MATH 722, COMPLEX ANALYSIS, SPRING 2009 PART 3 1. The calculus of residues 1.1. The Residue Theorem. Let Ω ⊂ C be a connected open set, and let Z ⊂ Ω be a discrete subset. Thus Z is either a finite set of points, or a countable set with no limit point in Ω. Write Z = fa1; a2;:::g. Let f :ΩnZ ! C be a holomorphic function. Then at each point aj 2 Z, the function f has one of three possible kinds of behavior: 0 (i) f can have a removable singularity, so that in some small punctured disk D (, aj), we can P1 n write f(z) = n=0 cn(z − aj) . In this case f extends to a holomorphic function on the whole disk D(, aj). 0 (ii) f can have a pole of order N ≥ 1, so that in some small punctured disk D (, aj), we can P1 n write f(z) = n=−N cn(z − aj) , with c−N 6= 0. 0 (iii) f can have an essential singularity at aj, so that in some small punctured disk D (, aj), we P1 n can write f(z) = n=−∞ cn(z − aj) , with infinitely many non-zero negative coefficients. Definition 1.1. If ρ < , the number 1 Z Rj = f(ζ) dζ 2πi jζ−aj j=ρ is called the residue of f at the point aj. Note that this number is independent of the radius ρ, and equals the coefficient a−1 in the Laurent expansion of f about the point aj. The residue is sometimes denoted by \Resz=aj f(z)". If the function f has a removable singularity at aj, then Resz=aj f(z) = 0. Let Γ be a piecewise differentiable, simple closed curve in Ω. Then CnΓ consists of two connected components, one of which is bounded, and the other unbounded. We call the bounded component the interior of Γ, and write it Int(Γ). If the interior of Γ is also a subset of Ω, then the curve Γ does not `go around' any hole in Ω. Let Z = fajg be a discrete set of points in Ω, and suppose that Γ does not pass through any of these points. Then Z \ Int(Γ) is a finite set of points fa1; : : : ; amg. We can find radii fρ1; : : : ; ρmg so that each disk D(aj; ρj) is contained in the interior of Γ, and so that the disks fD(aj; ρj)g are iθ mutually disjoint. Let γj be the closed curve in Ω parameteried by γj(t) = aj + ρje , 0 ≤ θ ≤ 2π, If f is holomorphic on Ω n Z, it follows that m Z X Z f(ζ) dζ = f(ζ) dζ: Γ j=1 γj From this, we easily obtain the following. 1 COMPLEX ANALYSIS 2 Theorem 1.2 (Residue Theorem). Let Ω ⊂ C be a connected open set, and suppose that f(z) is holomorphic for z 2 Ω except for z belonging to a discrete set Z = fajg ⊂ Ω. Let Γ be a simple closed curve in Ω with the property that Int(Γ) ⊂ Ω. If Γ does not pass through any of the points fajg, then 1 Z X f(ζ) dζ = Resz=aj f(z); 2πi Γ aj 2int(Γ) where the sum is over the finite number of points fa1; : : : ; amg contained in the interior of Γ. 1.2. The argument priniple. Recall that if Ω ⊂ C is a connected open set, a function f is meromorphic on Ω if there exists a discrete set E = fajg ⊂ Ω so that f is holomorphic on Ω n E, and f has either a removable singularity or a pole at each point aj. If f is not identically zero and is meromorphic on Ω, then g(z) = f(z)−1 is also meromorphic. In fact, g(z) is holomorphic at every point where f(z) 6= 0, g has poles at the points where f has zeros, and g has removable singularities (with value 0) at the poles of f. The following is then clear. Proposition 1.3. The set of meromorphic functions M(Ω) on a connected open set is a field. Moreover, if f 2 M(Ω), then f 0 2 M(Ω). We also have the following easy calculation. Proposition 1.4. Suppose that f is meromorphic on Ω. Then if a 2 Ω, 8 0 if f is holomorphic and non-zero at a; f 0 <> Res (z) = N if f has a zero of order N at a; z=a f :>−N if f has a pole of order N at a: Proof. If f is holomorphic and non-zero at z = a, then f −1(z) is also holomorphic near a, and so f 0=f is holomorphic near a. It follows that the residue is zero. If f has a zero of order N, then near a we can write f(z) = (z − a)N g(z) with g holomorphic near a and g(a) 6= 0. Then f 0(z) = N(z − a)N−1g(z) + (z − a)N g0(z), so near z = a we have f 0 N(z − a)N−1g(z) + (z − a)N g0(z) N g0(z) (z) = = + : f (z − a)N g(z) (z − a) g(z) Since g0=g is holomorphic near a, it follows that f 0=f has a simple pole at z = a and the residue is N. If f has a pole of order N, then near a we can write f(z) = (z − a)−N g(z) with g holomorphic near a and g(a) 6= 0. Then f 0(z) = −N(z − a)−N−1g(z) + (z − a)−N g0(z), so near z = a we have f 0 −N(z − a)−N−1g(z) + (z − a)−N g0(z) −N g0(z) (z) = = + : f (z − a)−N g(z) (z − a) g(z) Since g0=g is holomorphic near a, it follows that f 0=f has a simple pole at z = a and the residue is −N. If we combine this Proposition with the Residue Theorem, we obtain: Theorem 1.5 (The Argument Principle). Let Ω ⊂ C be a connected open set, and let f be mero- morphic on Ω. Let Γ be a closed curve in Ω which does not pass through any zero or pole of f, and such that the interior of Γ is contained in Ω. Then 1 Z f 0(ζ) dζ = Z(Γ) − P (Γ) 2πi Γ f(ζ) where Z(Γ) is the number of zeros of f (counted with multiplicity) inside Γ, and P (Γ) is the number of poles of f (counted with multiplicity) inside Γ. COMPLEX ANALYSIS 3 d f 0 If we could define a branch of the logarithm on the range of f, then dz (log(f)) = f . Thus R f 0(ζ) Γ f(ζ) dζ is the change in log(f) as we go around the curve Γ. However, the real part of the logarithm does not change as we go around a closed curve, but the imaginary part gives the total 1 R f 0(ζ) change in the angle. Thus 2πi Γ f(ζ) dζ gives the total number of times that the curve t ! f(Γ(t)) winds around the origin. 3/4/09 As an application, we have Theorem 1.6 (Rouch´e'sTheorem). Let Ω ⊂ C be a connected open set, and let f and g be holomorphic on Ω. Let Γ ⊂ Ω be a closed curve such that the interior of Γ is contained in Ω. If jf(z) − g(z)j < jf(z)j for all z 2 Γ, then f and g have the same number of zeros on Int(Γ). Proof. It follows from the assumption that neither f(z) nor g(z) can equal zero for z 2 Γ. In particular, f is not identically zero, and so the function F (z) = g(z)=f(z) is meromorphic on Ω. Moreover, jF (z) − 1j < 1 for z 2 Γ. Thus the image under F of the curve Γ is a compact subset of the open disk D(1; 1) centered at 1 with radius 1. Let log(w) be a branch of the logarithm defined on D(1; 1). Then log(F (z)) is defined and holomorphic in a connected open neighborhood U of Γ. Moreover, for z 2 U, we have d F 0(z) log(F (z) = : dz F (z) Thus Z F 0(ζ) Z d dζ = log(F (ζ) dζ = 0: Γ F (ζ) Γ dζ But then it follows from Theorem 1.5 that the number of zeros of F inside Γ equals the number of poles of F inside Γ. Since the zeros of F are just the zeros of g, and the poles of F are just the zeros of f, this gives the desired equality, and completes the proof. n n n−1 Example: Let f(z) = z and let g(z) = z + a1z + ··· + an−1z + an, so that g is a monic polynomial. If jzj = R > 1 we have n−1 n−1 jf(z) − g(z)j = ja1z + ··· + an−1z + anj ≤ ja1jR + ··· + jan−1jR + janj n−1 n−1 ≤ maxfjajjg (R + ··· + R + 1) ≤ n maxfjajjg R n maxfja jg n maxfja jg = j Rn = j jf(z)j R R Thus if R > n maxfjajjg, the functions f and g have the same number of zeros in the disk of radius R. But since f clearly has n zeros, it follows that the monic polynomial g(z) also has n zeros.