(w) Second, since lim z w z w z w δ. Thus, i r δ, then z w =r (w) z w = (w), there exist δ, M > 0 such that (w) z w M i dz ML({ z w = r}) = M2πr, which tends to 0 as r 0. This shows that g = 2πi(w), which proves (3.2). Theorem 3.5.3. Let and be as in the previous theorem. Then or any w 0 Int(), (w) = (w w 0 ) n 1 2πi (z w 0 ) n+1 dz, w w 0 < dist(w 0, ). (3.3) In particular, is analytic in Int(), and we have the ollowing Cauchy s Formula: (n) (w 0 ) = n! 2πi (z w 0 ) n+1 dz, w 0 Int(), n N {0}. (3.4) Proo. Fix w 0 Int(). Let R = dist(w 0, ) > 0. I w w 0 < R z w 0, then z w = (z w 0 ) (w w 0 ) = /(z w 0 ) 1 (w w 0 )/(z w 0 ) = z w 0 From (3.2) we see that, i w D(w 0, R), then w Int(), and For z, we have (w) = 1 2πi (w w 0 ) n (z w 0 ) n. (z w 0 ) n+1 (w w 0) n dz. (3.5) (z w 0 ) n+1 (w w 0) n w w 0 n R n+1. Since is compact and is continuous on, we have <. Since w w 0 < R, we have w w 0 n R n+1 <. Thus, (z w 0 (w w ) n+1 0 ) n converges uniormly on. This means that we may exchange the with in (3.5) to get (3.3). Thus, is analytic in Int(). Finally, since (n) (w 0 ) = n!a n i (w) = a n(w w 0 ) n near w 0, we get (3.4). Remark. I w 0 lies on the exterior o, then the right hand side o (3.4) is equal to 0 because the unction is holomorphic on Int(), and we may apply Cauchy s Theorem. (z w 0 ) n+1 Homework Compute e 3z γ dz or (i) γ = { z = 3} and (ii) γ = { z = 1}. (z 2) 3 46
Corollary 3.5.1. I is holomorphic in an open set U, then is analytic in U. Moreover, is ininitely many times complex dierentiable, and or every z 0 U, = (n) (z 0 ) (z z 0 ) n, (3.6) n! holds or any z D(z 0, R), where R = i U = C; or R = dist(z 0, U) i U C. Proo. The irst statement ollows rom the previous theorem by choosing = { z z 0 = r} where r > 0 is such that D(z 0, r) U. Since is analytic, it is ininitely many times complex dierentiable. The previous theorem also shows that (3.6) holds or any z D(z 0, r). Since we may choose r to be arbitrarily close to R, we conclude that (3.6) holds or any z D(z 0, R). Theorem 3.5.4. [Morera s Theorem] Suppose is continuous on a domain U, and satisies that or any closed curve γ in U, γ = 0. Then is analytic. Proo. Since U is connected, the assumption implies that has a primitive F in U. Such F is holomorphic, which is also analytic. So = F is also analytic in U. Remarks. 1. So ar we have seen that holomorphic is equivalent to analytic. Thus, i is complex dierentiable in an open set, then it is ininitely many times complex dierentiable in that set. This phenomena does not exist in Real Analysis. In the rest o this course, we will use the words analytic and holomorphic interchangeably. 2. The corollary tells us that the radius o the power series (n) (z 0 ) n! (z z 0 ) n is greater than or equal to dist(z 0, U) i U C. The equality may not hold since could be a restriction o a holomorphic unction deined on a bigger domain. I U = C, i.e., is an entire unction, then R = or any z 0. So we have = (n) (z 0 ) (z z 0 ) n, z 0, z C. n! 3. Recall the ormulas o Cauchy s Theorem and Cauchy s Formula (with certain conditions): = 0, 0 = n k=1 k, (n) (z 0 ) = n! 2πi dz. (z z 0 ) n+1 Theorem 3.5.5. [Liouville s Theorem] Every bounded entire unction is constant. 47
Proo. Let be an entire unction. Suppose that there is M R such that M or any z C. Then or any z C and R > 0 (z) = 1 (w) 2πi (w z) 2 dw. w z =R (w) For w { w z = R}, = (w) (w z) 2 R 2 M R 2. Thus, (z) 1 M 1 L({ z w = R}) = 2π R2 2π M R 2 2πR = M R. Since this holds or any R > 0, we get (z) = 0 or any z C. Thus, is constant. Theorem 3.5.6. [Fundamental Theorem o Algebra] Every non-constant complex polynomial has a zero in C. Proo. Let P (z) = a 0 + a 1 z + + a n z n be a non-constant polynomial with a n 0. Suppose P has no zero in C. Then Q(z) = 1/P (z) is holomorphic in C. We will show that Q is bounded on C. This requires a careul estimation. We ind that P (z) a n z n = 1 + a n 1 a n z 1 + + a 0 z n 1, as z. Thus, as z, P (z), which implies that Q(z) 0. So there is R > 0 such Q(z) 1 i z > R. Since Q is continuous on the compact set D(0, R), it is bounded on this set. So there is M 0 > 0 such that Q(z) M 0 i z R. Let M = max{m 0, 1}. Then Q(z) M or any z C. So Q is a bounded entire unction. Applying Liouville s theorem, we see that Q is constant, which implies that P is constant, which is a contradiction. We may apply FTA to conclude that every polynomial can be actorized into C n k=1 (z z k). Suppose z 1 is a zero o P (z), then P (z) = P (z) P (z 1 ) = n a k (z k z1 k ) = (z z 1 ) k=0 n k=1 a k (z k 1 + z k 1 1 ) =: (z z 1 )Q(z). Note that Q is a polynomial o degree n 1. I deg(q) = 0, then Q is constant; i deg(q) 1, we may ind a zero o Q and actorize Q. The conclusion ollows rom an induction. Homework III, 7: 3 Additional problems: 1. Let be an entire unction. Suppose that there exists r > 0 such that r or every z C. Prove that is constant. 48
3.6 Dierentiability o the Logarithm Function Let U be an open set with 0 U. Recall that L(z) is called a branch o log z in U, i it is continuous in U, and satisies e L(z) = z or any z U. In this section, we will show that a branch o log z is not only continuous, but also holomorphic, and its derivative is the unction 1/z. Theorem 3.6.1. Suppose is a primitive o 1/z in a domain U. Then there is some C C such that + C is a branch o log z in U. Proo. Let g(z) = e z. Then g is holomorphic in U. We compute g (z) = e (z) z e z 2 = e z 2 e z 2 = 0, z U. Since U is connected, g is constant C 0, which is not 0. Thus, e = C 0 z or any z U. Let C C be such that e C = 1/C 0. Then e +C = z in U. Since is holomorphic in U, + C is continuous in U, and so is a branch o log z in U. Theorem 3.6.2. Let U C \ {0} be a simply connected domain. Then there is a branch o log z in U, which is a primitive o 1/z in U. Proo. Since 1/z is holomorphic in U, which has a primitive in U as U is simply connected. Let denote this primitive. From the above theorem, there is a constant C such that g := + C is a branch o log z. The g is what we need since g = = 1/z in U. Lemma 3.6.1. I is a continuous unction on a domain U such that Z or every z U, then is constant. Proo. Let z 0 U and n 0 = (z 0 ) Z. Let A = 1 ({n 0 }) z 0. Since is continuous on U, A is relatively closed in U. On the other hand, since only takes integer values, A = 1 ((n 0 1/2, n 0 + 1/2)). Since is continuous on U, A is relatively open in U. Since U is connected, and A is not empty, we have A = U. So is constant n 0. Theorem 3.6.3. I L is a branch o log z in an open set U C \ {0}, then L is a primitive o 1/z in U. Proo. Let z 0 U. Let r > 0 be such that D(z 0, r) U. Since D(z 0, r) is simply connected, rom Theorem 3.6.2, there is a branch M(z) o log z in D(z 0, r), which is a primitive o 1/z in that disc. Since L(z) and M(z) are both branches o log z in D(z 0, r), we ind that := (L(z) M(z))/(2πi) is an integer valued continuous unction on the disc. From the above lemma, is constant. So L = M = 1/z in D(z 0, r). Since z 0 U is arbitrary, we conclude that L is a primitive o 1/z in U. 49
Let be an analytic unction in an open set U, which does not take value 0. We say that g is a branch o log in U, i g is continuous and satisies that e g =. Note that i g is holomorphic, then e g g =, which gives g =. Following the above proos, we can prove that g is a primitive o ollowing the order below: 1. Suppose h is a primitive o in U. Then there is some C C such that h+c is a branch o log in U. For this statement, note that d e h(z) dz = eh(z) h (z) eh(z) (z) 2 = 0. 2. I U is a simply connected domain, then there exists a branch o log, which is also a primitive o. To prove this, we irst ind a primitive o, and then add a constant to it. 3. For general U, we restrict our attention to discs in U, which are simply connected domain, and prove that g is a primitive o on each discs. Homework. Chapter III, 6: 6. Additional problems. 1. Fix α C. The unction (1 + z) α in D(0, 1) is deined as e αl(z), where L is a branch o log(1 + z) that satisies L(0) = 0. Prove that (1 + z) α is holomorphic in D(0, 1), and where ( α n) was deined earlier. (1 + z) α = ( ) α z n, z D(0, 1), n 2. Let g be holomorphic in a simply connected domain U. Show that there is, which is holomorphic in U without zero, such that g = in U. 3.7 The Maximum Modulus Principle Theorem 3.7.1. [Mean Value Theorem or Holomorphic Functions] Let be holomorphic on a closed disc D(z 0, r). Then (z 0 ) = 1 2π (z 0 + re iθ )dθ; 2π 0 (z 0 ) = 1 πr 2 dxdy. (3.7) z z 0 r 50
Proo. From Cauchy s Formula, (z 0 ) = 1 dz = 1 2π (z 0 + re iθ ) 2πi z z 0 =r z z 0 2πi 0 re iθ ire iθ dθ = 1 2π (z 0 + re iθ )dθ. 2π 0 This is also true i r is replaced by any s (0, r). Thus, 2πs(z 0 ) = Integrating s rom 0 to r, we get πr 2 (z 0 ) = 2π r 0 0 2π where in the last step we used rdrdθ = dxdy. 0 (z 0 + se iθ )sdθ, 0 s r. (z 0 + se iθ )sdsdθ = z z 0 r dxdy, Theorem 3.7.2. [Maximum Modulus Principle or Holomorphic Functions] Let be holomorphic in a domain U. Suppose attains local maximum at some z 0 U, i.e., there is r > 0 such that (z 0 ) or any z D(z 0, r). Then is constant in U. Proo. First, we show that is constant on D(z 0, r). I not, there is z 1 D(z 0, r) such that (z 1 ) < (z 0 ). Let ε = (z 0 ) (z 1 ) > 0. Since is continuous, we may ind r 1 > 0 such that D(z 1, r 1 ) D(z 0, r) and (z 1 ) + ε/2 = (z 0 ) ε/2 or z D(z 1, r 1 ). Let D 0 = D(z 0, r) and D 1 = D(z 1, r 1 ). From (3.7), we get πr 2 (z 0 ) dxdy = dxdy + dxdy D 0 D 1 D 0 \D 1 ( (z 0 ) ε/2)dxdy + (z 0 ) dxdy = (z 0 ) dxdy ε D 1 D 0 \D 1 D 0 2 πr2 1 < πr 2 (z 0 ), which is a contradiction. Thus, is constant in D(z 0, r). Thus, is constant in D(z 0, r) by a homework problem. By uniqueness theorem, is constant in the entire domain U. Corollary 3.7.1. Let U be a bounded domain and U be its closure. Suppose is a continuous on U, and holomorphic in U. Then there is z 0 U such that (z 0 ) or any z U. In short, the maximum o on U is attained at U. Proo. Since U is bounded, U is a compact set. Since is continuous on U, it attains its maximum at some w 0 U. I w 0 U, then the proo is done by taking z 0 = w 0. I w 0 U, then rom the above corollary, is constant in U. From the continuity, that is also constant in U. In this case, any z 0 U works. 51