Lecture 44:ResiduesDan SloughterFurman UniversityMathematics 39May 21, 200444.1 Some terminologyRecall that we say a point z0is a singular point of a function f if f isnot analytic at z0but is analytic at some point in every neighborho od ofz0. We will say that z0is an isolated singular point if it is a singular pointand there exists  > 0 such that f is analytic in the deleted neighborhood0 < |z − z0| < .Example 44.1. Both z = i and z = −i are isolated singular points off(z) =11 + z2.Example 44.2. z = 0 is a singular point, but not an isolated singular point,of f (z) = Log(z).If z0is an isolated singular point of f, then there exists R > 0 such thatf is analytic in D = {z ∈ C : 0 < |z − z0| < R}. It follows that f(z) has aLaurent s eries representation for all z ∈ D:f(z) =∞Xn=0an(z − z0)n+∞Xn=1bn(z − z0)n.In particularb1=12πiZCf(z)dz,1where C is any positively oriented, simple closed contour which lies in D andhas z0in its interior. In other words,ZCf(z)dz = 2πib1.We call b1the residue of f at the isolated singular point z0, which we willdenoteResz=z0f(z).Example 44.3. Let C be the circle |z − 1| = 1 and consider the integralZC1(z + 1)(z − 1)3dz.Now1z + 1=12 − (−(z − 1))=12·11 −−z−12=12∞Xn=0(−1)n(z − 1)n2n=∞Xn=0(−1)n2n+1(z − 1)nfor all z with |z − 1| < 2, and so1(z + 1)(z − 1)3=∞Xn=0(−1)n2n+1(z − 1)n−3=12(z − 1)3−14(z − 1)2+18(z − 1)+ −116++132(z − 1) −164(z − 1)2+ · · ·for all z with 0 < |z − 1| < 2. HenceResz=11(z + 1)(z − 1)3=18,2and soZC1(z + 1)(z − 1)3dz =π4i.Another approach to evaluating this integral begins with finding the partialfraction decomposition of1(z + 1)(z − 1)3.That is, there are constants A, B, C, and D such that1(z + 1)(z − 1)3=Az + 1+Bz − 1+C(z − 1)2+D(z − 1)3=A(z − 1)3+ B(z + 1)(z − 1)2+ C(z + 1)(z − 1) + D(z + 1)(z + 1)(z − 1)3,which implies that1 = A(z − 1)3+ B(z + 1)(z − 1)2+ C(z + 1)(z − 1) + D(z + 1).Evaluating at z = 1, we find that D =12and at z = −1 that A = −18.Evaluating at z = 2 then gives1 = −18+ 3B + 3C +32and at z = −2 gives1 =278− 9B + 3C −12.Thus3B + 3C = −38−9B + 3C = −158.from which it follows that B =18and C = −14. Hence1(z + 1)(z − 1)3= −18(z + 1)+18(z − 1)−14(z − 1)2+12(z − 1)3.Since the term −18(z+1)is analytic at z = 1, its contribution to the Laurentseries at z = 1 will have only positive powers of z − 1; it follows that the3remaining three terms contribute all the negative powers of z − 1 to theLaurent s eries. Thus we see once again thatResz=11(z + 1)(z − 1)3=18.Alternatively, with the partial fraction decomposition we may observe thatZC1(z + 1)(z − 1)3dz = −ZC18(z + 1)dz +ZC18(z − 1)dz −ZC14(z − 1)2+12(z − 1)3dz=18ZC1z − 1dz=18·