NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II

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1 NATIONAL UNIVERSITY OF SINGAPORE Department of Mathematics MA4247 Complex Analysis II Lecture Notes Part II Chapter 2 Further properties of analytic functions 21 Local/Global behavior of analytic functions; The Identity Theorem Defnition A point z C is said to be a zero of f if f(z ) = A point z is said to be an isolated zero of f if f(z ) =, and there exists some r > such that f(z) for all z satisfying < z z < r r z f(z) Proposition 211 Suppose that f is analytic at z and f(z ) =, then either (i) z is an isolated zero of f, or (ii) f is identically zero in some open ball centered at z (ie, δ > such that f(z) = for all z B(z, δ)) [Roughly speaking, zeros of non-constant analytic functions are always isolated] Proof Since f is analytic at z, there exists an open set, which can be taken to be an open ball B(z, r) with r >, such that f is differentiable (and hence analytic) everywhere in B(z, r) By Taylor s theorem, we have f(z) = k= a k (z z ) k for all z z < r, where a k = f (k) (z ) (1) k! If all the Taylor coefficients a k s are zero, then by (1), f(z) = for all z z < r, and thus (ii) holds Otherwise, let m be the smallest positive integer such that a m Then by (1), we have f(z) = a m (z z ) m + a m+1 (z z ) m+1 + = (z z ) m[ a m + a m+1 (z z ) + a m+1 (z z ) 2 + ] = (z z ) m φ(z) for all z z < r, (2) 1

2 2 where φ(z) := a m + a m+1 (z z ) + a m+1 (z z ) 2 + The power series φ(z) has the same radius of convergence as the Taylor series of f(z) about z, which is r, hence it represents an analytic (and hence continuous) function on B(z, r), with φ(z ) = a m By continuity, there exists δ > such that φ(z) for all z z < δ See tutorial 1 for details of argument Hence, there exists δ > such that φ(z) for all z B(z, δ) Shrinking δ if necessary, we may assume that δ < r (so that both (2) and (3) hold on B(z, δ)) Together with the fact that (z z ) m if z z, it follows from (2) that f(z) for all z B (z, δ) = B(z, δ) \ {z }, and thus (i) holds Thus, either (i) or (ii) has to hold, and this finishes the proof of the proposition Exercise Find an infinitely differentiable non-constant function from R to R which has a non-isolated zero y x Solution Consider the function { e 1 x, if x > ; f(x) = if x

3 3 As a consequence of the above, we have the following theorem: Theorem 212 (Locally zero implies globally zero) Suppose that f is analytic in a domain D and f on a non-empty open subset of D Then f on D Proof Suppose f is analytic on D and f on a non-empty open subset U of D Fix a point z U so that f(z ) = Take any other point w D, we want to show that f(w) = Since D is connected, there exists a polygonal line L = L L k in D joining z to w, ie, the initial point of L 1 is z and the terminal point of L k is w First we show that f along L 1 Let the terminal point of L 1 be z 1 (without loss of generality, we may assume z 1 z ) so that L 1 may be parametrized by Consider the set L 1 : γ(t) = z + t(z 1 z ), t 1 I := {s [, 1] f(γ(t)) = for all t s} (*) Since f(z ) =, it follows that I, and thus I Let s o := sup I Claim 1: s o > Proof of Claim 1 Since U is open, there exists an open ball B(z, r) U r with r > Then f on B(z, r) Then for all t 2 z 1 z, we must have f(γ(t)) =, since for such t, Therefore, s o γ(t) z = z + t(z 1 z ) z r 2 z 1 z > = t z 1 z r 2 z 1 z z 1 z = r < r (1) 2 Claim 2: f(γ(t)) = for all t s o Proof of Claim 2 For any t < s o = sup I, there exists s > t such that s I Then f(γ(t)) = since t < s (see (*)) Thus, f(γ(t)) = for all < t s o It remains to show that f(γ(s o )) = Since f is continuous at γ(s o ) and γ(t) γ(s o ) as t s o, it follows that f(γ(s o )) = lim t s o f(γ(t)) = lim t s o =

4 4 (Note t s o implies that t < s o and thus f(γ(t)) = ) This finishes the proof of Claim 2 y z 1 z x Claim 3: s o = 1 Proof of Claim 3 We prove the claim by contradiction Suppose s o < 1, so that by Claim 1, < s o < 1 By Claim 2, f(γ(t)) = for all t s o Also, for any r >, B(γ(s o ), r) contains some point γ(t) with t < s o (and thus f(γ(t)) = ) (Exercise, find an explicit t using a calculation similar to (1)) Thus, γ(s o ) is a zero but not an isolated zero of f Hence by Proposition 211, there exists δ > such that f on B(γ(s o ), δ) Shrinking δ if necessary, we may assume that δ B(γ(s o ), δ) D Then for all s o t s o +, one can check 2 z 1 z that γ(t) B(γ(s o ), δ) and thus f(γ(t)) = (Exercise) Together with δ Claim 2, it follows that f(γ(t)) = for all t s o + 2 z 1 z, and δ thus sup I s o + 2 z 1 z, contradicting that sup I = s o Therefore, we must have sup I = 1, and this finishes the proof of Claim 3 Completion of proof of Theorem 212 From Claim 2 and Claim 3, it follows that f on L 1 In particular, z 1 is a zero of f, but not an isolated zero of f Then since z 1 is also the initial point of L 2, one can repeat the above argument to show that f on L 2 By repeating the argument again and again, one can show that f on the entire L, and thus f(w) = Since w is an arbitrary point of D, it follows that f on D This finishes the proof of the theorem Exercise: The usual definition for D to be connected is that if D = V 1 V 2 where V 1 V 2 = and both V 1 and V 2 are open, then either V 1 or V 2 is the empty set (you cannot partition a connected set into two disjoint non-empty open sets) Use this definition to prove the theorem Definition 213 Consider a set T C A point z (not necessarily in T ) is said to be an accumulation point of T if T B(z, r) \ {z } = for any r >, ie, any open ball centered at z contains a point in T other than z Remark 214 By considering a sequence of open balls B(z, r n ) with r n decreasing to, one can find a sequence of distinct points {z n } in T such that lim n z n = z (Exercise)

5 Example Let T = B(, 1) \ {} Then is an accumulation point of T (note that T ) Theorem 215 (Zeros do not have accumulation point in D if f is not identically zero) Let f be an analytic function in a domain D, and let T D be such that T has an accumulation point z in D If f on T, then f on D Proof Since z D is an accumulation point of T, it follows from Remark 214 that one can find a sequence of distinct points {z n } T such that lim z n = z Then by continuity of f and since each z n T, n we have f(z ) = f( lim n z n) = lim n f(z n) = lim n = Thus z is a zero of f But z is not an isolated zero of f (Exercise) Thus by Proposition 211, there exists δ > such that f(z) = for all z B(z, δ) Shrinking δ if necessary, we may assume that B(z, δ) D Thus by Theorem 212, f on D Remark (i) The result is not necessarily true if z D (ii) Examples of a set T in a domain D with an accumulation point in D are given below: (1) T is a non-empty open subset of D; (2) T is a curve or line segment contained in D; (3) T consists of a sequence of distinct points {z n } D which converges to a point z D For example, D = B(, 2) and T = {1/n : n N} has the accumulation point in D 5

6 6 Theorem 216 (Identity theorem for analytic functions) Let f and g be analytic in a domain D If f g on a subset T D which has an accumulation point in D (such as an non-empty open subset of D or a line segment in D), then f g on D Proof: Consider f g on D Since f g on T D, and T has an accumulation point D, it follows from Theorem 215 that f g on D, hence f g on D Example 217 Consider the functions f(z) = sin(2z) and g(z) = 2 sin z cos z

7 7 22 Maximum modulus principle and applications Theorem 221(Gauss Mean Value Theorem) Suppose that f is analytic everywhere within and on the circle C : z z = r Then f(z ) = 1 f(z + re iθ )dθ Remark The formula says that the value of f at the centre of the circle is the arithmetic mean of the values of f on the circle r z f(z ) f(z) Proof By the Cauchy integral formula, f(z ) = 1 i C f(z) z z dz Take the parametrization of C given by γ(θ) = z + re iθ, θ, so that γ (θ) = ire iθ Hence f(z ) = 1 i = 1 f(z + re iθ ) z + re iθ z ire iθ dθ f(z + re iθ )dθ Example 222 Let f(z) = 2z 2 + z + 1 Then f() = 1 = 1 2e 2iθ + e iθ + 1 dθ

8 8 Proposition 223 (Local version of maximum modulus principle) Suppose that f is analytic on an open ball B(z, r) := {z C : z z < r} centered at z and of radius r > If f(z) f(z ) for each point z B(z, r), then f(z) f(z ) on B(z, r) Remark The above can be rephrased as If f is analytic on B(z, r) and f(z) attains its maximum in B(z, r) at z, then f is constant on B(z, r) f(z ) z r f(z) Proof Suppose f satisfies the conditions of the Proposition, and let < ρ < r Consider the circle z 1 z = ρ By Gauss MVT, f(z ) = 1 = f(z ) = 1 1 On the other hand, by the assumption, = By (2) and (5), f(z + ρe iθ )dθ (1) f(z + ρe iθ )dθ f(z + ρe iθ ) dθ (2) f(z + ρe iθ ) f(z ), θ (3) f(z + ρe iθ ) dθ = f(z ) 1 f(z ) = 1 1 = f(z ) 1 dθ = 1 = 1 f(z ) dθ = f(z ) (4) f(z + ρe iθ ) dθ (5) f(z + ρe iθ ) dθ f(z + ρe iθ ) dθ ( f(z ) f(z + ρe iθ ) ) dθ = (6)

9 9 By assumption, the integrand ( f(z ) f(z + ρe iθ ) ) for all θ (7) If there is a value of θ for which the inequality in (7) is strict, then by continuity, we have strict inequality for an open interval of θ and ( f(z ) f(z + ρe iθ ) )dθ >, which is a contradiction to (6) Hence we must have ( f(z ) f(z + ρe iθ ) ) =, for all θ, ie, f(z) = f(z ) for all z on the circle z z = ρ By letting ρ vary from to r, it follows that f(z) = f(z ) for all z B(z, r) Since f is an analytic function such that f(z) is constant on B(z, r), it follows that that f(z) itself is constant on B(z, r) (see eg [Churchill, 7th ed, p 74, Question 7]) Theorem 224 (Maximum modulus principle) If f is analytic and not constant on the domain D, then f(z) has no maximum value in D Proof: We prove the theorem by contradiction Suppose that f(z) attains its maximum at some point z D Since D is open, there exists r > such that z B(z, r) D Then f(z) f(z ) for all z B(z, r) Thus by the local version of the MMP (Proposition 223), f is constant on B(z, r) Together with the identity theorem for analytic functions (Theorem 216), it follows that f(z) is constant on D, contradicting the assumption that f is non-constant on D Thus, f(z) has no maximum value in D Remark The maximum modulus does not hold in the real case! Consider the function f(x) = 1 x 2, 1 < x < 1 Clearly, the function is non-constant and differentiable on ( 1, 1) But the maximum value of f(x) is attained at the interior point x = when f() = 1 y x

10 1 Corollary 225 Let R C be a closed bounded set whose interior is a domain Suppose f is continuous on R and analytic and not constant in the interior of R Then the maximum value of f(z) in R occurs on the boundary of R and never in the interior Proof Note that f(z) is a real-valued continuous function on the closed and bounded region R Then it follows from the Extreme Value Theorem that f(z) attains its maximum value at some point z of R By the maximum modulus principle, such z cannot lie in the interior of R Therefore, z must lie in the boundary of R Example 226 (minimum modulus principle) Let R C be a closed bounded set whose interior is a domain Suppose f is continuous on R and analytic and not constant in the interior of R If f(z) for any z R, then f(z) attains its minimum value at the boundary of R but not in the interior of R (To be done in tutorial) Example 227 Consider the function f(z) = e z on the closed disk B(1, 2) = {z C : z 1 2} Then the maximum and minimum values of f(z) in B(1, 2) occurs on the boundary, and not on the interior (Exercise: Find the points where the maximum and the minimum values for f(z) occurs)

11 11 An important application of the MMP is the following Theorem 228 (Schwarz s Lemma) Consider the unit ball B(, 1) := {z C : z < 1} Suppose that f is analytic on B(, 1) such that (i) f(z) 1 for each z B(, 1), and (ii) f() = Then (a) f(z) z for each z B(, 1), and (b) f () 1 Moreover, if the equality in (b) holds or the equality in (a) holds for at least one non-zero point in B(, 1), then f(z) az on B(, 1) for some complex constant a with a = 1 (that is, f(z) = e iα z is a rotation by some angle α about the origin) f(z) Proof (Key step:) Let g(z) := f(z) Clearly, g(z) analytic on B(, 1) \ z {}, and it has an isolated singular point at z = From (ii), we have f() = By Taylor s theorem, f(z) = + f ()z + f () z 2 +, for z < 1 2! Thus the Laurent series of g(z) at z = is given by g(z) = 1 z [ f ()z + f () 2 ] z 2 + = f () + f () z +, < z < 1 2 In particular, z = is a removable singular point of g(z), and we can define g() = f () so that the extended function g(z) becomes analytic on B(, 1) (see (163)) Now, since f(z) 1 for z B(, 1), we see that g(z) 1, for < z < 1 (1) z Now for any z 1 B(, 1), we choose an r such that z 1 < r < 1 The function g(z) is analytic on the closed disk B(, r) := {z C : z r}

12 12 and g(z) 1 on the circle z = r, which is the boundary of B(, r) r Hence, by the MMP, g(z 1 ) 1 r (2) Then by taking the limit as r 1, one gets from (2) that g(z 1 ) = lim g(z 1 1) lim r 1 r 1 r = 1 (3) Hence, upon renaming z 1 as z (since z 1 was an arbitrary point in B(, 1)), we have g(z) 1 for all z < 1 (4) Therefore, f(z) z for < z < 1 The inequality clearly holds when z =, since f() = Thus, f(z) z for all z < 1, which gives (a) Recall that g() = f () Thus by (4), we also have f () = g() 1, which gives (b) Finally, if the equality in (b) holds (ie, f () = g() = 1) or the equality in (a) holds at some non-zero point in B(, 1) (which implies g(z) = 1 for some non-zero z B(, 1)), then together with (3), it follows g(z) attains its maximum value 1 at an interior point z B(, 1) Thus by MMP, g(z) is a constant function on B(, 1), ie, g(z) a = e iα on B(, 1) Note also that a = g(z ) = 1 Thus, f(z) az on B(, 1) \ {} with a = 1 Since f() =, the equality also holds at z = This finishes the proof of Schwarz s Lemma