Lecture 46:PolesDan SloughterFurman UniversityMathematics 39May 25, 200446.1 Types of singular pointsIf z0is an isolated singular point of f , then, for some R > 0,f(z) =∞Xn=0an(z −z0)n+∞Xn=1bn(z −z0)nfor all z with 0 < |z − z0| < R. We call∞Xn=1bn(z − z0)nthe principal part of f at z0. If for some positive integer m, bm6= 0 andbm+1= bm+2= ··· = 0, that is,f(z) =∞Xn=0an(z − z0)n+mXn=1bn(z − z0)nwith bm6= 0, then we say z0is a pole of order m. If m = 1, we say z0is asimple pole. If an infinite number of the coefficients bnare nonzero, we say z0is an essential singular point of f. If bn= 0 for all n, we say z0is a removablesingular point.Example 46.1. Sincesin(z)z3=1z2−13!+z25!− ··· ,1f(z) =sin(z)z3has a pole of order m = 2 at z = 0.Example 46.2. Sincesin(z)z= 1 −z23!+z45!− ··· ,f(z) =sin(z)zhas a removable singular point at z = 0.Example 46.3. Sincee1z=∞Xn=01n!zn,f(z) = e1zhas an essential singular point at z = 0.46.2 Residues at polesProposition 46.1. An isolated singular point of a function f is a pole oforder m if and only if there is a function ϕ such that ϕ is analytic at z0,ϕ(z0) 6= 0, andf(z) =ϕ(z)(z − z0)m.Moreover, in this caseResz=z0f(z) =ϕm−1(z0)(m − 1)!.Example 46.4. Iff(z) =1(z + 1)(z − 1)3,then we may writef(z) =ϕ(z)(z − 1)3whereϕ(z) =1z + 1.Hence z = 1 is a pole of order m = 3 andResz=1f(z) =ϕ00(1)2=18,2as we have seen before. For the residue at z = −1, we writef(z) =ϕ(z)z + 1whereϕ(z) =1(z − 1)3.Hence z = −1 is a simple pole and we haveResz=−1f(z) = ϕ(−1) = −18.Example 46.5. Letf(z) =1z4+ 1=1z −1+i√2z −−1+i√2z −−1−i√2z −1−i√2.ThenResz=1+i√2f(z) =11+i√2−−1+i√21+i√2−−1−i√21+i√2−1−i√2=−1 − i4√2andResz=−1+i√2f(z) =1−1+i√2−1+i√2−1+i√2−−1−i√2−1+i√2−1−i√2=1 − i4√2Hence if R > 1 and C is the contour, with positive orientation, consisting ofthe upper half of the circle |z| = R and the segment along the real axis from−R to R, thenZC1z4+ 1dz = 2πi−12√2i=π√2.Now if CRis the upper half of the circle |z| = R, thenZCR1z4+ 1dz ≤1R4− 1· 2πR =2πRR4+ 1.HencelimR→∞ZCR1z4+ 1dz = 0.3Butπ√2=ZC1z4+ 1dz =ZR−R1x4+ 1dx +ZCR1z4+ 1dz,and solimR→∞ZR−R1x4+ 1dx =π√2− limR→∞ZCR1z4+ 1dz =π√2.It f ollows thatZ∞−∞1x4+ 1dx