Lecture 37:Laurent SeriesDan SloughterFurman UniversityMathematics 39May 10, 200437.1 Laurent’s theoremThe following result is known as Laurent’s theorem.Theorem 37.1. Suppose z0∈ C, f is analytic in the domainD = {z ∈ C : R1< |z − z0| < R2},and C is any positively oriented, simple closed contour in D, with z0in theinterior of C. Then, for any z ∈ D,f(z) =∞Xn=0an(z − z0)n+∞Xn=1bn(z − z0)n,wherean=12πiZCf(z)(z − z0)n+1dz,n = 0, 1, 2, . . ., andbn=12πiZCf(z)(z − z0)−n+1dz,n = 1, 2, 3, . . ..Note that we could write the series above asf(z) =∞Xn=−∞cn(z − z0)n,1wherecn=12πiZCf(z)(z − z0)n+1,n = 0, ±1, ±2, . . .. Moreover, note if f is analytic on the entire disk |z −z0| <R2, then bn= 0 f or all n andan=f(n)(z0)n!for n = 0, 1, 2, . . .. That is, in this case Laurent’s theorem reduces to Taylor’stheorem.Proof. We will assume z0= 0. The general case follows from a translation,as it did in the proof of Taylor’s theorem. Choose R1< r1< R2andr1< r2< R2so that the annular region r1< |w| < r2contains both z andC. Let C1be the circle |z| = r1and let C2be the circle |z| = r2. Let γ be acircle centered at z with radius smaller than both r2− |z| and |z| − r1. GiveC1, C2, and γ positive orientations.We now have, from extensions to the Cauchy-Goursat theorem, thatZC2f(s)s − zds −ZC1f(s)s − zds −Zγf(s)s − zds = 0.By the Cauchy integral formula,Zγf(s)s − zds = 2πif (z),and soZC2f(s)s − zds −ZC1f(s)s − zds = 2πif (z).That is,f(z) =12πiZC2f(s)s − zds −12πiZC2f(s)s − zds=12πiZC2f(s)s − zds +12πiZC2f(s)z − sds.We now proceed as we did in the proof of Taylor’s theorem. We first notethat, for any positive integer N,1s − z=1s11 −zs2=1s N−1Xn=0znsn+zNsN1 −zs!=N−1Xn=0znsn+1+zN(s − z)sN,from which it follows that12πiZC2f(s)s − zds =N−1Xn=012πiZC2f(s)sn+1dszn+zN2πiZC2f(s)(s − z)sNds=N−1Xn=0anzn+ ρN(z),wherean=12πiZC2f(s)sn+1dsandρN(z) =zN2πiZC2f(s)(s − z)sNds.Similarly, interchanging z and s, we see that, for any positive integer N ,1z − s=N−1Xn=0snzn+1+sN(z − s)zN=NXn=1sn−1zn+sN(z − s)zN,from which it follows that12πiZC1f(s)z − sds =NXn=112πiZC1f(s)s−n+1ds1zn+12πizNZC1sNf(s)z − sds=NXn=1bnzn+ σN(z),wherebn=12πiZC1f(s)s−n+1dsandσN(z) =12πizNZC1sNf(s)z − sds.3Now let M1be the maximum value of |f(s)| on C1, let M2be the maximumvalue of |f (s)| on C2, and le t r = |z|. Then, for s on C2,|s − z| ≥ ||s| − |z|| = r2− r,and sof(s)(s − z)sN≤M2(r2− r)rN2.Hence|ρN(z)| =zN2πiZC2f(s)(s − z)sNds≤rN2π·M2(r2− r)rN2· 2πr2=M2r2r2− r·rr2N.Sincerr2< 1,limN→∞ρN(z) = 0.For s on C2,|z − s| ≥ ||z| − |s|| = r − r1,and sosNf(s)z − s≤rN1M1r − r1.Hence|σN(z)| =12πizNZC1sNf(s)(z − s)ds≤12πrN·rN1M1r − r1· 2πr1=M1r1r − r1·r1rN.Sincer1r< 1,limN→∞σN(z) = 0.Thus we havef(z) =∞Xn=0anzn+∞Xn=1bnzn.4Finally, we note that, since f is analytic in D,an=12πiZC2f(s)sn+1ds =12πiZCf(s)sn+1dsfor n = 0, 1, 2, . . ., andbn=12πiZC1f(s)s−n+1ds =12πiZCf(s)s−n+1dsfor n = 1, 2, 3, . . .. This completes the proof when z0= 0. When z06= 0,define g(z) = f(z + z0) and proceed as in the proof of Taylor’s