Lecture 29:Simply Connected DomainsDan SloughterFurman UniversityMathematics 39April 27, 200429.1 Simply connected comainsDefinition 29.1. We say a domain D is simply connected if, whenever C ⊂D is a simple closed contour, every point in the interior of C lies in D. Wesay a domain which is not simply connected is multiply connectedExample 29.1. The domainU = {z ∈ C : |z| < 1}is simply connected. The domainA = {z ∈ C : 1 < |z| < 2}is not simply connected.Theorem 29.1. If D is a simply connected domain and f is analytic in D,thenZCf(z)dz = 0for every closed contour C in D.Proof. If C is a simple closed contour, then the conclusion follows from theCauchy-Goursat theorem. If C is not simple, but intersects itself only afinite number of times, then the conclusion follows by writing C as a sum ofsimple closed contours. We will omit the more difficult situation in which Cintersects itself an infinite number of times.1Corollary 29.1. If D is a simply connected domain and f is analytic in D,then f has an antiderivative at all points of D.Note that, in particular, entire functions have antiderivatives on all of C.29.2 Multiply connected domainsTheorem 29.2. Suppose C is a positively oriented, simple closed curve andthat C1, C2, . . . Cnare negatively oriented, simple closed c ontours, all ofwhich are in the interior of C, are disjoint, and have disjoint interiors. LetR be the region consisting of C, C1, C2, . . . , Cn, and all points which are inthe interior of C and the exterior of each Ck. If f is analytic in R, thenZCf(z)dz +nXk=1ZCkf(z)dz = 0.Proof. Let L1be a polygonal path connecting C to C1, Lka polygonal pathconnecting Ckto Ck+1, k = 1, 2, . . . , n − 1, and Ln+1a polygonal path con-necting Cnto C. Let B1be the part of C from where Ln+1joins C to whereL1joins C, B2the remaining part of C, αkthe part of Ckbetween where Lkand Lk+1join Ck, and βkthe remaining part of Ck. LetΓ1= B1+ L1+ α1+ L2+ α2+ · · · + αn+ Ln+1andΓ2= B2− Ln+1+ βn− Ln+ βn−1− · · · + β1− L1.Then, by the Cauchy-Goursat theorem,ZΓ1f(z)dz = 0 =ZΓ2f(z)dz.Hence0 =ZΓ1f(z)dz +ZΓ2f(z)dz =ZCf(z)dz +nXk=1ZCkf(z)dz.2Corollary 29.2. Suppose C1and C2are positively oriented, simply closedcontours with C2lying in the interior of C1. Let R be the region consistingof C1, C2, and the part of the interior of C1which is in the exterior of C2. Iff is analytic in R, thenZC1f(z)dz =ZC2f(z)dz.Proof. From the previous theorem, we haveZC1f(z)dz −ZC2f(z)dz = 0.Example 29.2. By a homework exercise, if C0is any positively orientedcircle with center at the origin, thenZC01zdz = 2πi.It now follows that if C is any positively oriented, simple closed contour withthe origin in its interior, thenZC1zdz =