Lecture 27:AntiderivativesDan SloughterFurman UniversityMathematics 39April 23, 200427.1 Equivalent conditionsTheorem 27.1. Suppose D ⊂ C is a domain and f : D → C is continuouson D. Then f has an antiderivative F on D if and only ifZC1f(z)dz =ZC2f(z)dzwhenever C1, C2⊂ D have the same initial point z1and the same final pointz2.Proof. Suppose f has an antiderivative F on D and let C be a smooth arcwith parametrization z(t), a ≤ t ≤ b. Let z1= z(a) and z2= z(b). ThenZCf(z)dz =Zbaf(z(t))z0(t)dt = F (z(t))ba= F (z2) − F (z1),and so would be the same for any smooth arc from z1to z2. If C is acontour consisting of smooth arcs Ck, with initial point zk−1and final pointzk, k = 1, 2, . . . , n, thenZCf(z)dz =nXk=1ZCkf(z)dz =nXk=1(F (zk) − F (zk−1)) = F (zn) − F (z0),a value which, again, depends only on the the initial and final points of C.1Now suppose the value ofZCf(z)dzdepends only on the initial and final points of C. Let z0∈ D and defineF (z) =ZCf(s)dsfor any contour C in D with intial point z0and final point z. Since this valuedoes not depend the particular contour C, we will denote the integral byZzz0f(s)ds.We need to show that F0(z) = f (z) for any z ∈ D. Choose a γ neighborhoodof z lying in D and a ∆z with 0 < |∆z| < γ. ThenF (z + ∆z) − F (z) =Zz+∆zz0f(s)ds −Zzz0f(s)ds=Zzz0f(s)ds +Zz+∆zzf(s)ds −Zzz0f(s)ds=Zz+∆zzf(s)ds.NowZz+∆zzds = zz+∆zz= ∆z,and sof(z) = f (z)Rz+∆zzds∆z=1∆zZz+∆zzf(z)ds.HenceF (z + ∆z) − F (z)∆z− f(z) =1∆zZz+∆zzf(s)ds −Zz+∆zzf(z)ds=1∆zZz+∆zz(f(s) − f(z))ds.2Now given  > 0, choose an α > 0 such that|f(s) − f(z)| < whenver |s − z| < α. Let δ be the smaller of γ and α. Then, whenever|∆z| < δ, we haveF (z + ∆z) − F (z)∆z− f(z)<1|∆z|(|∆z|) = .Hencelim∆z→0F (z + ∆z) − F (z)∆z− f(z)= 0,and soF0(z) = lim∆z→0F (z + ∆z) − F (z)∆z= f(z).Theorem 27.2. Suppose D ⊂ C is a domain and f : D → C is continuouson D. ThenZC1f(z)dz =ZC2f(z)dzwhenever C1, C2⊂ D have the same initial point z1and the same final pointz2if and only ifZCf(z)dz = 0whenever C ⊂ D is a closed contour.Proof. Suppose the value ofZCf(z)dzdepends only on the initial and final points of C. Given a closed contour C,let z1and z2be distinct points on C. Write C = C1− C2, where C1and C2are the two parts of C having initial point z1and final point z2. ThenZC1f(z)dz =ZC2f(z)dz,and soZCf(z)dz =ZC1f(z)dz −ZC2f(z)dz = 0.3Now supposeZCf(z)dz = 0for any closed contour C ∈ D. Let C1and C2be two contours in D, bothhaving initial point z1and final point z2. Then C = C1− C2, and so0 =ZCf(z)dz =ZC1f(z)dz −ZC2f(z)dz.ThusZC1f(z)dz =ZC2f(z)dz.27.2 ExamplesExample 27.1. For any contour C with initial point 0 and final point 1 + i,ZCzdz =Z1+i0zdz =12z21+i0=12(1 + i)2= i.Example 27.2. SinceF (z) = −1zis an antiderivative off(z) =1z2on the domain D = {z ∈ C : z 6= 0}, it follows thatZC1z2dz = 0for any closed contour C in D.Example 27.3. Let C1be the right half of the circle |z| = 4, extending from−4i to 4i. ThenZC11zdz = Log(z)4i−4i=ln(4) + iπ2−ln(4) − iπ2= πi.4Now let C2be the lefthand side of the same circle, starting at 4i and endingat −4i. Although we could not use Log(z) to evaluateZC21zdz,we could use another branch of log(z), for example,log(z) = ln(r) + iθ, 0 < θ < 2π.Using this branch, we haveZC21zdz = log(z)−4i4i=ln(4) + i3π2−ln(4) + iπ2= πi.Note that C = C1+ C2is the circle |z| = 4, and we haveZC1zdz =ZC11zdz +ZC21zdz = πi + πi =