COMPLEX ANALYSIS Spring 2014

Transcription

1 COMPLEX ANALYSIS Spring 24 Homework 4 Solutions Exercise Do and hand in exercise, Chapter 3, p. 4. Solution. The exercise states: Show that if a <, then 2π log ae iθ dθ =, Then prove the above result remains true if we assume only a. We did this in class by now. Assume a < a, The function z az is holomorphic and never in D r () where r = / a > ; since D r () is simply connected there is a holomorphic function h such that az = e h (z) for z D r (); that is, there is a holomorphic determination of log( az) = h(z). The function h is determined uniquely up to a multiple of 2πi; in particular, Reh(z) = log az. By Cauchy s formula, since the positively oriented unit circle is contained in D r () h() = h(z) dz = 2π h(e iθ ) 2πi z = z 2πi e iθ ie iθ dθ = 2π Taking real parts we get log = 2π 2π log ae iθ dθ. 2π h(e iθ ) dθ. The result follows. Assuming now a =, a simple change of variables allows ua to assume that a =. Let ɛ > and consider the curve γ = σ ɛ + τ ɛ where σ ɛ (th) = e iθ, ɛ < θ < 2π ɛ ad τ ɛ is a curve of length of order ɛ closing the gap in σ ɛ and leaving outside. The easiest example, and the one I ll use, is τ ɛ (t) = e iɛ + t(e iɛ e iɛ ), t. Notice that Re τ ɛ (t) = cos ɛ < for all t [, ] and we see that γ is a closed curve contained in an open simply connected set in which there is an analytic determination of log( z). As before (since W γ () = ), = 2π ɛ ɛ log e iθ dθ + Re i Re log( τ ɛ (t)) τ τ ɛ (t) ɛ(t) dt. (The /i in front of the second integral is what remains after everything was multiplied by 2π.)

2 It all reduces to proving lim Re ɛ i Re log( τ ɛ (t)) τ τ ɛ (t) ɛ(t) dt =. Now i τ ɛ(t) = (e iɛ e iɛ )/i = 2 sin ɛ ɛ; we also see that τ ɛ = e iɛ + 2it sin ɛ 2 sin ɛ /2 for ɛ > small enough. On the other hand, τ ɛ (t) = e iɛ 2it sin ɛ = cosɛ i sin ɛ 2it sin ɛ. For small values of ɛ, cos ɛ ɛ 2 /4, sin ɛ is of order ɛ; one can get a bound of the form τ ɛ (t) cɛ 2 for ɛ > small, c a positive constant. Then log τ ɛ (t) = log τ ɛ (t) log(cɛ 2 ) = 2 log ɛ + log c. Putting all this together one sees that Re i Re log( τ ɛ (t)) τ τ ɛ (t) ɛ(t) dt (C + C 2 log ɛ )ɛ as ɛ. Exercise 2 Exercise 3, Chapter 3, p. 5. Solution. To prove: Suppose f is holomorphic in a punctured disk D r(z ) = {z C : < z z < r and there exist constants A, ɛ > such that f(z) A z z +ɛ for all z near z. Then the singularity at z is removable. Proof. Let g(z) = (z z )f(z) for z D r(z ). Then g is bounded near z ; in fact lim z z g(z) = ; thus defining g() =, g is holomorphic in all of D r (z ). Since g(z ) =, there is k N such that g(z) = (z z ) k h(z) for z D r (z ), with h holomorphic in D r (z ). It follows that f(z) = (z z ) k h(z) for z D r(z ); defining f(z ) = lim z z (z z ) k h(z), f becomes holomorphic also at z. Exercise 3 Exercise 4, Chapter 3, p. 5. Solution. To Prove: If f is entire and injective, then f(z) = az + b for some a, b C, a. Proof. The hint suggests using the Casorati-Weierstrass Theorem, applied to f(/z). So define g by g(z) = f(/z) if z. What the Casorati-Weierstrass Theorem tells us is that g cannot have an essential singularity at. In fact, assume it did. Then U n = g(d /n ()) is dense in C for every n N. But by the open mapping theorem, U n is also open. By Baire s Theorem, n= U n is dense; in particular, not empty. Let w n= U n. Then w = f(z n ) for some z n C, < z n < /n, n=,2,.... Clearly z n z if m > / z, hence g is not injective.. But if g is not injective, then f is not injective. 2

3 Since the behavior of f at infinity is, by definition, that of g at, we see that f does not have an essential singularity at infinity. By Theorem 3.4 in the textbook, it has to be a rational function thus, being entire, a polynomial. The only injective polynomials are those of the form f(z) = az + b with a. This fact may require a proof. Clearly a constant polynomial is not injective, so the alternative is f(z) is a polynomial of degree n >. If f has two distinct zeros it can t be injective, so we are reduced to the case f(z) = a(z z ) n, where a, some z C. Such a polynomial is trivially not injective. Exercise 4 Exercise 5, Chapter 3, p Solution. Use the Cauchy inequalities or the maximum modulus principle, or whatever, to prove the following statements.. If f is an entire function satisfying sup f(z) AR k + B z =R for all R >, some constants A, B, some integer k, then f is a polynomial of degree k. Proof. Let f(z) = n= a nz n be the power series expansion of f at. By the Cauchy inequalities, a n ARk + B R n for all n =,, 2 ldots; letting R we see that a n = for n > k. 2. If f is holomorphic and bounded in the unit disk D = D () and converges uniformly to in the sector θ < arg z < ϕ as z, then f is identically. Proof. Without loss of generality we may assume that the sector is centered about the positive x-axis, thus given by S = {z : z <, arg z < θ} for some θ > ; we may also assume θ π/2 (though this last assumption is not really going to be used). A first approach could be to let δ > be small enough so that D δ (), the disk of center, radius δ satisfies D δ () D S. Let Ω = D D δ () and define g : Ω C by g(z) = f(z) if z D, g9z) = if z D δ ()\D. Now use Morera s Theorem, or the Cauchy formula, to prove g is holomorphic, and presto! 3

4 A picture of Ω But it is simpler than that. I start again. As before we assume the sector S is centered around the x-axis. For ɛ >, ɛ < r, let w ±r be the intersections of the circle of center radius r with the circle of center, radius ɛ. It is not hard to find precise values for w ±r but for our purposes it will suffice to see that w ±r = re ±iϕr, ϕ r ϕ < θ. Extend f to all of C by f(z) = if z D. This is a somewhat violent extension, we really only need it around, enough to say that the restriction of this new f to the circle of center, radius is continuous. We can thus define g : D ɛ () C by g(z) = f(ζ) dζ 2πi z =ɛ ζ z, and g is holomorphic in D ɛ (). It is however easy to prove that g(z) = f(z) if z D ɛ () D, g(z) = in D ɛ ()\D, and since the second set is open, non-empty, we must have g =, thus f = in D ɛ () D, hence in D. It is a technicality to prove all this. For ɛ < r let γ r be the closed path that starts, say, at w r, follows the circle z = ɛ in the positive (counterclockwise) direction to w r, then goes back to w r along the circle z = r in the negative direction. 4

5 The path γ r in red So far we really only needed that lim z f(z) = for z S; uniform continuity was not needed. We will now need the uniform convergence to say for all z D ɛ (): lim r + f(ζ) dζ 2πi γ r ζ z = f(ζ) dζ 2πi γ ζ z = g(z). The first equality is true because of the uniform convergence, the second one because f is zero on the parts where γ differs from the circle z = ɛ. If z D ɛ () D, then f(ζ) dζ 2πi γ r ζ z = f(z) for r > z ; if z D ɛ ()\D, then f(ζ) dζ 2πi γ r ζ z = for all r ( ɛ, ). Notice that we did not need to use that f is bounded. On the other hand, with the hypothesis that f is bounded, one does not need to have uniform convergence to in the sector; one could have used Lebesgue dominated (bounded) convergence instead. 3. Solution. This is a cute exercise: Let w,..., w n be points on the unit circle in the complex plane. Prove that there exists a point z on the 5

6 unit circle such that the product of the distances of z to the points w j, j n is at least. Conclude that there exists a point z on the unit circle such that the product of the distances of z to the points w j, j n is exactly. Proof. Consider f(z) = n j= (z w j). We have f() = n j= w j = thus, by the maximum modulus principle there has to exist z with z = and f(z). Finally, since θ f(e iθ : [.2π] R is continuous and assumes the value and a value larger than od equal, it has to assume the value. 4. Solution. Here we are required to show that if the real part of an entire function is bounded, then the function is constant. There are many approaches, here is a quick one. Let f(z) = u(z) + iv(z) be entire and assume merely that u(z) M for all z C. Let g be defined by g(z) = e f(z). Then g(z) = e u(z) e M. Thus g is entire, bounded, hence g is constant. This forces f to be also constant. Exercise 5 Exercise 6, Chapter 3, p. 6. Suppose f, g are holomorphic in an open set containing the closed disk D (). Suppose f has a simple zero at zero and vanishes nowhere else in D (). For ɛ C let f ɛ = f + ɛg. Show that if ɛ > is sufficiently small then. f ɛ has a unique zero in D (), and 2. if z ɛ is this zero, then the mapping ɛ z ɛ is continuous. Solution. I think this exercise screams for Rouché s Theorem. Let a = min z = f(z), b = max z = g(z), then a >. Taking ɛ so that ɛ < a/b we see that ɛg(z) < f(z) for z = hence, by Rouché s Theorem, f + ɛg has the exact same number of zeros in D () as f does; namely, one zero. Call that zero z ɛ. To prove continuity we apply Rouché once again. Let ɛ < a/b. Let η > be given. Since f ɛ (z) if z z ɛ, we have that if we define c by c = min f ɛ(z), z z ɛ =η then c >. Let d = max z zɛ =η g(z). If ɛ ɛ < c/d, then for z z ɛ = η we have that (ɛ ɛ)g(z) < f ɛ (z), hence f ɛ = f ɛ + (ɛ ɛ)g has the same number of zeros as f ɛ in z z ɛ < η. Since the only zero of f ɛ is z ɛ this proves that z ɛ z ɛ < ɛ if ɛ ɛ < c/d. Exercise 6 Exercise 7, Chapter 3, p. 6. 6

7 Solution. We are asked to prove: Assume f is not constant and holomorphic in an open set containing the closed unit disc. Then. If f(z) = whenever z =, then the image of f contains the unit disc. 2. If f(z) whenever z = and there exists a point z with z < and f(z ) <, then the image of f contains the unit disc. One would assume that the first part is needed to prove the second part. I ll prove both sort of simultaneously. In the first place, in both cases there is z D such that f(z ) <. In case 2, by hypothesis. In case, by the maximum modulus principle since f(z) = on the boundary, and is not constant. Claim: In both cases, there is z D such that f(z ) =. In fact, otherwise h = /f is holomorphic in some open set containing the closed unit disc and h(z) for z = but h(z ) > in violation of maximum modulus. The claim is established. Let w D. The constant function g(z) = w for all z is, of course, very holomorphic and satisfies g(z) = w < f(z) for all z with z = ; By Rouché s Theorem, both f and f + g have the same number of zeros in D. Since f has at least one, so does f + g; i.e., there is z D such that f(z) = w. Laurent Series Note: Solutions to these rather simple exercises can be found in a new set of notes posted unedr the title of Laurent Series. I may even have corrected some of the obvious typos. It may be convenient to have a notation for an annulus. If z C, r < r, I ll denote by A(z, r, r ) the annulus defined by A(z, r, r ) = {z C : r < z z < r }. If r =, then A(z,, r ) = D r (z). Exercise 7 Let z C, r < r and assume that f : A(z, r, r ) C is holomorphic.. Use the general form of Cauchy s Formula (as presented in my notes on Cauchy and Runge under one roof) to prove: Lemma Let ρ, ρ be such that r < ρ < ρ < r. Then f(z) = f(ζ) 2πi ζ z dζ f(ζ) 2πi ζ z dζ ζ z =ρ for all z such that ρ < z z < ρ. ζ z =ρ 7

8 Note: By f(z) dz we will understand the integral over the positively oriented circle of center z and radius z z =r r. 2. With ρ, ρ as in Lemma, consider ζ z when ζ z = ρ or ρ and z A(z, ρ, ρ ). If ζ z = ρ show that ζ z = (ζ z ) n (z z ) n, n= with absolute convergence for z z > ρ ; for each fixed z, convergence being uniform in ζ, for ζ z = ρ. (The Weierstrass M-test does it all in one fell swoop.) On the other hand, if ζ z = ρ show that ζ z = (ζ z ) n (z z ) n, n= with absolute convergence for z z < ρ ; for each fixed z, convergence being uniform in ζ, for ζ z = ρ 3. Prove the following theorem: Theorem 2 Let z C, r < r and let f : A(z, r, r ) C be holomorphic. Then there exist a n C for n =, ±, ±2,... such that f(z) = n= a n (z z ) n for all z A(z, r, r ). The series converges absolutely in A(z, r, r ). In fact, defining () (2) S + (z) = S (z) = a n (z z ) n, n= a n (z z ) n = a n (z z ) n, n= n= the series for S + converges for z z < r ; the one for S for z z > r, so A(z, r, r ) is in the region where both series converge, and f = S + + S. Moreover, the coefficients a n are uniquely determined and satisfy: If r < r < r, then a n = 2πi z z =r f(z)(z z ) n dz, n Z. 8

9 The series constructed/defined in Theorem 2 is known as the Laurent series of f. The following is a rather trivial exercise, I hope: Exercise 8 Let z C, r < r and let f : A(z, r, r ) C be holomorphic. Let f(z) = a n (z z ) n n= be the Laurent expansion of f in A(z, r, r ). Prove: f extends to an analytic function in the whole disk D r (z ) if and only if a n = for all n <. An interesting case, perhaps the most interesting one, is the case in which r = so that z is either a removable or an isolated singularity of f. Exercise 9 Let z C, < r and let f : A(z,, r) C be holomorphic. Let f(z) = a n (z z ) n = S ( z) + S + (z) n= be the Laurent expansion of f in A(z,, r). Prove:. f has a pole of order k at z if and only if a k and a n = if n < k. 2. f has an essential singularity at z if and only if the set {n Z : n <, a n } is infinite. And now we should get our hands slightly dirty: Exercise expand into a Laurent series the following functions, in the given annuli.. z e /z, in A(,, ). 2. z z 2, in A(,, ) z (z )(z 2) in A(,, 2). 9