CHAPTER 6 Residue Theory We developed three methods of evaluating contour integrals around closed curves: (1) The Cauchy-Goursat Theorem - but this requires the integrand to be ana- lytic in and on the curve; (2) Cauchy’s integral formulas - but only applies to integrands that have a very special form; (3) Theorem 4.8 - but replaces a contour integral along a curve containing a number of singularities with contour integrals along curves each of which encloses only one singularity. Cauchy’s residue theorem, developed in this chapter, encompasses all of these. 6.1. Cauchy’s Residue Theorem Cauchy’s residue theorem is a simple consequence of the equations for coeffi- cients of a Laurent series. 1 To illustrate, suppose f(z) = a (z z )n is the Laurent series of a function n − 0 n= f valid in some annulus 0 < z X1z < R about an isolated singularily z . When | − 0| 0 C is a simple, closed, piecewise smooth curve in this annulus that contains z0 in its interior, coefficient a 1 is defined by − 146 6.1. CAUCHY’S RESIDUE THEOREM 147 1 a 1 = f(z) dz = f(z)dz = 2⇡ia 1. − 2⇡i C ) c − Cauchy’s residue theorem Iis a simple extensionI of this result to the case when C encloses an arbitrary number of isolated singularities. Definition (6.1). The residue of a function f at an isolated singularity z0 is denoted by Res[f, z0]. It is the coefficient a 1 in the Laurent expansion of − f about z0. From the examples on pages 141-142 of these notes, sin z Res , 0 = 0, Res[e1/z, 0] = 1 z h z i sin z 1 Res − , 0 = z6 −120 Theorem (6.1 – Cauchy’sh Residue Theorem)i . Suppose a function f is analytic inside and on a simple, closed, piecewise smooth curve C, except at singularies z1, . , zn in its interior. Then n f(z) dz = 2⇡i Res[f, zj]. C j=1 I X Proof. Construct around the z disjoint circles C : z z = r such j j | − j| j that each Cj is interior to C (see the diagram at the top of the next page). By Theorem 4.8, the contour integral around C can be replaced by contour integrals around the Cj, 148 6. RESIDUE THEORY n f(z) dz = f(z) dz. C j=1 Cj I X I If f is expanded in a Laurent series about zj, 1 n 1 f(z) = an(z zj) where a 1 = f(z) dz = − − 2⇡i ) n= Cj X1 I f(z) dz = 2⇡ia 1 = 2⇡i Res[f, zj] = − ) ICj n f(z) dz = 2⇡i Res[f, zj]. ⇤ C j=1 I X 1 Problem (Page 268 #13). Evaluate dz where C : z = 1. z2 + 4z + 16 | | IC Solution. 4 p16 64 z2 + 4z + 16 = 0 z = − ± − = 2 2p3 i. () 2 − ± Both of these values are exterior to C, as seen in the diagram at the top of the next page. Thus 1 dz = 0. z2 + 4z + 16 IC ⇤ 6.1. CAUCHY’S RESIDUE THEOREM 149 1 Example. Evaluate sin dz where C : z = 2. C z | | I ⇣ ⌘ 1 Solution. Since the Laurent series for sin around z = 0 is z 1 1 1 1⇣ ⌘ sin = + , z z − 3!z3 5!z5 − · · · 1 ⇣ ⌘ 1 the residue of sin is Res sin , 0 = 1. Thus z z ⇣ ⌘ ⇣ ⌘ 1 sin dz = 2⇡i(1) = 2⇡i. z ⇤ IC The following theorem giv⇣es ⌘a formula for calculating residues at poles. Theorem (6.2). If f has a pole of order m at z0, then m 1 1 d − m Res[f, z0] = lim m 1 (z z0) f(z) z z0 (m 1)! dz − ! − − Proof. f has a pole of order m at z = (Lauren⇥ t expansion)⇤ 0 ) a m a m+1 a 1 f(z) = − + − + + − + a + , (z z )m (z z )m 1 · · · (z z ) 0 · · · − 0 − 0 − − 0 valid in some annulus 0 < z z < R. Thus | − 0| m m 1 (z z0) f(z) = a m + a m+1(z z0) + + a 1(z z0) − + = − − − − · · · − − · · · ) m 1 d − m m 1 (z z0) f(z) = (m 1)!a 1 + m!a0(z z0) + = dz − − − − − · · · ) ⇥ ⇤ 150 6. RESIDUE THEORY m 1 1 d − m lim m 1 (z z0) f(z) = a 1 = Res[f, z0] ⇤ z z0 (m 1)! dz − − ! − − Problem (Page 268 # 1).⇥ Calculate the ⇤residue for the isolated singularity (z + 3)6 of at z = 4i. (z 4i)4 − Solution. The isolated singularity at z = 4i is a pole of order 4 = ) (z + 3)6 1 d3 (z 4i)4(z + 3)6 Res , 4i = lim − (z 4i)4 3! z 4i dz3 (z 4i)4 − ! ⇢ − 1 3 = lim 6(5)(4)(z + 3) = 2340 + 880i. ⇤ 6 z 4i ! − ⇥ 1 ⇤ Problem (Page 268 # 21). Evaluate dz where C : z = 3. z2 z + 2 | | IC − 1 1 Solution. = has simple poles at z = 2 and z2 z + 2 (z + 2)(z 1) − z = 1, both interior−to C. − 1 z 1 1 Res , 1 = lim − = , and z2 z + 2 z 1 (z + 2)(z 1) 3 − ! − 1 z + 2 1 Res , 2 = lim = = z2 z + 2 − z 2 (z + 2)(z 1) −3 ) − ! − 1 1 1 dz = 2⇡i = 0. z2 z + 2 3 − 3 ⇤ IC − ✓ ◆ 6.1. CAUCHY’S RESIDUE THEOREM 151 An important conequence of the residue theorem is the principle of the ar- gument. Theorem (6.3 – Principle of the Argument). Let f be analytic in a do- main D except at a finite number of poles. If C is a simple, closed, piece- wise smooth curve in D which does not pass through any poles or zeros of f, then f (z) ( ) 0 dz = 2⇡i(N P ) ⇤ f(z) − IC where N and P are the sums of the orders of the zeroes and poles of f within C. To see why this is called the principle of the argument,we transform the contour integral ( ) from the z-plane to the w-plane with w = f(z). It follows that ⇤ 1 f (z) 1 1 N P = 0 dz = dw, − 2⇡i f(z) 2⇡i w IC If(C) where f(C) is the image of C in the w-plane. But according to text Example 4.22 (page 182), the contour integral on the right is the number of times that f(C) encircles w = 0. In other words, N P measures how many times the curve f(C) encircles w = 0 as z traverses C− in the counterclockwise direction; that is N P measures how many times the argument of f(z) increases by 2⇡ as z traverses− C. 152 6. RESIDUE THEORY Problem (Page 269 # 39). Use the principle of the argument to evaluate z sinh z + 3 cosh z dz where C : z i = 7. z cosh z | − | IC Solution. Letting f(z) = z cosh z, we have z sinh z + 3 cosh z z sinh z + cosh z 2 dz = + dz z cosh z z cosh z z IC IC ✓ ◆ f (z) 1 = 0 dz + 2 dz = 2⇡i(N P ) + 2(2⇡i), f(z) z − IC IC where N and P are the sums of the orders of the zeroes and poles of f(z) inside C. There are 6 zeroes of order 1 and no poles within this circle, so N P = 6. Then − z sinh z + 3 cosh z dz = 2⇡i(6) + 4⇡i = 16⇡i. z cosh z ⇤ IC 6.1. CAUCHY’S RESIDUE THEOREM 153 Theorem (6.4 – Rouche’s Theorem). Let fand g be analytic in a domain D. If C us a simple, closed, piecewise smooth curve in D which contains in its interior only points of D, and if g(z) < f(z) on C, then sums of the orders of the zeros of f + g and f inside| | C |are the| same. Proof. g < f on C = f(z) = 0 and f(z)+g(z) f(z) g(z) > 0 | | | | f)(z) + g6(z) | | ≥ | |−| | on C. The function F (z) = , then, does not have any zeros or poles f(z) on C. Applying the principle of the argument to F and C, 1 F (z) 0 dz = N P, 2⇡i F (z) − IC and each side may be interpreted as the number of times w = F (z) encircles w = 0 as z traverses C. But on C, f(z) + g(z) g(z) F (z) 1 = 1 = < 1 = | − | f(z) − f(z) ) the curve F (C) cannot encircle w = 0, and th us N = P . Since the zeros of F occur at the zeros of f + g and the poles of F occur at the zeros of f, it follows that the sum of the orders of the zeros of f and f + g inside C must be the same. ⇤ 154 6. RESIDUE THEORY Problem (Page 269 # 45). Show that if g is analytic inside and on C : z = 1 and if g(z) < 1, then there is exactly one point z inside C for which g| (|z) = z. | | Solution. Let f(z) = z. Then, on C, f(z) = 1. Since g(z) < f(z) on C, it follows from Rouche’s− theorem that|f(z)|+ g(z) = g(|z) |z has| the| same number of zeros inside C as f(z) = z, namely one. Thus,− g(z) z has a single zero inside C = there is exactly− one solution for g(z) = z inside− ) C. ⇤ 6.2.