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 analyt- icity 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✏.