Process of wave refraction

SINGULARITIES AND ZEROS OF HOLOMORPHIC FUNCTIONSContents1. Singularities: an introduction12. Zeros of holomorphic functions23. Poles of holomorphic functions34. Distinguishing poles from removable singularities7So far, our focus of study has been holomorphic functions. We will now concentrateon understanding points where functions are not holomorphic. In particular, we wantto generalize our understanding of the behavior off(z) = 1=znearz= 0 to a broaderrange of functions, and eventually prove interesting theorems about those functions.In particular, we will prove a generalization of Cauchy's Theorem to functions whichare more general than holomorphic functions.1.Singularities: an introductionLetfbe a function de ned on an open set . Iffis not de ned atz0, butis de ned in somepunctured disc(sometimes also called adeleted neighborhood)02SINGULARITIES AND ZEROS OF HOLOMORPHIC FUNCTIONSIfz0is a singularity off(z) and it is possible to de nef(z0) in such a way to makefholomorphic atz0, then we callz0aremovable singularityoff. In some ways,removable singularities are the least interesting type of singularity, because they arisefrom holomorphic functions with a few points in the domain deleted. Neverthelessone can prove interesting theorems about removable singularities, which we will dolater.2.Zeros of holomorphic functionsAs the examplef(z) = 1=znmight indicate, a good strategy for understandingsingularities might be to understand zeros of holomorphic functions rst, since thesingularities of 1=znarise at the points where the denominatorznhas zeros.The main theorem on zeros which we will use is the following:Theorem 1.Supposef(z)is a holomorphic function on an open setwhich isnot identically zero. Letz02be a point withf(z0) = 0. Then there exists aunique positive integernsuch that there exists an open discUcontainingz0and aholomorphic functiong(z)de ned onUsuch thatg(z)is nonzero onUandf(z) =(zz0)ng(z)onU.Proof.Becausefis holomorphic atz0, we can nd a power series expansion forfatz0:f(z) =1Xk=0ak(zz0)k:The assumption thatfis not identically 0 guarantees that not all coecientsakequal0. (If all the coecients did equal 0, thenf= 0 on some open disc, which impliesfis 0 on all of .) Furthermore, sincef(z0) = 0,a0= 0. Letnbe the smallest positiveinteger such thatan= 0. We claim thatnis the integer in the theorem.In the disc of convergence of the above power series, we havef(z) = (zz0)n1Xk=nak(zz0)kn:Notice that after factoring out (zz0)nwe have another power series which convergesin the same disc as the original power series. Letg(z) =P1k=nak(zz0)knbe thisnew power series; evidentlygis holomorphic in its open disc of convergence. We needto show that there exists some open discUcontainingz0such thatg(z) is nonzeroon all ofU.Sincean6= 0, we must have thatg(z0)6= 0. Sinceg(z) is continuous on its domain,there exists some open disc centered atz0such thatgis never 0 on that open disc.TakeUto be this open disc.Finally, we need to prove that thendescribed above is unique. Supposemwereanother positive integer such that there existed an open discVand functionh(z) suchthatf(z) = (zz0)mh(z) onV. Then we havef(z) = (zz0)ng(z) = (zz0)mh(z)onU\V, which is an open disc containingz0. Without loss of generality we mayassumenm; then (zz0)nmg(z) =h(z) at all points exceptz0. Ifn > m, then(zz0)nmg(z)!0 asz!z0, which by continuity ofhatz0implies thath(z0) = 0,contradicting the assumption thath(z0)6= 0. Thereforem=n, as desired.