that maps only one number z in its domain into each f(z) in its range is said to be one to one.A 1 one-to-one function f(z) has a well-defined inverse function f (z). The exponential function is one to one when restricted to the real numbers. It maps every real number x into a positive number exp(x). It has an inverse function ln(x) that maps every positive number exp(x) back into x.Butthe exponential function is not one to one on the complex numbers because exp(z +2⇡ni)=exp(z) for every integer n. Because it is many to one,the exponential function has no inverse function on the complex numbers. Its would-be inverse function ln maps it to ln(exp(z)) or z +2⇡ni which is not unique. It has in it an arbitrary integer n. In other words, when exponentiated, the logarithm of a complex number z returns exp(ln z)=z.Soifz = r exp(i✓), then a suitable logarithm is ln z = ln r +i✓. But what is ✓? In the polar representation of z, the argument ✓ can just as well be ✓ +2⇡n because both give z = r exp(i✓)=r exp(i✓ + i2⇡n). So ln r + i✓ + i2⇡n is a correct value for ln[r exp(i✓)] for every integer n. People usually want one of the correct values of a logarithm, rather than all of them. Two conventions are common. In the first convention, the angle ✓ is zero along the positive real axis and increases continuously as the point z moves counterclockwise around the origin, until at points just below the positive real axis, ✓ =2⇡ ✏ is slightly less than 2⇡. In this convention, the value of ✓ drops by 2⇡ as one crosses the positive real axis moving counterclockwise. This discontinuity on the positive real axis is called a cut. The second common convention puts the cut on the negative real axis. Here the value of ✓ is the same as in the first convention when the point z is in the upper half-plane. But in the lower half-plane, ✓ decreases from 0to ⇡ as the point z moves clockwise from the positive real axis to just below the negative real axis, where ✓ = ⇡ + ✏. As one crosses the negative real axis moving clockwise or up, ✓ jumps by 2⇡ while crossing the cut. The two conventions agree in the upper half-plane but di↵er by 2⇡ in the lower half-plane. Sometimes it is convenient to place the cut on the positive or negative imaginary axis—or along a line that makes an arbitrary angle with the real axis. In any particular calculation, we are at liberty to define the polar angle ✓ by placing the cut anywhere we like, but we must not change from one convention to another in the same computation. 224 Complex-Variable Theory 5.16 Powers and Roots The logarithm is the key to many other functions to which it passes its arbitrariness. For instance, any power a of z = r exp(i✓) is defined as za =exp(a ln z)=exp[a (ln r + i✓ + i2⇡n)] = ra eia✓ ei2⇡na. (5.190) So za is not unique unless a is an integer. The square root, for example, has a sign ambiguity pz =exp 1 (ln r + i✓ + i2⇡n) = prei✓/2 ein⇡ =( 1)n prei✓/2 (5.191) 2 and changes⇥ sign when we change⇤ ✓ by 2⇡ as we cross a cut. The m-th root ln z mpz = z1/m =exp (5.192) m ✓ ◆ changes by exp( 2⇡i/m) when we cross a cut and change ✓ by 2⇡. And ± when a = u + iv is a complex number, za is a a ln z (u+iv)(ln r+i✓+i2⇡n) u+iv ( v+iu)(✓+2⇡n) z = e = e = r e (5.193) which changes by exp[2⇡( v + iu)] as we cross a cut. Example 5.32 (ii). The number i =exp(i⇡/2+i2⇡n) for any integer n.So the general value of ii is ii =exp[i(i⇡/2+i2⇡n)] = exp( ⇡/2 2⇡n). One can define a sequence of mth-root functions ln r + i(✓ +2⇡n) z1/m =exp (5.194) n m ⇣ ⌘ ✓ ◆ one for each integer n. These functions are the branches of the mth-root function. One can merge all the branches into one multivalued mth-root function. Using a convention for ✓, one would extend the n = 0 branch to the n = 1 branch by winding counterclockwise around the point z = 0. One would encounter no discontinuity as one passed from one branch to another. The point z = 0, where any cut starts, is called a branch point because by winding around it, one passes smoothly from one branch to another. Such branches, introduced by Riemann, can be associated with any multivalued analytic function not just with the mth root. Example 5.33 (Explicit Square Roots). If the cut in the square root pz is on the negative real axis, then an explicit formula for the square root of x + iy is

x2 + y2 + x x2 + y2 x x + iy = + i sign(y) (5.195) sp 2 sp 2 p 5.16 Powers and Roots 225 in which sign(y) = sgn(y)=y/ y . On the other hand, if the cut in the | | square root pz is on the positive real axis, then an explicit formula for the square root of x + iy is

x2 + y2 + x x2 + y2 x x + iy = sign(y) + i (5.196) sp 2 sp 2 p (exercise 5.29). Example 5.34 (Cuts). Cuts are discontinuities, so people place them where they do the least harm. For the function f(z)= z2 1= (z 1)(z + 1) (5.197) two principal conventions workp well. Wep could put the cut in the deﬁnition of the angle ✓ along either the positive or the negative real axis. And we’d get a bonus: the sign discontinuity (a factor of 1) from pz 1 would cancel the one from pz + 1 except for 1 z 1. So the function f(z) would   have a discontinuity or a cut only for 1 z 1.   But now suppose we had to work with the function f(z)= z2 +1= (z i)(z + i). (5.198) If we used one of the usual conventions,p p we’d have two semi-inﬁnite cuts. So we put the ✓-cut on the positive or negative imaginary axis, and the function f(z) now has a cut running along the imaginary axis only from i to i. Example 5.35 (Square-root cut). To evaluate the integral dx I = 1 ,a>0, (5.199) s (x + a)2px Z0 we put the cut on the positive real axis. The integral backwards along and just below the positive real axis 0 dx 0 dx I = = = Is (5.200) (x + a)2px i✏ (x + a)2px + i✏ Z1 Z1 is the same as Is since a minus sign from the square root cancels the minus sign due to the backwards direction. Since z lim | | =0, (5.201) z z + a 2 pz | |!1 | | | | the integrals of f(z)=1/[(z + a)2pz ] along the contours z = R exp(i✓) for 0 <✓<⇡and for ⇡<✓<2⇡ vanish as R . So these contours are !1 ghost contours. We then add a pair of cancelling integrals along the negative 226 Complex-Variable Theory real axis up to the pole at z = a and then add a clockwise loop around C it. As in Fig. 5.8, the integral along this collection of contours encloses no singularity and therefore vanishes

0=Is + I + I + + I + I . (5.202) G G C

Thus 2Is = I , and so from Cauchy’s integral formula (5.40) for n = 1, C we have

1 1 1 d 1/2 ⇡ Is = I = dz = i⇡ z = (5.203) 2 C 2 (z + a)2pz dz 2a3/2 I z= a C which one may check with the Mathematica command Assuming[a>0, Integrate[ 1/((x + a)2 Sqrt[x]), x,0,Inﬁnity ]]. ⇤ { } Example 5.36 (Contour integral with a cut). Let’s compute the integral xa I = 1 dx (5.204) (x + 1)2 Z0 for 1

Figure 5.8 The integrals of f(z)=1/[(x + a)2 pz ] as well as that of a 2 f(z)=z /(z + 1) along the ghost contours + and and the contours G G , , and + vanishes because the combined contour encircles no poles of eitherC I f(z).I The cut (solid line) runs from the origin to inﬁnity along the positive real axis.

Now the sum of all these contour integrals is zero because it is a closed contour that encloses no singularity. So we have 0= 1 e2⇡ai I +2⇡i a e⇡ai (5.209) or xa ⇡a I = 1 dx = (5.210) (x + 1)2 sin(⇡a) Z0 as the value of the integral (5.204). Example 5.37 (Euler’s reﬂection formula). The beta function (4.76) for x = z and y =1 z is the integral 1 z 1 z B(z,1 z)=(z)(1 z)= t (1 t) dt. (5.211) Z0 228 Complex-Variable Theory Setting t = u/(1 + u), so that u = t/(1 t) and dt =1/(1 + u)2,wehave uz 1 B(z,1 z)= 1 du. (5.212) 1+u Z0 z 1 We integrate f(u)=u /(1+u) along the contour of the preceding example (5.36) which includes the ghost contour = + and runs down both G G [G sides of the cut along the positive real axis. Since f(u) is analytic inside the contour, the integral vanishes

0= f(u) du + f(u) du + f(u) du + f(u) du. (5.213) + ZI ZG ZC ZI The clockwise contour is C z 1 u z 1 i⇡(z 1) i⇡z du = 2⇡i( 1) = 2⇡ie =2⇡ie . (5.214) 1+u ZC The contour runs just above the positive real axis, and the integral of f(u) I+ along it is the desired integral B(z,1 z). The contour runs backwards I and just below the cut where u = u i✏ | | 2⇡i ✏ z 1 z 1 1 ( u e ) 1 u f(u) du = | | du = e2⇡iz du. (5.215) 0 1+u 0 1+u ZI Z Z Thus the vanishing (5.213) of the contour integral

0=B(z,1 z)+2⇡i ei⇡z e2⇡iz B(z,1 z) (5.216) gives us Euler’s reﬂection formula ⇡ B(z,1 z)=(z)(1 z)= . (5.217) sin ⇡z

Example 5.38 (Integral with a square root). Consider the integral 1 1 I = dx (5.218) 2 1 (x k) p1 x Z in which the constant k lies anywhere in the complex plane but not on the interval [ 1, 1]. We promote x to a complex variable z, put the cut on the negative real axis, and write the square root as p1 x2 = ipx2 1= i (z 1)(z + 1) so as ✏ 0withx = z = i✏ we have 1 = p1 x2 = ! ip 1= ii = 1. The function f(z)=1/[(z k)( i) (z 1)(z + 1)] is p analytic everywhere except along a cut on the interval [ 1, 1] and at z = k. p The circle z = Rei✓ for 0 ✓ 2⇡ is a ghost contour as R .Ifwe   !1 5.17 Conformal Mapping 229 shrink-wrap this counterclockwise contour around the pole at z = k and the interval [ 1, 1], then we get 0 = 2I +2⇡i/[( i)pk 1pk + 1] or ⇡ I = . (5.219) pk 1pk +1 So if k = 2, then I = ⇡/p3, while if k = 2, then I = ⇡/p3.

5.17 Conformal Mapping An analytic function f(z) maps curves in the z plane into curves in the f(z) plane. In general, this mapping preserves angles. To see why, we consider the angle d✓ between two tiny complex lines dz = ✏ exp(i✓) and dz0 = ✏ exp(i✓0) that radiate from the same point z. This angle d✓ = ✓ ✓ is the phase of 0 the ratio i✓0 dz0 ✏e i(✓ ✓) = = e 0 . (5.220) dz ✏ei✓ Let’s use w = ⇢ei for f(z). Then the analytic function f(z) maps dz into

dw = f(z + dz) f(z) f 0(z) dz (5.221) ⇡ and dz0 into

dw0 = f(z + dz0) f(z) f 0(z) dz0. (5.222) ⇡ The angle d = between dw and dw is the phase of the ratio 0 0 dw ei0 f (z) dz dz ei✓0 0 0 0 0 i(✓0 ✓) = i = = = i✓ = e . (5.223) dw e f 0(z) dz dz e

So as long as the derivative f 0(z) does not vanish, the angle in the w-plane is the same as the angle in the z-plane d = d✓. (5.224) Analytic functions preserve angles. They are conformal maps. What if f (z) = 0? In this case, dw f 00 (z) dz2/2 and dw f 00 (z) dz 2/2, 0 ⇡ 0 ⇡ 0 and so the angle d = d d between these two tiny complex lines is the 0 phase of the ratio

i0 00 2 2 dw0 e f (z) dz0 dz0 2i(✓ ✓) = = = = e 0 . (5.225) dw ei f 00 (z) dz2 dz2 So angles are doubled, d =2d✓. 230 Complex-Variable Theory In general, if the ﬁrst non-zero derivative is f (n)(z), then

i0 (n) n n dw0 e f (z) dz0 dz0 ni(✓ ✓) = = = = e 0 (5.226) dw ei f (n)(z) dzn dzn and so d = nd✓. The angles increase by a factor of n. Example 5.39 (zn). The function f(z)=zn has only one nonzero derivative f (k)(0) = n! at the origin z = 0. So at z = 0 the map z zn scales nk ! angles by n, d = nd✓, but at z = 0 its the ﬁrst derivative f (1)(z)=nzn 1 6 is not equal to zero. So zn is conformal except at the origin.

Example 5.40 (The M¨obius transformation). The function az + b f(z)= (5.227) cz + d maps (straight) lines into lines and circles and circles into circles and lines, unless ad = bc in which case it is the constant b/d.

5.18 Cauchy’s principal value Suppose that f(x) is di↵erentiable or analytic at and near the point x = 0, and that we wish to evaluate the integral b f(x) K =lim dx (5.228) ✏ 0 a x i✏ ! Z for a>0 and b>0. First we regularize the pole at x =0byusingamethod devised by Cauchy f(x) f(x) b f(x) K =lim lim dx + dx + dx . (5.229) 0 ✏ 0 a x i✏ x i✏ x i✏ !  ! ✓Z Z Z ◆ In the first and third integrals, since x ,wemayset✏ =0 | | f(x) b f(x) f(x) K =lim dx + dx +lim lim dx . (5.230) 0 a x x 0 ✏ 0 x i✏ ! ✓Z Z ◆ ! ! Z We’ll discuss the first two integrals before analyzing the last one. The limit of the first two integrals is called Cauchy’s principal value b f(x) f(x) b f(x) P dx lim dx + dx . (5.231) a x ⌘ 0 a x x Z ! ✓Z Z ◆ 5.18 Cauchy’s principal value 231 If the function f(x) is nearly constant near x = 0, then the large negative values of 1/x for x slightly less than zero cancel the large positive values of 1/x for x slightly greater than zero. The point x = 0 is not special; Cauchy’s principal value about x = y is defined by the limit

b f(x) y f(x) b f(x) P dx lim dx + dx . (5.232) a x y ⌘ 0 a x y y+ x y Z ! ✓Z Z ◆ Using Cauchy’s principal value, we may write the quantity K as

b f(x) f(x) K = P dx +lim lim dx . (5.233) a x 0 ✏ 0 x i✏ Z ! ! Z To evaluate the second integral, we use di↵erentiability of f(x) near x =0 to write f(x)=f(0) + xf 0(0) and then extract the constants f(0) and f 0(0)

f(x) f(0) + xf (0) lim lim dx =limlim dx 0 0 ✏ 0 x i✏ 0 ✏ 0 x i✏ ! ! Z ! ! Z dx xdx = f(0) lim lim + f 0(0) lim lim 0 ✏ 0 x i✏ 0 ✏ 0 x i✏ ! ! Z ! ! Z dx = f(0) lim lim + f 0(0) lim 2 0 ✏ 0 x i✏ 0 ! ! Z ! dx = f(0) lim lim . (5.234) 0 ✏ 0 x i✏ ! ! Z Now since 1/(z i✏) is analytic in the lower half-plane, we may deform the straight contour from x = to x = into a tiny semicircle that avoids the point x =0bysettingz = ei✓ and letting ✓ run from ⇡ to 2⇡

b f(x) 1 K = P dx + f(0) lim lim dz . (5.235) a x 0 ✏ 0 z i✏ Z ! ! Z We now can set ✏ = 0 and so write K as b 2⇡ f(x) i✓ 1 K = P dx + f(0) lim ie d✓ i✓ a x 0 ⇡ e Z ! Z b f(x) = P dx + i⇡f(0). (5.236) a x Z Recalling the deﬁnition (5.228) of K,wehave

b f(x) b f(x) lim dx = P dx + i⇡f(0) (5.237) ✏ 0 a x i✏ a x ! Z Z 232 Complex-Variable Theory for any function f(x) that is di↵erentiable at x = 0. Physicists write this as

1 1 1 1 = P + i⇡(x) and = P i⇡(x) (5.238) x i✏ x x + i✏ x or as 1 1 = P i⇡(x y). (5.239) x y i✏ x y ⌥ ± Example 5.41 (An application of Cauchy’s trick). We use (5.238) to evaluate the integral 1 1 I = 1dx (5.240) x + i✏ 1+x2 Z1 as 1 1 (x) I = P 1dx i⇡ 1dx . (5.241) x 1+x2 1+x2 Z1 Z1 Because the function 1/x(1 + x2) is odd, the principal part is zero. The integral over the delta function gives unity, so we have I = i⇡. Example 5.42 (Cubic form of Cauchy’s principal value). Cauchy’s principal value of the integral b f(x) P 3 dx (5.242) a x Z is ﬁnite as long as f(z) is analytic at z = 0 with a vanishing ﬁrst derivative there, f 0(0) = 0. In this case Cauchy’s integral formula (5.39) says that b b 2⇡ i✓ f(x) f(x) i✓ f(e ) dx 3 = P dx 3 +lim ie d✓ 3 a (x i✏) a x 0 ⇡ (ei✓) Z Z ! Z (5.243) b f(x) ⇡ = P dx 3 + i f 00(0). a x 2 Z

Example 5.43 (Cauchy’s Principal Value). By explicit use of the formula dx 1 x + a = ln (5.244) x2 a2 2a x a Z one may show (exercise 5.31) that dx a dx dx P 1 = + 1 = 0 (5.245) x2 a2 x2 a2 x2 a2 Z0 Z0 Za+ a result we’ll use in section 5.21. 5.18 Cauchy’s principal value 233 Example 5.44 (sin k/k). To compute the integral dk I = 1 sin k (5.246) s k Z0 which we used to derive the formula (3.113) for the Green’s function of the laplacian in three dimensions, we ﬁrst express Is as an integral along the whole real axis

1 dk ik ik 1 dk ik Is = e e = e (5.247) 0 2ik 2ik Z ⇣ ⌘ Z1 by which we actually mean the Cauchy principal part

ik ik ik e 1 e 1 e Is =lim dk + dk = P dk . (5.248) 0 2ik 2ik 2ik ! ✓Z1 Z ◆ Z1 Using Cauchy’s trick (5.238), we have

eik eik eik I = P 1 dk = 1 dk + 1 dk i⇡ (k) . (5.249) s 2ik 2i(k + i✏) 2i Z1 Z1 Z1 To the ﬁrst integral, we add a ghost contour around the upper half-plane. For the contour from k = L to k = L + iH and then to k = L + iH and then down to k = L, one may show (exercise 5.34) that the integral of exp(ik)/k vanishes in the double limit L and H . With this ghost !1 !1 contour, the ﬁrst integral therefore vanishes because the pole at k = i✏ is in the lower half plane. The delta function in the second integral then gives ⇡/2, so that eik ⇡ ⇡ I = dk + = (5.250) s 2i(k + i✏) 2 2 I as stated in (3.112).

Example 5.45 (The Feynman Propagator). Adding i✏ to the denominator ± of a pole term of an integral formula for a function f(x) can slightly shift the pole into the upper or lower half plane, causing the pole to contribute if a ghost contour goes around the upper half-plane or the lower half-plane. Such an i✏ can impose a boundary condition on a Green’s function. The Feynman propagator F (x) is a Green’s function for the Klein- Gordon di↵erential operator (Weinberg, 1995, pp. 274–280)

(m2 2) (x)=4(x) (5.251) F 234 Complex-Variable Theory in which x =(x0, x) and @2 @2 2 = = (5.252) 4@t2 4@(x0)2 is the four-dimensional version of the laplacian . Here 4(x)isthe 4⌘r·r four-dimensional Dirac delta function (3.36) d4q d4q 4(x)= exp[i(q x q0x0)] = eiqx (5.253) (2⇡)4 · (2⇡)4 Z Z in which qx = q x q0x0 is the Lorentz-invariant inner product of the 4- · vectors q and x. There are many Green’s functions that satisfy Eq.(5.251). Feynman’s propagator F (x) is the one that satisﬁes boundary conditions that will become evident when we analyze the e↵ect of its i✏

0 0 d4q exp(iqx) d3q dq0 eiq x iq x (x)= = 1 · . (5.254) F (2⇡)4 q2 + m2 i✏ (2⇡)3 2⇡ q2 + m2 i✏ Z Z Z1 2 2 The quantity Eq = q + m is the energy of a particle of mass m and momentum q in natural units with the speed of light c = 1. Using this p abbreviation and setting ✏0 = ✏/2Eq, we may write the denominator as

2 2 0 2 2 0 0 2 q + m i✏ = q q q + m i✏ = E i✏0 q E i✏0 + q + ✏0 · q q (5.255) 2 0 in which ✏0 is negligible. Dropping the prime on ✏,wedotheq integral 0 1 dq iq0x0 1 I(q)= e 0 0 . (5.256) 2⇡ [q (Eq i✏)] [q ( Eq + i✏)] Z1 As shown in Fig. 5.9, the integrand

iq0x0 1 e (5.257) [q0 (E i✏)] [q0 ( E + i✏)] q q has poles at E i✏ and at E + i✏.Whenx0 > 0, we can add a ghost q q contour that goes clockwise around the lower half-plane and get

0 iEqx 1 0 I(q)=ie x > 0. (5.258) 2Eq When x0 < 0, our ghost contour goes counterclockwise around the upper half-plane, and we get

0 1 I(q)=ieiEqx x0 < 0. (5.259) 2Eq 5.18 Cauchy’s principal value 235 Ghost Contours and the Feynman Propagator

x0 < 0

− E q + iϵ >

E q − iϵ

x0 > 0 <

0 Figure 5.9 In equation (5.257), the function f(q ) has poles at (Eq i✏), and the function exp( iq0x0) is exponentially suppressed in the± lower half plane if x0 > 0 and in the upper half plane if x0 < 0. So we can add a ghost contour (. . . ) in the LHP if x0 > 0 and in the UHP if x0 < 0.

Using the step function ✓(x)=(x + x )/2, we combine (5.258) and (5.259) | | 0 0 1 0 iEqx 0 iEqx iI(q)= ✓(x ) e + ✓( x ) e . (5.260) 2Eq h i In terms of the Lorentz-invariant function 1 d3q (x)= exp[i(q x E x0)] (5.261) + (2⇡)3 2E · q Z q 236 Complex-Variable Theory and with a factor of i, Feynman’s propagator (5.254) is i (x)=✓(x0) (x)+✓( x0) (x, x0). (5.262) F + + The integral (5.261) defining +(x) is insensitive to the sign of q, and so 1 d3q ( x)= exp[i( q x + E x0)] (5.263) + (2⇡)3 2E · q Z q 1 d3q = exp[i(q x + E x0)] = (x, x0). (2⇡)3 2E · q + Z q Thus we arrive at the Standard form of the Feynman propagator i (x)=✓(x0) (x)+✓( x0) ( x). (5.264) F + + The annihilation operators a(q) and the creation operators a†(p) of a scalar field (x) satisfy the commutation relations 3 [a(q),a†(p)] = (q p) and [a(q),a(p)] = [a†(q),a†(p)] = 0. (5.265) Thus the commutator of the positive-frequency part d3p +(x)= exp[i(p x p0x0)] a(p) (5.266) (2⇡)32p0 · Z + of a scalar field = p+ with its negative-frequency part 3 d q 0 0 (y)= exp[ i(q y q y )] a†(q) (5.267) (2⇡)32q0 · Z is the Lorentz-invariantp function (x y) + 3 3 + d pd q ipx iqy [ (x),(y)] = e [a(p),a†(q)] 3 0 0 Z (2⇡) 2 q p 3 d p ip(x y) = p e = (x y) (5.268) (2⇡)32p0 + Z in which p(x y)=p (x y) p0(x0 y0). · At points x that are space-like, that is, for which x2 = x2 (x0)2 r2 > 0, ⌘2 the Lorentz-invariant function +(x) depends only upon r =+px and has the value (Weinberg, 1995, p. 202) m (x)= K (mr) (5.269) + 4⇡2r 1 in which the Hankel function K1 is ⇡ 1 z z 1 K (z)= [J (iz)+iN (iz)] = + ln + + ... (5.270) 1 2 1 1 z 2 2 2  ⇣ ⌘ 5.19 Dispersion Relations 237 where J1 is the first Bessel function, N1 is the first Neumann function, and =0.57721 ... is the Euler-Mascheroni constant. The Feynman propagator arises most simply as the mean value in the vacuum of the time-ordered product of the fields (x) and (y)

(x)(y) ✓(x0 y0)(x)(y)+✓(y0 x0)(y)(x). (5.271) T{ }⌘ The operators a(p) and a (p) respectively annihilate the vacuum ket a(p) 0 = † | i 0 and bra 0 a (p) = 0, and so by (5.266 & 5.267) do the positive- and h | † negative-frequency parts of the ﬁeld +(z) 0 = 0 and 0 (z) = 0. Thus | i h | the mean value in the vacuum of the time-ordered product is

0 (x)(y) 0 = 0 ✓(x0 y0)(x)(y)+✓(y0 x0)(y)(x) 0 h |T { }| i h | 0 0 + 0 0 + | i = 0 ✓(x y ) (x)(y)+✓(y x ) (y)(x) 0 h | 0 0 + | i = 0 ✓(x y )[ (x),(y)] h | 0 0 + + ✓(y x )[ (y),(x)] 0 . (5.272) | i But by (5.268), these commutators are (x y) and (y x). Thus the + + mean value in the vacuum of the time-ordered product

0 (x)(y) 0 = ✓(x0 y0) (x y)+✓(y0 x0) (y x) h |T { }| i + + = i (x y) (5.273) F is the Feynman propagator (5.262) multiplied by i.

5.19 Dispersion Relations In many physical contexts, functions occur that are analytic in the upper half-plane. Suppose for instance that fˆ(t) is a transfer function that deter- mines an e↵ect e(t) due to a cause c(t)

1 e(t)= dt0 fˆ(t t0) c(t0). (5.274) Z1 If the system is causal, then the transfer function fˆ(t t ) is zero for t t < 0, 0 0 and so its Fourier transform dt dt f(z)= 1 fˆ(t) eizt = 1 fˆ(t) eizt (5.275) p2⇡ 0 p2⇡ Z1 Z will be analytic in the upper half-plane and will shrink as the imaginary part of z = x + iy increases. 238 Complex-Variable Theory So let us assume that the function f(z) is analytic in the upper half-plane and on the real axis and further that

lim f(rei✓) = 0 for 0 ✓ ⇡. (5.276) r !1 | |   By Cauchy’s integral formula (5.36), if z0 lies in the upper half-plane, then f(z0) is given by the closed counterclockwise contour integral 1 f(z) f(z )= dz (5.277) 0 2⇡i z z I 0 in which the contour runs along the real axis and then loops over the semicircle lim rei✓ for 0 ✓ ⇡. (5.278) r !1   Our assumption (5.276) about the behavior of f(z) in the upper half plane implies that this contour (5.278) is a ghost contour because its modulus is bounded by 1 f(rei✓) r lim | | d✓ =lim f(rei✓) =0. (5.279) r 2⇡ r r | | !1 Z !1 So we may drop the ghost contour and write f(z0) as

1 1 f(x) f(z0)= dx. (5.280) 2⇡i x z0 Z1 Letting the imaginary part y0 of z0 = x0 + iy0 shrink to ✏

1 1 f(x) f(x0)= dx (5.281) 2⇡i x x0 i✏ Z1 and using Cauchy’s trick (5.239), we get

1 1 f(x) i⇡ 1 f(x0)= P dx + f(x) (x x0) dx (5.282) 2⇡i x x0 2⇡i Z1 Z1 or 1 1 f(x) 1 f(x0)= P dx + f(x0) (5.283) 2⇡i x x0 2 Z1 which is the dispersion relation

1 1 f(x) f(x0)= P dx. (5.284) ⇡i x x0 Z1 5.20 Kramers-Kronig Relations 239 If we break f(z)=u(z)+iv(z) into its real u(z) and imaginary v(z) parts, then this dispersion relation (5.284)

1 1 u(x)+iv(x) u(x0)+iv(x0)= P dx (5.285) ⇡i x x0 Z1 1 v(x) i u(x) = P 1 dx P 1 dx ⇡ x x0 ⇡ x x0 Z1 Z1 breaks into its real and imaginary parts

1 1 v(x) 1 1 u(x) u(x0)= P dx and v(x0)= P dx (5.286) ⇡ x x0 ⇡ x x0 Z1 Z1 which express u and v as Hilbert transforms of each other. In applications of dispersion relations, the function f(x) for x<0 sometimes is either physically meaningless or experimentally inaccessible. In such cases, there may be a symmetry that relates f( x)tof(x). For instance, if f(x) is the Fourier transform of a real function fˆ(k), then by Eq.(3.25) it obeys the symmetry relation

f ⇤(x)=u(x) iv(x)=f( x)=u( x)+iv( x), (5.287) which says that u is even, u( x)=u(x), and v odd, v( x)= v(x). Using these symmetries, one may show (exercise 5.37) that the Hilbert transfor- mations (5.286) become 2 xv(x) 2x u(x) u(x )= P 1 dx and v(x )= 0 P 1 dx (5.288) 0 ⇡ x2 x2 0 ⇡ x2 x2 Z0 0 Z0 0 which do not require input at negative values of x.

5.20 Kramers-Kronig Relations i!t If we use E for the current density J and E(t)=e E for the electric ﬁeld, then Maxwell’s equation B = µJ + ✏µE˙ becomes r⇥ B = i!✏µ 1+i E i!n2✏ µ E (5.289) r⇥ ✏! ⌘ 0 0 ⇣ ⌘ in which the squared index of refraction is ✏µ n2(!)= 1+i . (5.290) ✏0µ0 ✏! ⇣ ⌘ The imaginary part of n2 represents the scattering of light mainly by electrons. At high frequencies in nonmagnetic materials n2(!) 1, and so ! 240 Complex-Variable Theory Kramers and Kronig applied the Hilbert-transform relations (5.288) to the function n2(!) 1 in order to satisfy condition (5.276). Their relations are 2 ! Im(n2(!)) Re(n2(! )) = 1 + P 1 d! (5.291) 0 ⇡ !2 !2 Z0 0 and 2 2! 1 Re(n (!)) 1 Im(n2(! )) = 0 P d!. (5.292) 0 ⇡ !2 !2 Z0 0 What Kramers and Kronig actually wrote was slightly di↵erent from these dispersion relations (5.291 & 5.292). H. A. Lorentz had shown that the index of refraction n(!) is related to the forward scattering amplitude f(!) for the scattering of light by a density N of scatterers (Sakurai, 1982)

2⇡c2 n(!)=1+ Nf(!). (5.293) !2 They used this formula to infer that the real part of the index of refraction approached unity in the limit of inﬁnite frequency and applied the Hilbert transform (5.288)

2 1 !0 Im[n(!0)] Re[n(!)] = 1 + P d!0. (5.294) ⇡ ! 2 !2 Z0 0 The Lorentz relation (5.293) expresses the imaginary part Im[n(!)] of the index of refraction in terms of the imaginary part of the forward scattering amplitude f(!) Im[n(!)] = 2⇡(c/!)2NIm[f(!)]. (5.295)

And the optical theorem relates Im[f(!)] to the total cross-section 4⇡ 4⇡c = Im[f(!)] = Im[f(!)]. (5.296) tot k ! | | Thus we have Im[n(!)] = cNtot/(2!), and by the Lorentz relation (5.293) Re[n(!)] = 1 + 2⇡(c/!)2NRe[f(!)]. Insertion of these formulas into the Kramers-Kronig integral (5.294) gives a dispersion relation for the real part of the forward scattering amplitude f(!) in terms of the total cross-section

2 ! 1 tot(!0) Re[f(!)] = P d!0. (5.297) 2⇡2c ! 2 !2 Z0 0 5.21 Phase and Group Velocities 241 5.21 Phase and Group Velocities Suppose A(x,t) is the amplitude

i(p x Et)/~ 3 i(k x !t) 3 A(x,t)= e · A(p) d p = e · B(k) d k (5.298) Z Z where B(k)=~3A(~k) varies slowly compared to the phase exp[i(k x !t)]. · The phase velocity vp is the linear relation x = vp t between x and t that keeps the phase = p x Et constant as a function of the time · E ! 0=p dx Edt=(p v E) dt v = pˆ = kˆ (5.299) · · p () p p k in which p = p , and k = k . For light in the vacuum, v = c =(!/k) kˆ. | | | | p For a particle of mass m>0, the phase velocity exceeds the speed of light, v = c2p2 + m2c4/p c. p The more physical group velocity v is the linear relation x = v t p g g between x and t that maximizes the amplitude A(x,t) by keeping the phase = p x Et constant as a function of the momentum p · (p x Et)=x E(p) t = 0 (5.300) rp · rp at the maximum of A(p). This condition of stationary phase gives the group velocity as v = E(p)= !(k). (5.301) g rp rk 2 If E = p /(2m), then vg = p/m. For a relativistic particle with E = c2p2 + m2c4, the group velocity is v = c2p/E, and v c. g g  When light traverses a medium with a complex index of refraction n(k), p the wave vector k becomes complex, and its (positive) imaginary part represents the scattering of photons in the forward direction, typically by the electrons of the medium. For simplicity, we’ll consider the propagation of light through a medium in one dimension, that of the forward direction of the beam. Then the (real) frequency !(k) and the (complex) wave number k are related by k = n(k) !(k)/c, and the phase velocity of the light is ! c v = = . (5.302) p Re(k) Re(n(k)) If we regard the index of refraction as a function of the frequency !, instead of the wave-number k, then by di↵erentiating the real part of the relation !n(!)=ck with respect to !,weﬁnd dn (!) dk n (!)+! r = c r (5.303) r d! d! 242 Complex-Variable Theory in which the subscript r means real part. Thus the group velocity (5.301) of the light is d! c vg = = . (5.304) dkr nr(!)+!dnr/d! Optical physicists call the denominator the group index of refraction

dn (!) n (!)=n (!)+! r (5.305) g r d! so that as in the expression (5.302) for the phase velocity vp = c/nr(!), the group velocity is vg = c/ng(!). In some media, the derivative dnr/d! is large and positive, and the group velocity vg of light there can be much less than c (Steinberg et al., 1993; Wang and Zhang, 1995)—as slow as 17 m/s (Hau et al., 1999). This e↵ect is called slow light. In certain other media, the derivative dn/d! is so negative that the group index of refraction ng(!) is less than unity, and in them the group velocity vg exceeds c ! This e↵ect is called fast light. In some media, the derivative dn /d! is so negative that dn /d!< n (!)/!, and then r r r ng(!) is not only less than unity but also less than zero. In such a medium, the group velocity vg of light is negative! This e↵ect is called backwards light. Sommerfeld and Brillouin (Brillouin, 1960, ch. II & III) anticipated fast light and concluded that it would not violate special relativity as long as the signal velocity—deﬁned as the speed of the front of a square pulse— remained less than c. Fast light does not violate special relativity (Stenner et al., 2003; Brunner et al., 2004) (L´eon Brillouin 1889–1969, Arnold Som- merfeld 1868–1951). Slow, fast, and backwards light can occur when the frequency ! of the light is near a peak or resonance in the total cross-section tot for the scattering of light by the atoms of the medium. To see why, recall that the index of refraction n(!) is related to the forward scattering amplitude f(!) and the density N of scatterers by the formula (5.293)

2⇡c2 n(!)=1+ Nf(!) (5.306) !2 and that the real part of the forward scattering amplitude is given by the Kramers-Kronig integral (5.297) of the total cross-section

!2 (! ) d! Re(f(!)) = P 1 tot 0 0 . (5.307) 2⇡2c ! 2 !2 Z0 0 5.22 The Method of Steepest Descent 243 So the real part of the index of refraction is cN (! ) d! n (!)=1+ P 1 tot 0 0 . (5.308) r ⇡ ! 2 !2 Z0 0 If the amplitude for forward scattering is of the Breit-Wigner form /2 f(!)=f (5.309) 0 ! ! i/2 0 then by (5.306) the real part of the index of refraction is ⇡c2Nf (! !) n (!)=1+ 0 0 (5.310) r !2 [(! ! )2 +2/4] 0 and by (5.304) the group velocity is 1 2 2 (! ! ) 2/4 ⇡c Nf0 !0 0 vg = c 1+ . (5.311) 2 !2 [(h ! ! )2 +2/4]i2 3 0 4 5 This group velocity v is less than c whenever (! ! )2 > 2/4. But we get g 0 fast light v >c,if(! ! )2 < 2/4, and even backwards light, v < 0, if g 0 g ! ! with 4⇡c2Nf /! 1. Robert W. Boyd’s papers explain how to ⇡ 0 0 0 make slow and fast light (Bigelow et al., 2003) and backwards light (Gehring et al., 2006). We can use the principal-part identity (5.245) to subtract

cN 1 1 0= (!) P d!0 (5.312) ⇡ tot ! 2 !2 Z0 0 from the Kramers-Kronig integral (5.308) so as to write the index of refraction in the regularized form

cN 1 tot(!0) tot(!) n (!)=1+ P d!0 (5.313) r ⇡ ! 2 !2 Z0 0 which we can di↵erentiate and use in the group-velocity formula (5.304)

2 2 1 cN 1 [tot(!0) tot(!)] (!0 + ! ) v (!)=c 1+ P d!0 . (5.314) g ⇡ (! 2 !2)2  Z0 0

5.22 The Method of Steepest Descent Integrals like b I(x)= dz h(z)exp(xf(z)) (5.315) Za 244 Complex-Variable Theory often are dominated by the exponential. We’ll assume that x is real and that the functions h(z) and f(z) are analytic in a simply connected region in which a and b are interior points. Then the value of the integral I(x)is independent of the contour between the end points a and b but is sensitive to the real part u(z) of f(z)=u(z)+iv(z). But since f(z) is analytic, its real and imaginary parts u(z) and v(z) are harmonic functions which have no minima or maxima, only saddle points (5.52). For simplicity, we’ll assume that the real part u(z) of f(z) has only one saddle point between the points a and b. (If it has more than one, then we must repeat the computation that follows.) If w is the saddle point, then ux = uy = 0 which by the Cauchy-Riemann equations (5.10) implies that vx = vy = 0. Thus the derivative of the function f also vanishes at the saddle point f 0(w) = 0, and so near w we may approximate f(z) as 1 f(z) f(w)+ (z w)2f 00 (w). (5.316) ⇡ 2 Let’s write the second derivative as f 00 (w)=⇢ei and choose our contour through the saddle point w to be a straight line z = w + yei✓ with ✓ ﬁxed for z near w. As we vary y along this line, we want (z w)2f 00 (w)=y2 ⇢e2i✓ ei < 0 (5.317) so we keep 2✓ + = ⇡ so that near z = w 1 f(z) f(w) ⇢y2. (5.318) ⇡ 2 Since z = w + yei✓, its di↵erential is dz = ei✓ dy, and the integral I(x)is

1 1 I(x) h(w)exp x f(w)+ (z w)2f 00 (w) dz (5.319) ⇡ 2 Z1 ⇢  2⇡ = h(w) ei✓ exf(w) 1exp 1 x⇢y2 dy = h(w) ei✓ exf(w) . 2 x⇢ Z r 1 Moving the phase ei✓ inside the square root 2⇡ I(x) h(w) exf(w) (5.320) ⇡ x⇢e 2i✓ r and using f 00 (w)=⇢ei and 2✓ + = ⇡ to show that 2i✓ i i⇡ i ⇢e = ⇢e = ⇢e = f 00 (w) (5.321) we get our formula for the saddle-point integral (5.315) 2⇡ 1/2 I(x) h(w) exf(w). (5.322) ⇡ xf 00 (w) ✓ ◆ 5.23 The Abel-Plana Formula and the Casimir E↵ect 245 Example 5.46 (Stirling’s formula for n!). An exact formula for n! dny 1 n!= ( 1)n (5.323) dyn y=1 is the integral n n d 1 yz 1 n z 1 n ln z z n!= ( 1) e dz = z e dz = e dz. dyn Z0 y=1 Z0 Z0 (5.324) Comparing it to the integral (5.315) for I(x), we set f(z)=n ln z z and x = h(z) = 1. The saddle point z = w is where f 0(w) = 0, or w = n.Since f (n)= 1/n, our steepest-descent approximation (5.322) to n! is Stirling’s 00 formula n n n! p2⇡n n e . (5.325) ⇡

If there are n saddle points wj for j =1,...,n,thenthesteepest-descent approximation to the integral I(x)isthesum

N 1/2 2⇡ xf(wj ) I(x) h(wj) e . (5.326) ⇡ xf 00 (wj) Xj=1 ✓ ◆

5.23 The Abel-Plana Formula and the Casimir E↵ect This section is optional on a ﬁrst reading. Suppose the function f(z) is analytic and bounded for n Re z n . Let 1   2 + and be two contours that respectively run counterclockwise along the C C rectangles with vertices n ,n ,n +i ,n +i and n ,n ,n i ,n i 1 2 2 1 1 1 1 2 2 1 1 1 indented with tiny semi-circles and quarter-circles so as to avoid the integers z = n1, n1 + 1, n1 + 2, . . . , n2 while keeping Imz>0 in the upper rectangle and Imz<0 in the lower one (and n1 < Re z

n2 1 1 1 n2 T = f(n )+ f(n)+ f(n ) f(x) dx. (5.331) x 2 1 2 2 n=n +1 n1 X1 Z Since exp( 2⇡in ) = 1, the di↵erence between the integrals along the 1 imaginary axes above and below n1 is (exercise 5.41)

n1 n1 i f(z) 1 f(z) T1 = 2⇡iz dz 2⇡iz dz (5.332) n1+i e 1 n1 e 1 Z 1 Z 1 f(n + iy) f(n iy) = i 1 1 dy. (5.333) e2⇡y 1 Z0 5.23 The Abel-Plana Formula and the Casimir E↵ect 247 Similarly, the di↵erence between the integrals along the imaginary axes above and below n2 is (exercise 5.42)

n2+i n2 1 f(z) f(z) T2 = 2⇡iz dz 2⇡iz dz (5.334) n2 e 1 n2 i e 1 Z Z 1 1 f(n + iy) f(n iy) = i 2 2 dy. (5.335) e2⇡y 1 Z0 Since + = Tx + T1 + T2 = 0, we can use (5.331) and (5.333–5.335) I I to build the Abel-Plana formula (Whittaker and Watson, 1927, p. 145)

n2 1 1 1 n2 f(n )+ f(n)+ f(n ) f(x) dx (5.336) 2 1 2 2 n=n +1 n1 X1 Z 1 f(n + iy) f(n iy) f(n + iy)+f(n iy) = i 1 1 2 2 dy e2⇡y 1 Z0 (Niels Abel 1802–1829 and Giovanni Plana 1781–1864). In particular, if f(z)=z, the integral over y vanishes, and the Abel-Plana formula (5.336) gives

n2 1 1 1 n2 n + n + n = xdx (5.337) 2 1 2 2 n=n +1 n1 X1 Z which is an example of the trapezoidal rule.

Example 5.47 (The Casimir E↵ect). The Abel-Plana formula provides one of the clearer formulations of the Casimir e↵ect. We will assume that the hamiltonian for the electromagnetic ﬁeld in empty space is a sum over two polarizations and an integral over all momenta of a symmetric product

2 1 3 H0 = ~ !(k) a†(k)as(k)+as(k)a†(k) d k (5.338) 2 s s s=1 X Z h i of the annihilation and creation operators as(k) and as†(k) which satisfy the commutation relations

[as(k),a† (k0)] = ss (k k0) and [as(k),as (k0)] = 0 = [as†(k),a† (k0)]. s0 0 0 s0 (5.339) The vacuum state 0 has no photons, and so on it a (k) 0 = 0 (and | i s | i 0 as†(k) = 0). But because the operators in H are symmetrically ordered, h | 0 the energy E0 of the vacuum as given by (5.338) is not zero; instead it is 248 Complex-Variable Theory quarticly divergent

2 3 1 3 d k E0 = 0 H0 0 = ~ !(k) (0) d k = V ~ !(k) (5.340) h | | i 2 (2⇡)3 s=1 X Z Z in which we used the delta-function formula 3 i(k k0) x d x (k k0)= e± · (5.341) (2⇡)3 Z to identify (0) as the volume V of empty space divided by (2⇡)3.Sincethe photon has no mass, its (angular) frequency !(k)isc k , and so the energy | | density E0/V is

4 E0 3 dk K ~c 1 = ~c k = ~c = (5.342) V 2⇡2 8⇡2 8⇡2 d4 Z 1 in which we cut o↵the integral at some short distance d = K below which the hamiltonian (5.338) and the commutation relations (5.339) are no longer valid. But the energy density of empty space is

2 3H0 ~c 1 ⌦⇤⇢c =⌦⇤ (5.343) 8⇡G ⇡ 8⇡2 (2.8 10 5 m)4 ⇥ which corresponds to a distance scale d of 28 micrometers. Since quantum 18 electrodynamics works well down to about 10 m, this distance scale is too big by thirteen orders of magnitude. If the universe were inside an enormous, perfectly conducting, metal cube of side L, then the tangential electric and normal magnetic ﬁelds would vanish on the surface of the cube Et(r,t)=0=Bn(r,t). The available wave-numbers of the electromagnetic ﬁeld inside the cube then would be kn =2⇡(n1,n2,n3)/L, and the energy density would be

E0 2⇡~c = pn2. (5.344) V L4 n X The Casimir e↵ect exploits the di↵erence between the continuous (5.342) and discrete (5.344) energy densities for the case of two metal plates of area A separated by a short distance ` pA. ⌧ If the plates are good conductors, then at low frequencies the boundary conditions Et(r,t)=0=Bn(r,t) hold, and the tangential electric and normal magnetic ﬁeld vanish on the surfaces of the metal plates. At high frequencies, above the plasma frequency !p of the metal, these boundary 5.23 The Abel-Plana Formula and the Casimir E↵ect 249 conditions fail because the relative electric permittivity of the metal

!2 i ✏(!) 1 p 1 (5.345) ⇡ !2 !⌧ ✓ ◆ has a positive real part. Here ⌧ is the mean time between electron collisions. The modes that satisfy the low-frequency boundary conditions Et(r,t)= 0=Bn(r,t) are (Bordag et al., 2009, p. 30)

⇡n 2 !(k ,n) c k2 + (5.346) ? ⌘ ? ` r ⇣ ⌘ where n k = 0. The di↵erence between the zero-point energies of these · ? modes and those of the continuous modes in the absence of the two plates per unit area would be

1 2 2 2 2 E(`) ⇡~c 1 k dk ` k 2 1 ` k 2 `k = ? ? ? + n ? + x dx ? A ` 2⇡ 2 s ⇡2 s ⇡2 2⇡ 3 0 n=0 0 Z X Z 4 (5.347)5 if the boundary conditions held at all frequencies. With p = `k /⇡,wewill ? represent the failure of these boundary conditions at the plasma frequency 2 2 2 4 !p by means of a cuto↵function like c(n, p)=(1+(n + p )/np) where np = !p`/⇡c. In terms of such a cuto↵function, the energy di↵erence per unit area is

2 E(`) ⇡ ~c 1 1 1 p = pdp c(n,p) p2 + n2 c(x,p) p2 + x2 dx . A 2`3 2 Z0 "n=0 Z0 # X p p (5.348) Since c(n,p) falls o↵as (n2/(n2 + p2))4 for n,p n , we may neglect terms p p in the sum and integral beyond some integer M that is much larger than np

2 M M E(`) ⇡ ~c 1 p = pdp c(n,p) p2 + n2 c(x,p) p2 + x2 dx . A 2`3 2 Z0 "n=0 Z0 # X p p (5.349) The function

2 2 2 2 p + z f(z)=c(z,p) p + z = 2 2 2 4 (5.350) (1 + (pz + p )/np) p is analytic in the right half-plane Re z = x>0 (exercise 5.43) and tends to zero limx f(x + iy) 0 as Re z = x . So we can apply the !1 | |! !1 Abel-Plana formula (5.336) with n1 = 0 and n2 = M to the term in the 250 Complex-Variable Theory square brackets in (5.349) and get 2 E(`) ⇡ ~c 1 c(M,p) 2 2 = 3 pdp p + M (5.351) A 2` 0 2 Z ⇢ p +i 1 c(iy,p) p2 +(✏ + iy)2 c( iy,p) p2 +(✏ iy)2 0 Z h c(M + iyp,p) p2 +(M + iy)2 p dy +c(M iy,p) pp2 +(M iy)2 e2⇡y 1 p i in which the inﬁnitesimal ✏ reminds us that the contour lies inside the right half-plane. We now take advantage of the properties of the cuto↵function c(z,p). Since M n , we can omit the term c(M,p) p2 + M 2/2. The denominator p exp(2⇡y) 1letsusdrop the terms c(M iy,p) p2 +(M iy)2. We are ⌥ ⌥p ⌥ left with p 2 E(`) ⇡ ~c 1 = pdp (5.352) A 2`3 Z0 1 2 2 2 2 dy i c(iy,p) p +(✏ + iy) c( iy,p) p +(✏ iy) 2⇡y . ⇥ 0 e 1 Z h p p i Since the y integration involves the factor 1/(exp(2⇡y) 1), we can neglect the detailed behavior of the cuto↵functions c(iy,p) and c( iy,p) for y>n p where they di↵er appreciably from unity. The energy now is

2 2 2 2 2 E(`) ⇡ ~c 1 1 p +(✏ + iy) p +(✏ iy) = pdp i dy. (5.353) A 2`3 e2⇡y 1 Z0 Z0 p p When yp, they are p2 y2 2i✏y = i y2 p2. (5.354) ± ± Their di↵erence is 2i py2 p2, and so E(`)isp 2 2 2 E(`) ⇡ ~pc 1 1 2 y p ✓(y p) = pdp dy (5.355) A 2`3 e2⇡y 1 Z0 Z0 p in which the Heaviside step function ✓(x) (x + x )/(2 x )keepsy>p ⌘ | | | | 2 y 2 2 E(`) ⇡ ~c 1 y p = pdp dy. (5.356) A `3 e2⇡y 1 Z0 Z0 p The p-integration is elementary, and so the energy di↵erence is 2 3 2 2 E(`) ⇡ ~c 1 y dy ⇡ ~c B2 ⇡ ~c = = = (5.357) A 3`3 e2⇡y 1 3`3 8 720 `3 Z0 5.24 Applications to String Theory 251 in which B2 is the second Bernoulli number (4.122). The pressure pushing the plates together then is

1 @E(`) ⇡2~c p = = (5.358) A @` 240 `4 a result due to Casimir (Hendrik Brugt Gerhard Casimir, 1909–2000). Although the Casimir e↵ect is very attractive because of its direct connec- tion with the symmetric ordering of the creation and annihilation operators in the hamiltonian (5.338), the reader should keep in mind that neutral atoms are mutually attractive, which is why most gases are diatomic, and that Lifshitz explained the e↵ect in terms of the mutual attraction of the atoms in the metal plates (Lifshitz, 1956; Milonni and Shih, 1992) (Evgeny Mikhailovich Lifshitz, 1915–1985).

5.24 Applications to String Theory This section is optional on a first reading. String theory may or may not have anything to do with physics, but it does provide many amusing applications of complex-variable theory. The coordinates and ⌧ of the world sheet of a string form a complex variable 2(⌧ i) z = e . The product of two operators U(z) and V (w) often has poles in z w as z w but is well defined if z and w are radially ordered ! U(z)V (w) U(z) V (w) ✓( z w )+V (w) U(z) ✓( w z ) (5.359) R{ }⌘ | || | | || | in which ✓(x)=(x + x )/2 x is the step function. Since the modulus of 2(⌧ i) | | | | z = e depends only upon ⌧, radial order is time order in ⌧z and ⌧w. The modes Ln of the principal component of the energy-momentum tensor T (z) are defined by its Laurent series

1 L T (z)= n (5.360) zn+2 n= X1 and the inverse relation 1 L = zn+1 T (z) dz. (5.361) n 2⇡i I Thus the commutator of two modes involves two loop integrals 1 1 [L ,L ]= zm+1 T (z) dz, wn+1 T (w) dw (5.362) m n 2⇡i 2⇡i  I I 252 Complex-Variable Theory

Radial Order

< U (z ) V (w ) < |z | > |w | w < V (w ) U (z ) |w | > |z |

Figure 5.10 The two counterclockwise circles about the origin preserve radial order when z is near w by veering slightly to z > w for the product U(z)V (w) and to w > z for the product V (w)U|(z| ). | | | | | | which we may deform as long as we cross no poles. Let’s hold w ﬁxed and deform the z loop so as to keep the T ’s radially ordered when z is near w as in Fig. 5.10. The operator-product expansion of the radially ordered product T (z)T (w) is R{ } c/2 2 1 T (z)T (w) = + T (w)+ T 0(w)+... (5.363) R{ } (z w)4 (z w)2 z w in which the prime means derivative, c is a constant, and the . . . denote terms that are analytic in z and w. The commutator introduces a minus sign that cancels most of the two contour integrals and converts what remains into an integral along a tiny circle Cw about the point w as in Fig. 5.10 dw dz c/2 2T (w) T (w) [L ,L ]= wn+1 zm+1 + + 0 . m n 2⇡i 2⇡i (z w)4 (z w)2 z w Cw  I I (5.364) After doing the z-integral, which is left as a homework exercise (5.44), one Exercises 253 may use the Laurent series (5.360) for T (w)todothew-integral, which one may choose to be along a tiny circle about w = 0, and so ﬁnd the commutator c [L ,L ]=(m n) L + m(m2 1) (5.365) m n m+n 12 m+n,0 of the Virasoro algebra. Example 5.48 (Using ghost contours to sum series). Consider the integral csc ⇡z I = dz (z a)2 IC along the counterclockwise rectangular contour C from z = N +1/2 iY to z = N +1/2+iY to z = N 1/2+iY to z = N 1/2 iY and back to z = N +1/2 iY in which N is a positive integer, and a is not an integer. In the twin limits N and Y , the integral vanishes because on !1 !1 the contour 1/ z a 2 1/N 2 or 1/Y 2 while csc ⇡z 1. We now shrink | | ⇡ | | the contour down to tiny circles about the poles of csc ⇡z at all the integers, z = n and about the nonintegral value, z = a. By Cauchy’s integral formula (5.38), the tiny contour integral around z = a is csc ⇡z d csc ⇡z cos ⇡a dz =2⇡i = 2⇡2i . (z a)2 dz sin2 ⇡a Ia z=a In the twin limits N and Y , the tiny counterclockwise integrals !1 !1 around the poles of 1/ sin ⇡z at z = n⇡ are (exercise 5.47)

1 csc ⇡z 1 1 dz =2i ( 1)n . (z a)2 (n a)2 n= n n= X1 I X1 We thus have the sum rule

1 1 ( 1)n = ⇡2 cot ⇡a csc ⇡a. (n a)2 n= X1

Further reading For examples of conformal mappings see (Lin, 2011, section 3.5.7).

Exercises 5.1 Compute the two limits (5.6) and (5.7) of example 5.2 but for the function f(x, y)=x2 y2 +2ixy. Do the limits now agree? Explain. 254 Complex-Variable Theory 5.2 Show that if f(z) is analytic in a disk, then the integral of f(z) around a tiny (isosceles) triangle of side ✏ 1 inside the disk is zero to order ⌧ ✏2. 5.3 Derive the two integral representations (5.50) for Bessel’s functions Jn(t) of the ﬁrst kind from the integral formula (5.49). Hint: Think of the integral (5.49) as running from ⇡ to ⇡. 5.4 Do the integral dz z2 1 IC in which the contour C is counterclockwise about the circle z = 2. | | 5.5 The function f(z)=1/z is analytic in the region z > 0. Compute | | the integral of f(z) counterclockwise along the unit circle z = ei✓ for 0 ✓ 2⇡. The contour lies entirely within the domain of analyticity   of the function f(z). Did you get zero? Why? If not, why not? 5.6 Let P (z) be the polynomial

P (z)=(z a )(z a )(z a ) (5.366) 1 2 3

with roots a1, a2, and a3. Let R be the maximum of the three moduli a . (a) If the three roots are all di↵erent, evaluate the integral | k| dz I = (5.367) P (z) IC along the counterclockwise contour z =2Rei✓ for 0 ✓ 2⇡. (b) Same   exercise, but for a = a = a . 1 2 6 3 5.7 Compute the integral of the function f(z)=eaz/(z2 3z + 2) along the counterclockwise contour that follows the perimeter of a square C⇤ of side 6 centered at the origin. That is, find eaz I = dz. (5.368) z2 3z +2 IC⇤ 5.8 Use Cauchy’s integral formula (5.40) and Rodrigues’s expression (5.41) for Legendre’s polynomial Pn(x) to derive Schlaefli’s formula (5.42). 5.9 Use Schlaefli’s formula (5.42) for the Legendre polynomials and Cauchy’s integral formula (5.36) to compute the value of P ( 1). n 5.10 Evaluate the counterclockwise integral around the unit circle z =1 | | dz 3sinh2 2z 4 cosh3 z . (5.369) z I ⇣ ⌘ Exercises 255 5.11 Evaluate the counterclockwise integral around the circle z =2 | | z3 dz. (5.370) z4 1 I 5.12 Evaluate the contour integral of the function f(z)=sinwz/(z 5)3 along the curve z = 6 + 4(cos t + i sin t) for 0 t 2⇡.   5.13 Evaluate the contour integral of the function f(z)=sinwz/(z 5)3 along the curve z = 6 + 4(cos t + i sin t) for 0 t 2⇡.   5.14 Is the function f(x, y)=x2 + iy2 analytic? 5.15 Is the function f(x, y)=x3 3xy2+3ix2y iy3 analytic? Is the function x3 3xy2 harmonic? Does it have a minimum or a maximum? If so, what are they? 5.16 Is the function f(x, y)=x2 + y2 + i(x2 + y2) analytic? Is x2 + y2 a harmonic function? What is its minimum, if it has one? 5.17 Derive the first three nonzero terms of the Laurent series for f(z)= 1/(ez 1) about z = 0. 5.18 Assume that a function g(z) is meromorphic in R and has a Laurent series (5.99) about a point w R. Show that as z w, the ratio 2 ! g0(z)/g(z) becomes (5.97). 5.19 Use a contour integral to evaluate the integral ⇡ d✓ I = ,a>1. (5.371) a a + cos ✓ Z0 5.20 Find the poles and residues of the functions 1/ sin z and 1/ cos z. 5.21 Derive the integral formula (5.129) from (5.128). 5.22 Show that if Re w<0, then for arbitrary complex z

2 ⇡ 1 ew(x+z) dx = . (5.372) w Z1 r 5.23 Use a ghost contour to evaluate the integral x sin x 1 dx. x2 + a2 Z1 Show your work; do not just quote the result of a commercial math program. 5.24 For a>0 and b2 4ac < 0, use a ghost contour to do the integral dx 1 . (5.373) ax2 + bx + c Z1 256 Complex-Variable Theory 5.25 Show that 1 x2 1 a2/4 cos ax e dx = p⇡e . (5.374) 2 Z0 5.26 Show that dx ⇡ 1 = . (5.375) 1+x4 p2 Z1 5.27 Evaluate the integral cos x 1 dx. (5.376) 1+x4 Z0 5.28 Show that the Yukawa Green’s function (5.158) reproduces the Yukawa x potential (5.148) when n = 3. Use K1/2(x)= ⇡/2xe (9.111). 5.29 Derive the two explicit formulas (5.195) and (5.196) for the square root p of a complex number. 5.30 What is ( i)i? What is the most general value of this expression? 5.31 Use the indeﬁnite integral (5.244) to derive the principal-part formula (5.245). 5.32 The Bessel function Jn(x) is given by the integral

1 (x/2)(z 1/z) dz J (x)= e (5.377) n 2⇡i zn+1 IC along a counterclockwise contour about the origin. Find the generating function for these Bessel functions, that is, the function G(x, z) whose Laurent series has the Jn(x)’s as coecients

1 n G(x, z)= Jn(x) z . (5.378) n= X1 5.33 Show that the Heaviside function ✓(y)=(y + y )/(2 y ) is given by the | | | | integral 1 dx ✓(y)= 1 eiyx (5.379) 2⇡i x i✏ Z1 in which ✏ is an infinitesimal positive number. 5.34 Show that the integral of exp(ik)/k along the contour from k = L to k = L + iH and then to k = L + iH and then down to k = L vanishes in the double limit L and H . !1 !1 5.35 Use a ghost contour and a cut to evaluate the integral 1 dx I = (5.380) 2 2 1 (x + 1)p1 x Z by imitating example 5.38. Be careful when picking up the poles at Exercises 257 z = i. If necessary, use the explicit square root formulas (5.195) and ± (5.196). 5.36 Redo the previous exercise (5.35) by defining the square roots so that the cuts run from to 1 and from 1 to . Take advantage of the 1 1 evenness of the integrand and integrate on a contour that is slightly above the whole real axis. Then add a ghost contour around the upper half plane. 5.37 Show that if u is even and v is odd, then the Hilbert transforms (5.286) imply (5.288). 5.38 Show why the principal-part identity (5.245) lets one write the Kramers- Kronig integral (5.308) for the index of refraction in the regularized form (5.313). 5.39 Use the formula (5.304) for the group velocity and the regularized expression (5.313) for the real part of the index of refraction nr(!)to derive formula (5.314) for the group velocity. 5.40 Show that the quarter-circles of the Abel-Plana contours contribute 1 C± 2 (f(n1)+f(n2)) to the sum S in the formula Tx = Ix + S. 5.41 Derive the integral formula (5.333) from (5.332). 5.42 Derive the integral formula (5.335) from (5.334). 5.43 Show that the function (5.350) is analytic in the RHP Re z>0. 5.44 (a) Perform the z-integral in Eq.(5.364). (b) Use the result of the part (a) to find the commutator [Lm,Ln] of the Virasoro algebra. Hint: use the Laurent series (5.360). 5.45 Assume that ✏(z) is analytic in a disk that contains a tiny circular contour Cw about the point w as in Fig 5.10. Do the contour integral c/2 2T (w) T (w) dz ✏(z) + + 0 (5.381) (z w)4 (z w)2 z w 2⇡i ICw  and express your result in terms of ✏(w), T (w), and their derivatives. n 5.46 Show that if the coecients ak of the equation 0 = a0 +a1z +...+anz are real, then its n roots zk are real or come in pairs that are complex conjugates, z` and z`⇤, of each other. 5.47 Show that if a is not an integer, then the sum of the tiny counterclockwise integrals about the points z = n of example 5.48 is

1 csc ⇡z 1 1 dz =2i ( 1)n . (z a)2 (n a)2 n= n n= X1 I X1 258 Complex-Variable Theory 5.48 Use the trick of example 5.48 with csc ⇡z cot ⇡z to show that ! 1 1 ⇡2 = (n a)2 2 n= sin ⇡a X1 as long as a is not an integer.