Hence limz→∞f(z) =L.33.Ifz6=-1-3ithen the functionh(z) =z-iz+1+3iis continuous inz.The discontinuity atz0=-1-3iis not removable becausez0-i=-1-3i-i=-1-4iandz0+ 1 + 3i= 0. And wehavelimz→-1-3iz-iz+ 1 + 3i=(-1-4i)limz→-1-3i1z+ 1 + 3i=(-1-4i)∞=∞.37.We use the definition of sinzusing the exponential function. We havesinz=eiz-e-iz2i.From the Example 2.2.19 we know thatezis continuous everywhere inC. Sinceeiz=f(g(z)) forg(z) =izandf(z) =eizand both functions are continuous it follows from Theorem 2.2.13 thateizis also continuous. Similar we show that the functione-izis continuous. Then by Theorem 2.2.13sinz=eiz-e-iz2i=12ieiz-12ie-izis continuous inC.41.(a)First, assume thatfis continuous andAis open. Letz0be fromf-1[A]. This meansthatf(z0) is inA.SinceAis open there isε >0 such that theε-neighborhood is a subset ofA, orBε(f(z0))⊂A. Sincefis continuous atz0there is aδ >0 such that if|z-z0|< δthen|f(z)-f(z0)|< ε. Thusf(z) is inA. And thereforezwhich is inf-1[f(z)] is inf-1[A]. Thereforethe wholeδ-neighborhood ofz0is inf-1[A]. And it follows thatf-1[A] is open.Second, assume thatf-1[A] is open wheneverAis open. Letz0be any complex number andε >0 be any too. Then by assumption theε-neighborhoodBε(f(z0)) off(z0) is open too. Thereforef-1[Bε(f(z0))] is open. In particular, there isδ >0 such thatBδ(z0)⊂f-1[Bε(f(z0))]. It followsthat for anyzsuch that|z-z0|< δwe have|f(z)-f(z0)|< ε. Hencefis continuous.(b)First, assume thatf-1[A] is closed wheneverAis closed.To show thatfis continuousby part(a) it is enough to show thatf-1[B] is open wheneverBis open. So, let openBbe given.Since the complement of an open set is closed (problem 17 Section 2.1) we haveA=C\Bis closed.Therefore, by assumptionf-1[A] is closed too. Now we use the identityf-1[B] =C\f-1[A]which is true for any setBandA=C\B. Really, ifzis inf-1[B]. Thenf(z) is inB=C\A.Thereforef(z) is in the complement off-1[A]. On the other hand ifzis inC\f-1[A] thenf(z)cannot be inA=C\B. And, thusf(z) is inB. So,f-1(z) is inf-1[B]. Sincef-1[A] is closedit follows from (2) and Problem 17 Section 2.1 that the complement setf-1[B] is open. Thusfiscontinuous.Second, assume thatfis continuous andBis closed. ThenA=C\Bis open. Therefore bypart (a) it follows thatf-1[A] is open. By (2) it follows thatf-1[B] =C\f-1[A].f-1[A] is open.Then by Problem 17 Section 2.1 it follows that its complement,f-1[B] is closed.
56Chapter 2Analytic FunctionsSolutions to Exercises 2.31.If we takeg(z) = 3z2+ 2zandh(z) =z-1 then we havef(z) = 3(z-1)2+ 2(z-1) =g(h(z)).By the chain rule and the formula that (zn)0=nzn-1for a positive integernwe havef0(z)=g0(h(z))h0(z) = (6(z-1) + 2)1 = 6(z-1) + 2 = 6z-4.5.By the quotient rule and the formula for the derivative of a polynomial it follows1z3+ 10=(1)0(z3+ 1)-1(z3+ 1)0(z3+ 1)2=0-3z2(z3+ 1)2=-3z2(z3+ 1)2.