Exercise Solutions to Complex Analysis

Transcription

1 More solutios mul t Thks to the Willim's work! Exercise Solutios to Complex Alysis Complex Alysis is ot tht complex! Rel lysis is rel Mth! Note: Refereces refer to Greee & Krtz Fuctio Theory of Oe Complex Vrible. Most problems come from there. Exersize. Prove tht for y complex umbers z d w, z + w 2 = z 2 + w 2 + 2Re(z w) z + w 2 + z w 2 = 2 z w 2 Solutio. This follows by direct coputtio. Write, for coveiece (d buse of ottio) z = z + zi d w = w + wi, with z, z, w, w R. Now we hve, i the first cse, z + w 2 = (z + w) + ( z + w)i 2 = (z + w) 2 + ( z + w) 2 Now, fter expdig the squres, we get from bove (z 2 + z 2 ) + (w 2 + w 2 ) + 2(zw + z w) We kow tht (z 2 + z 2 ) = z d similrly for w. Hece from bove we obti O the other hd, z 2 + w 2 + 2(zw + z w) Re(z w) = Re((z + zi)(w w)) = Re(zw + z w + ( zw z w)i) = zw + z w The result ow follows. The secod cse ow follows from the first, by otig tht which equls The result ow follows. z w 2 = z + ( w) 2 = z 2 + w 2 + 2Re(z w) Exersize 2. Prove tht the fuctio z 2 + w 2 2Re(z w) φ(z) = i z + z mps the set D = {z C : z < } to d oto the set U = {z C : Imz > }. Solutio 2. Exersize 3. Compute ll fifth roots of + i, ll cube roots of i, ll sixth roots of d ll squre roots of 3/2 + i/2.

2 More solutios mul t Thks to the Willim's work! Solutio 3. Exersize 4. Compute ech of the followig derivtives: () z (x2 y) (b) z (x + y2 ) (c) 4 z z 3 (xy 2 ) (d) 2 z z (zz2 z 3 z + 7z) Solutio 4. Exersize 5. Prove tht if f is holomorphic o U C, the 2 ( f 2 ) = 4 f z Solutio 5. Exersize 6. Prove tht if f is C 2, holomorphic, d ovishig, the log f is hrmoic. Solutio 6. Exersize 7. () Let be the boudry curve of the uit disc, equipped with couterclockwise oriettio. Give exmple of C fuctio f o eighborhood of D(, ) such tht f(ζ)dζ = but such tht f is ot holomorphic o y ope set. (b) Suppose tht f is cotiuous fuctio o the disc D(, ) d stisfies f(ζ)dζ = D(,r) for ll < r <. Must f be holomorphic o D(, )? Solutio 7. () Cosider f(z) = zz. If z = x + iy, the we see tht f(z) = (x 2 + y 2 ). Clerly the lie itegrl of this fuctio over the uit circle is zero. However, the fuctio is ot holomorphic o y ope set (ideed, / z of f is ever zero o ope set). (b) The swer here is o. Cosider, for exmple, f(z) = /z 2. This fuctio is holomorphic o C \ {}. Further, (s show i exmple o p. 5), the itegrl of f(z) over the uit circle is. Now, sice D(, r) is homotopic to D(, ) where f is holomorphic, we must hve = f(z)dz = f(z)dz D(,) D(,r) But f is clerly ot holomorphic o D(, ) (there is sigulrity t the origi). Exersize 8. Clculte explicitly the itegrls z 3 dz D(8i,2) D(6+i,3) (z i) 2 dz

3 More solutios mul t Thks to the Willim's work! Solutio 8. For the first itegrl, our prmeteriztio is (t) = 8i + 2e it with t [, 2π] with t icresig. The we hve, by defiitio, z 3 dz = (8i + 2e it ) 3 (2ie it )dt D(8i,2) By substitutio with u = 8i + 2e it, oe immeditely obtis (8i + 2e it ) 3 (2ie it )dt = s expected. For the secod itegrl, our prmetriztio is Now, by defiitio, we hve (z i) 2 dz = D(6+i,3) (t) = 6 + i + 3e it with t [, 2π] After expdig d simplifyig, oe gets 3i(6 2i) 2 e it dt + 6i(6 2i) (6 + i + 3e it i) 2 (3ie it )dt dt + 27i e it dt The oly itegrls requirig ttetio re those with the itegrd e it. These re computed by itegrtig the rel d complex prts of e it = cos(t) + i si(t). Exersize 9. Clculte D(,) D(,2) ζ + 2 dζ ζ + dζ Solutio 9. For the first itegrl, let f(z) = z z+2. It is cler tht f is holomorphic o D(, ). I fct, f is holomorphic o ope set cotiig D(, ). Now, with this, we see tht D(,) ζ + 2 dζ = f(ζ) D(,) ζ dζ Ad by Cuchy s Itegrl Formul, the ltter is equl to f() =. The situtio is differet for the secod itegrl, sice the sigulrity of ζ+ lies iside D(, 2). Observe tht f(ζ) = +ζ is holomorphic o D(, 2) \ { }. Thus, sice D(, ) is homotopic to D(, 2) o D(, 2) \ { }, we hve f(ζ)dζ = f(ζ)dζ D(,2) D(,) O the other hd, the itegrl is ivrit uder trsltio. Hece, we must hve f(ζ)dζ = f(ζ )dζ = ζ dζ Ad s we kow, D(,) D(,) D(,) dζ = ζ D(,)

4 More solutios mul t Thks to the Willim's work! Hece we hve D(,2) ζ + dζ = Exersize. Let U = {z C : /2 < z < 2}. Cosider the two pths (t) = e t d 2 (t) = e 4πit, t. Prove tht d 2 re ot homotopic. Solutio. The fudmetl group of the give ulus is isomorphic to the fudmetl group of the circle (the circle is retrct of the ulus). I tur, the fudmetl group of the circle is isomorphic to Z. It c be show, by the coverig lemm (or the stdrd costructio of the isomorphism of π (S ) with Z), tht e 4πt = 2e 2πt whe viewed s elemets i Z. So clerly e 4πt is ot homotopic to e 2πt (ituitively, oe is trced twise, the other is trced oce). Exersize. Let U U 2 C be topologiclly simply coected ope sets. Defie U = j U j. Prove tht U is the topologiclly simply coected. Wht if the hypothesis of opeess is removed? Solutio. Sy is loop i U. The j U j, where by we me the trce of. But is compct, so there exists m < such tht m j= U j. O the other hd, m j= U m+. So is loop i U m+. Give tht U m+ is simply coected, is homotopic to poit i U m+. Hece is homotopic to poit i U. We re doe. Exersize 2. Let S be topologiclly simply coected subset of C d let f be complex-vlued, cotiuous fuctio o S. Does it follow tht f(s) is topologiclly simply coected? Solutio 2. The swer is o. Let S = C d let f(z) = e z. Clerly C is simply coected d f(c) is ot (it is missig the origi). Exersize 3. Prove tht the Cuchy itegrl formul is vlid if we ssume oly tht F C (D), d F is holomorphic o D. Solutio 3. We stte d prove the clim s theorem: Suppose tht f is holomorphic o the ope disk D(z, r) d cotiuous o D. Let be D, with couterclockwise oriettio. The for ll z D(z, r), we hve f(z) = f(ζ) ζ z dζ Proof: Let < r < r, such tht z D(z, r ). Let s : [, ] be the stdrd prmetriztio of D(z, s) (couterclockwise). Now, sice f is cotiuous o the ulus A = {w : r w r}, the so is F (ζ) = f(ζ) ζ z So F (ζ) is uiformly cotiuous o A. Hece for ll ɛ >, there exists r(ɛ) with r r(ɛ) r such tht F (ζ)dζ F (ζ)dζ r(ɛ) f(z + re it ) z + re it z rieit f(z + r(ɛ)e it ) z + r(ɛ)e it r(ɛ)ie it dζ < ɛ

5 More solutios mul t Thks to the Willim's work! But by Cuchy s Itegrl Formul, we hve Hece we must hve r(ɛ) F (ζ)dζ = f(z) F (ζ)dζ f(z) < ɛ Sice the left side is idepedet of ɛ, we coclude tht F (ζ)dζ = f(z) s desired. Exersize 4. Let f be cotiuous fuctio o {z : z = }. Defie, with = the uit circle trversed couterclockwise, { f(z) if z = F (z) = f(ζ) ζ=z dζ if z < Is F cotiuous o D(, )? Solutio 4. Tke f(z) = z. The we hve f(ζ) ζ z dζ = If z =, the we obviously get ζ(ζ z) dζ = ζ(ζ z) dζ O the other hd, if z, by prtil frctio decompozitio, we hve Hece we get ζ(ζ z) = /z ζ z /z ζ f(ζ) ζ z dζ = z ζ z dζ z ζ dζ By Cuchy s Itegrl Formul, both itegrls bove re equl to. Hece we get bove. So we hve for ll z i the ope disk D(, ), F (z) =. O the other hd, by defiitio, for z =, F (z) = f(z) = z Clerly the F (z) is ot cotiuous o D(, ). Exersize 5. Clculte (ζ )(ζ 2i) dζ where is the circle with ceter, rdius 4, d couterclockwise oriettio. Solutio 5. This itegrl is esily computed by prtil frctio decompositio d Cuchy s Itegrl Formul. Ideed, there re costts A, B (we skip the tedious computtio) such tht (ζ )(ζ 2i) = A ζ + B ζ 2i

6 More solutios mul t Thks to the Willim's work! Hece we hve (ζ )(ζ 2i) dζ = A ζ dζ + B ζ 2i dζ Settig f (ζ) = A d f 2 (ζ) = B d employig Cuchy s Itegrl Formul, we get A ζ dζ + B ζ 2i dζ = A + B Exersize 6. Clculte ζ 2 + ζ (ζ 2i)(ζ + 3) dζ where is the circle with ceter, rdius 5, d couterclockwise oriettio. Solutio 6. This too is esily computed by Cuchy s itegrl formul. Ideed, ζ 2 + ζ (ζ 2i)(ζ + 3) = ζ 2 (ζ 2i)(ζ + 3) + ζ (ζ 2i)(ζ + 3) Hece we hve ζ 2 + ζ (ζ 2i)(ζ + 3) dζ = ζ 2 (ζ 2i)(ζ + 3) dζ + ζ (ζ 2i)(ζ + 3) dζ Now oe c use prtil frctio decompositio for both itegrls o the right, d employ Cuchy s Itegrl Formul, much like i the previous solutio. We perform the computtio for oly oe of the itegrls. We hve: This leds to Hece we get A(ζ + 3) + B(ζ 2i) = ζ 2 A = 4 2i + 3 d B = i ζ 2 (ζ 2i)(ζ + 3) dζ = A ζ 2i dζ + B ζ + 3 dζ with the vlues A d B determied bove. Cuchy s itegrl formul yields the coclusio. Exersize 7. Verify tht dζ (ζ )(ζ + ) = 2 dζ (ζ )(ζ + ) where is D(, ) equipped with couterclockwise oriettio d 2 is D(, ) equipped with clockwise oriettio. Solutio 7. Let 2 trce 2 i couterclockwise directio. The we hve dζ (ζ )(ζ + ) = dζ (ζ )(ζ + ) 2 By buse of ottio, write 2 for 2. So we must verify tht dζ (ζ )(ζ + ) dζ = dζ (ζ )(ζ + ) dζ where d 2 both hve couterclockwise oriettio. 2 2

7 More solutios mul t Thks to the Willim's work! Well, set f (ζ) = ζ+ d f 2(ζ) = ζ The dζ (ζ )(ζ + ) dζ = f (ζ) ζ dζ Similrly dζ 2 (ζ )(ζ + ) dζ = f (ζ) 2 ζ + dζ Now employig Cuchy s Itegrl Formul, we get d f (ζ) ζ dζ = πi f (ζ) 2 ζ + dζ = πi Exersize 8. Let be the uit circle equipped with clockwise oriettio. For ech rel umber λ, give exmple of ocostt holomorphic fuctio F o the ulus A = {z : /2 < z < 2} such tht F (z)dz = λ Solutio 8. Set F (z) = λ z. The we hve F (z)dz = λ z dz = λ Exersize 9. TRUE or FALSE: Let j >, j =, 2,.... If j z j is coverget o D(, r) d if ɛ > is sufficietly smll, the ( j + ɛ)z j is coverget o D(, r ) for some < r < r. Solutio 9. The swer is yes, d follows directly from defiitios. Ideed, m m m lim ( j + ɛ)z j = lim j z j + ɛ m j= m j= Now if we let < r < mi {, r}, we get m ɛ lim z j C m j= Sice by hypothesis j jz j coverges, we deduce tht lim m j= m ( j + ɛ)z j C Exersize 2. Let Ω = C \ {z : z }. Determie ll biholomorphic self-mps of Ω. j= z j

8 More solutios mul t Thks to the Willim's work! Solutio 2. For coveiece, set D = D(, ) \ {}. Observe tht f : D Ω give by f(z) = /z is coforml mp from D to Ω. This is esy to see: f is bijective, holomorphic whe viewed o D or o Ω, d it is its ow iverse. Now, let Aut( D), Aut(Ω) be the set of biholomorphic self-mps o D d Ω, respectively. These form group, d we kow tht f iduces group isomorphism ˆf: ˆf : Aut( D) Aut(Ω) vi ˆf : h f h So, we oly eed to describe mps of Aut( D). Let, for coveiece, D = D(, ), the ope uit disc cetered t the origi. We kow tht {φ Aut(D) : φ() = } = {ωz : D D : ω = } Aut(D) We clim tht Aut( D) = Aut(D). To prove this clim, observe tht if φ Aut( D), the φ, d its iverse, φ re holomorphic o D. Now, sice D is bouded domi, it is ot the cse tht lim φ(z) = z Agi, sice D is bouded domi, it is t the cse tht φ hs essetil sigulrity t zero, sice the φ will dmit rbitrrily lrge vlues er zero. Thus φ hs removble sigulrity t zero, hece φ c be exteded to holomorphic fuctio, sy Φ, o D. We prove tht Φ() =. Ideed, sice D is simply coected, Φ(D) = D Φ() must be simply coected. This is the cse oly if Φ() =. Tht is, Φ iduces surjective homomorphism ˆΦ : π (D) π (Φ(D)) d sice π (D) = {}, we must hve π (Φ(D)) = {}. But the Φ Aut(D). This proves tht φ is of promissed form. Thus ll biholomorphic mps o Ω re of the form z /(ω/z) with ω =. So we get z z/ω = zω. Sice ω is rbitrry, we get z ωz. Tht is, Aut(Ω) is the set of rottios. Exersize 2. Let Ω C be bouded, simply coected domi. Let φ : Ω D d φ 2 : Ω D be coforml mps. How re φ, φ 2 relted to ech other? Solutio 2. Observe tht φ φ 2 : D D is coforml mp. Let φ (z) = z z with D = D(, ). The we kow tht every coforml mp from D to D is of the form ωφ, with ω = (ω d deped o the mp). Hece we must hve φ φ 2 = ωφ for some ω, C s bove. Hece φ = ωφ φ 2 = ω φ 2 φ 2

9 More solutios mul t Thks to the Willim's work! Exersize 22. Let {f α } be orml fmily of holomorphic fuctios o domi U. Prove tht {f α} is orml fmily. Solutio { } 22. For coveiece, let A = {f α } d A = {f α}. Suppose tht f j A is sequece i A. Now, sice A is orml fmily, we kow tht for every sequece {f j } A, there is ormlly covergig subsequece {f jk }. Tht is, there is f defied o U such tht f jk f uiformly o compct. But the it follows from Theorem 3.5 d its corollry tht f j k f uiformly o compct (the compct sets here re subsets of U, of course). Hece A is orml fmily. (Of course, it is tke here for grted tht f is holomorphic - this follows immeditely, sice f jk re holomorphic). Exersize 23. Suppose tht {f } is uiformly bouded fmily of holomorphic fuctios o domi Ω. Let {z k } k= Ω d suppose tht z k z Ω. Assume tht lim f (z k ) exists for k =, 2,.... Prove tht the full sequece {f } coverges uiformly o compct subsets of Ω. Solutio 23. Sice there is M > such tht for ll, we hve f M, by Motel s theorem we do kow tht there exists ormlly covergig subsequece, sy {f m }, covergig to g. We show tht i fct {f m } = {f }. For cotrdictio let {f j } = {f }\{f m } d ssume tht {f j } is ifiite. But the there is subsequece {f jl } {f j }, gi by Motel s theorem, tht coverges ormlly, sy to h. Assume lso tht g h. Both re obviously holomorphic. Now, for ll k the limit lim f (z k ) exists. Hece g(z k ) = lim m f m (z k ) = lim f (z k ) = lim l f jl (z k ) = h(z k ) But z k z, so g d h gree o ccumultig sequece. Thus we must hve g = h. We re doe. Exersize 24. Costruct lier frctiol trsformtio tht seds the uit disc to the hlf ple tht lies below the lie x + 2y = 4. Solutio 24. Let φ : D(, ) CH + (where CH + is the complex upper hlf ple) be give by This is the Cyley trsform. Now let Filly, let φ(z) = i + iz z ψ(z) = z + 2i σ(z) = e (2π t (/2))i z (tht is, rottio couterclockwise by 2π t (/2)). Now let f = σ ψ φ It is esily verified tht f is the trsform we eed.

10 More solutios mul t Thks to the Willim's work! Exersize 25. Let L be prtil differetil opertor of the form L = 2 x 2 + b 2 y 2 + c 2 x y with, b, c costts. Assume tht L commutes with rottios i the sese tht wheever f is C 2 fuctio o C, the Lf ρ θ = L(f ρ θ ) for y rottio ρ θ = e iθ z. Prove tht L must be costt multiple of the Lplci. Solutio 25. Exersize 26. Compute formul logous to the Poisso itegrl formul, for the regio U = {z : Imz > } (the upper hlf ple), by mppig U coformlly to the uit disc. Solutio 26. Let φ(z) = z i z + i It s esy to cofirm tht φ : HC + D(, ), where HC + deotes the upper hlf ple, is coforml mp. We lredy hve the Poisso itegrl formul o the uit disc, give by u() = u(e iθ ) 2 2π e iθ 2 dθ Now, usig φ : D(, ) HC +, we obti, for u hrmoic o HC +, u() = (u φ )(φ()) for ll HC + So we oly eed to pply the Poisso itegrl formul to (u φ ). So we obti u() = (u φ )(φ()) = 2π Now, sice we obti u() = 2π φ (w) = (u φ )(e iθ ) φ() 2 φ() e iθ 2 dθ i( + w) w ( i( + e iθ ) ) i +i 2 u e iθ i +i 2 dθ eiθ Exersize 27. Compute Possio itegrl formul for U = {z : z <, Imz > } by mppig U coformlly to the disc. Solutio 27. Let ψ d ψ 2 be give by ( ) 2 i + z ψ (z) = d ψ 2 (z) = z i i z z + i The it is esily verified tht ψ : U HC + is coforml d, s we lredy kow from the previous exercise, ψ 2 : HC + D(, ) is coforml. So let φ = ψ 2 ψ be coforml mp from U to D(, ). Now pply the formul from the previous exercise to this φ.

11 More solutios mul t Thks to the Willim's work! Exersize 28. If u is cotiuous rel-vlued fuctio o the closure of the uit disc which is hrmoic o the iterior of the disc, the prove tht u is the rel prt of the holomorphic fuctio Solutio 28. Observe tht h(z) = 2π Re(h(z)) = 2π e iθ + z e iθ z u(eiθ )dθ ( e iθ ) + z Re e iθ z u(eiθ ) dθ Sice u is rel-vlued, we hve ( e iθ ) ( + z e iθ ) + z Re e iθ z u(eiθ ) = Re e iθ u(e iθ ) z A esy computtio shows tht e iθ + z e iθ z = z 2 + (e iθ z e iθ z) z e iθ 2 Observe tht for, b C, it is true (s is verified by direct computtio) tht b is pure imgiry. Hece from bove we get ( e iθ ) + z Re e iθ = + z z z e iθ 2 Hece we get Re(h(z)) = 2π u(e iθ ) + z z e iθ 2 dθ From the Poisso itegrl formul we kow tht this is precisely u(z). Exersize 29. Let u be cotiuous fuctio o ope set U C d let P U. Suppose tht u is hrmoic o U \ {P }. Prove tht u is i fct hrmoic o ll of U. Solutio 29. Assume tht U is coected (if ot, we c restrict our ttetio to the coected compoet of U tht cotis P ). Let R be squre cetered t P, so smll tht R U. We c cover R by four rectgles R i, i 4, such tht P R i for ll i. Further, R j R j+ for j 3 d R R 4. Now, u is hrmoic o ech R j. Thus, for ech j, u o R j is the rel prt of some holomorphic h j = u + iv j. But becuse of the itersectio property of R j s (bove), Re(h j ) = Re(h j+ ) o ope set. Hece, becuse of the ope mppig theorem, h j h j+ is imgiry costt. We my tke this costt to be zero. So we get h i = h j for ll i, j 4. Hece u is the rel prt of holomorphic fuctio h o R \ {P }. O the other hd, by costructio of v j (see Theorem.5.), v j re bouded. So, sice u is cotiuous, h is bouded o R \ {P }. So h c be exteded to holomorphic fuctio h o ll of R. By cotiuity of u, u = Re( h) o R. So u is hrmoic o ll of R. Exersize 3. If h is etire, the restrictio of h to the rel xis is rel, d the restrictio of h to the imgiry xis is imgiry, the prove tht h( z) = h(z) for ll z.

12 More solutios mul t Thks to the Willim's work! Solutio 3. By the Schwrtz s reflectio priciple we kow tht h(z) is holomorphic. Hece h(z) = h(z) h(z) is lso holomorphic. Sice h is rel-vlued o R, we see tht h(z) is ideticlly zero o R. So o C we get h(z) = h(z) O the other hd, sice h is pure imgiry for ll pure imgiry vlues z, we see tht for these vlues of z, we hve h( z) = h(z) or h( z) + h(z) = But z h( z) + h(z) is holomorphic, which is ideticlly zero o the imgiry xis, hece is zero everywhere. Thus h( z) = h(z) for ll z C. Exersize 3. Let G d H be fuctios tht re cotiuous o D d holomorphic o D. Prove tht if ReG d ReH gree o D, the F G is imgiry costt o D. Solutio 3. Set u G = Re(G) d u H = Re(H). The u G, u H re hrmoic o D d cotiuous o D. Hece u G u H is hrmoic o D d cotiuous o D. By the mi/mx priciple for hrmoic fuctios, u G u H ttis mi/mx o D. But u G = u H o D. Hece mi/mx of u G u H o D =. So u G = u H o D. But the F H : D ir d by the ope mppig theorem we coclude tht F H is ideticlly imgiry costt. Exersize 32. If U is domi, the boudry of which is the disjoit uio of fiitely my simple closed C curves, ech with the property tht the domi lies oly o oe side of the curve t ech poit, the prove tht every poit of U hs brrier. Solutio 32. U is ope, d U = m j= j where j is C simple, closed curve. Hece U some m (ot ecessrily equl to m) coected compoets, ech with C smooth, simple, closed boudry. Now, let P k, for some k m. From the ssumptios we immeditely deduce tht there exists Q U c d closed lie segmet I U c coected P d Q such tht I U = {P }. Now, followig verbtim Exmple 7.7. (we do t retype it here - there re o modifictios, everythig follows word-for-word), oe reduces the problem to the disc D(, ). Now followig verbtim exmple 7.7.9, we get the existece of brrier t P. Of course, this hs bee doe oly for the coected compoet bouded by k. We c exted the brrier b t P to ll of U by simply defiig b = ll the other (closures of) coected compoets of U, sice the closures of these re disjoit.

13 More solutios mul t Thks to the Willim's work! Exersize 33. Give exmple of domi U C, U C, such tht o boudry poit of U hs brrier. But prove tht, for y bouded domi U, there is poit i U tht hs brrier. Solutio 33. To do: Develop the ide i red otepd... complete lter. O the other hd, for bouded domis U, oe c lwys fid P U d Q U c such tht closed segmet I coects P d Q, I U c d I U = {P }. Now oe c follow the techique from solutio. Exersize 34. Let U be bouded, holomorphiclly simply coected domi with the property tht ech P U hs brrier. Let φ be positive, cotiuous fuctio o U. Prove tht there is holomorphic fuctio f o U such tht f is cotiuous o U d f restricted to U equls φ. Solutio 34. Let ψ = log(φ). The ψ is cotiuous fuctio o U (sice φ is positive). Give tht U hs brrier for ll P U, we c use Theorem 7.8. to coclude tht there is hrmoic fuctio h o U, cotiuous o U, such tht h U = ψ. Sice U is bouded d coected, we c cover U by fiitely my (closed), simply coected sets B i i such wy, tht the iterior of ech B i lies etirely i U. Clerly h is hrmoic o ech B i. But the h is rel prt of some holomorphic h i o B i (h is loclly rel prt of holomorphic fuctio o U, so tke B i smll eough to isure this locl property). Where B i itersects B j, h i h j is complex costt (by the ope mppig theorem) which we my tke to be zero. Hece h c be regrded s the rel prt of some holomorphic fuctio h o U, cotiuous o U. Let H = e h. The H is lso holomorphic o U d cotiuous o U. Furthermore, so o U we hve H = e h H U = e h U = e h U = e ψ = φ Exersize 35. Solve ech of the followig Dirichlet problems by usig coforml mppig to trsform the problem to oe o the disc. () Ω is the first qudrt. The boudry fuctio φ equls o the positive rel xis d y o the positive imgiry xis. (b) Ω is the upper hlf of the uit disc. The boudry fuctio is φ(e iθ ) = θ, θ π φ(x) = π x, x 2 (c) Ω is the strip {z : < Rez < }. The boudry fuctio is ideticlly o the imgiry xis d is ideticlly o the lie {z : Rez = }. Solutio 35. We c trsform Ω to D(, ) coformlly vi z z 2 z2 i z 2 + i

14 More solutios mul t Thks to the Willim's work! It is esy to verify tht this mp is coforml. We c ccordigly trsform the boudry coditios to get i( + w) ũ(w) = Im w for w D(, ). Oe c verify the tht ũ correspods to u(x, y) = y o Ω vi our coforml mp bove. Observe tht w = correspods to y = o the boudry of Ω. Of course, ll of this is too complicted. Oe could hve simply foud u(x, y) = y by observtio. (b) To do: complete this lter, follow computtio i red otepd. (c) This is essetilly the sme s (). u(x, y) = x will do. Oe c fid it either by observtio, or by mppig coformlly the strip to the puctured disc, d the boudry coditios ccordigly. Exersize 36. Prove tht if f is cotiuous, rel-vlued fuctio o the itervl [, b], < b, d if φ is covex fuctio o R, the ( ) b φ f(x)dx b (φ f)(x)dx. Solutio 36. Let m be lier, sy m(x) = αx + β. The ( ) b m f(t)dt = α b f(t)dt + β sice from bove we hve ( m b f(t)dt b ) βdt = β = b m(f(t))dt Now, by covexity of φ, φ(x), is the supremum over ll of its supportig tget lies m evluted t x. Hece we hve ( ) ( b ) b φ f(t)dt m f(t)dt = b m(f(t))dt where m is supportig tget lie. Tkig the supremum over ll such m we obti immeditely ( ) b φ f(t)dt φ(f(t))dt Exersize 37. If z < R, the prove tht ( ) R 2 + z 2 Solutio 37. = R 2 = R R z

15 More solutios mul t Thks to the Willim's work! Exersize 38. If b > for ll, the prove tht coverges if d oly if log(b ) < Solutio 38. Suppose first tht log(b ) < Observe the tht hece b ( ) exp log(b ) < e b = M < It is clerly true, for b >, tht e b > b hece we hve N N M > e b > for ll N, provig covergece of the product. b Coversely, suppose tht the product ( + (b )) = coverges. Observe tht if for ifiitely my, b > 2 the, keepig i mid tht for ll b >, the product will diverge. Hece for ll > N, N lrge eough, b < 2. Without loss of geerlity we my thus ssume tht for ll we hve b < 2. The b <, hece we kow tht ( ) N N exp (b ) b 2 provig tht b (b ) must lso coverge. Observe tht x log(x) is icresig fuctio o [, ), d is equl to zero t x =. This shows tht for ll b, b log(b ), provig covergece of log(b ) b

16 More solutios mul t Thks to the Willim's work! Exersize 39. Determie whether coverges. Do the sme for =2 =2 ( + ( ) ) ( + ( ) ) Solutio 39. Suppose tht N is eve, sy N = 2K. For coveiece, write ( b = + ( ) ) Observe tht A simple clcultio shows tht Hece O the other hd, N+ =2 N b = b 2 b 3 b 4 b 5 b K b K b = b N+ b b + = for ll 2 N =2 N b = =2 b = b N+ s N This shows tht the odd d the eve subsequeces both coverge to, showig tht the product coverges to. For the secod clcultio, we ll follow the sme techique. Deote by P N the eve subsequece of the prtil products (tht is, N = 2K). Hece P N+ is the odd subsequece. We show tht both coverge to the sme umber. A simple clcultio will show tht ( + ) ) ( < for ll 2. Hece the eve subsequece, P N is decresig i N, thus coverges. Also, P N P N+ = P N ( b N+ ) b N+ s N ( where b = + ( ) ). Hece P N P N+ d we lredy kow tht P N coverges. Hece so does P N+, hece so does the product.

17 More solutios mul t Thks to the Willim's work! Exersize 4. Clculte ( ) 2 explicitly. Solutio 4. =2 Exersize 4. Prove tht if j j < d if σ is y permuttio of the positive itegers, the ( + j ) = ( + σ(j) ) Solutio the j= j= 4. Sice j coverges, we kow tht σ(j) coverges. But ( + σ(j) ) coverges d hece the product ( + σ(j) ) lso coverges. But the for y give ɛ > we hve ( + j ) ( + σ(j) ) j=n j=m = ( + j ) ( + σ(j) ) j=n j=m ( + j ) + ( + σ(j) ) < 2ɛ j=n j=m for N, M lrge (this follows from covergece of both products). Now, for fixed M by tkig N so lrge tht {, 2,..., N} icludes {σ(), σ(2),..., σ(m)}, we re doe. Exersize 42. Prove tht, i geerl, the result of Exercise 8 fils if the hypothesis j j < is replced by weker coditio like j <. Solutio 42. Cosider j = ( ) j j for j 2. Let σ be such tht σ(j) gives the jth eve umber if j is ot multiple of, sy, 5, d σ(j) gives the j/5th odd umber if j is multiple of 5; keepig i mid tht the first eve umber is 2 d the first odd umber is 3 (sice j 2). So, for istce: σ(2) = 2; σ(3) = 4; σ(4) = 6; σ(5) = 3; σ(6) = ;... It is t hrd to see tht i this cse ( + σ(j) ) diverges. O the other hd, s we hve see bove, ( + j ) coverges. j j

18 More solutios mul t Thks to the Willim's work! Exersize 43. For which z does coverge? ( + z 3 ) = Solutio 43. Clerly the product coverges for ll z <. If z is rel d z >, the product clerly diverges. If z =, the limit does t exist. If z = the product diverges. Now, simple clcultio will show tht ( + z 3 ) = ( + z)( z + z 2 ) hece ( + z 3 ) = ( + z) ( z + z 2 ) d the ltter product coverges oly whe z + z 2 <. Exersize 44. If U is domi, the boudry of which is the disjoit uio of fiitely my simple closed C curves, ech with the property tht the domi lies oly o oe side of the curve t ech poit, the prove tht every poit of U hs brrier. Solutio 44. U is ope, d U = m j= j where j is C simple, closed curve. Hece U some m (ot ecessrily equl to m) coected compoets, ech with C smooth, simple, closed boudry. Now, let P k, for some k m. From the ssumptios we immeditely deduce tht there exists Q U c d closed lie segmet I U c coected P d Q such tht I U = {P }. Now, followig verbtim Exmple 7.7. (we do t retype it here - there re o modifictios, everythig follows word-for-word), oe reduces the problem to the disc D(, ). Now followig verbtim exmple 7.7.9, we get the existece of brrier t P. Of course, this hs bee doe oly for the coected compoet bouded by k. We c exted the brrier b t P to ll of U by simply defiig b = ll the other (closures of) coected compoets of U, sice the closures of these re disjoit. Exersize 45. Give exmple of domi U C, U C, such tht o boudry poit of U hs brrier. But prove tht, for y bouded domi U, there is poit i U tht hs brrier. Solutio 45. To do: Develop the ide i red otepd... complete lter. O the other hd, for bouded domis U, oe c lwys fid P U d Q U c such tht closed segmet I coects P d Q, I U c d I U = {P }. Now oe c follow the techique from solutio. Exersize 46. Let U be bouded, holomorphiclly simply coected domi with the property tht ech P U hs brrier. Let φ be positive, cotiuous fuctio o U. Prove tht there is holomorphic fuctio f o U such tht f is cotiuous o U d f restricted to U equls φ. Solutio 46. Let ψ = log(φ). The ψ is cotiuous fuctio o U (sice φ is positive). Give tht U hs brrier for ll P U, we c use Theorem 7.8.

19 More solutios mul t Thks to the Willim's work! to coclude tht there is hrmoic fuctio h o U, cotiuous o U, such tht h U = ψ. Sice U is bouded d coected, we c cover U by fiitely my (closed), simply coected sets B i i such wy, tht the iterior of ech B i lies etirely i U. Clerly h is hrmoic o ech B i. But the h is rel prt of some holomorphic h i o B i (h is loclly rel prt of holomorphic fuctio o U, so tke B i smll eough to isure this locl property). Where B i itersects B j, h i h j is complex costt (by the ope mppig theorem) which we my tke to be zero. Hece h c be regrded s the rel prt of some holomorphic fuctio h o U, cotiuous o U. Let H = e h. The H is lso holomorphic o U d cotiuous o U. Furthermore, so o U we hve H = e h H U = e h U = e h U = e ψ = φ Exersize 47. Solve ech of the followig Dirichlet problems by usig coforml mppig to trsform the problem to oe o the disc. () Ω is the first qudrt. The boudry fuctio φ equls o the positive rel xis d y o the positive imgiry xis. (b) Ω is the upper hlf of the uit disc. The boudry fuctio is φ(e iθ ) = θ, θ π φ(x) = π x, x 2 (c) Ω is the strip {z : < Rez < }. The boudry fuctio is ideticlly o the imgiry xis d is ideticlly o the lie {z : Rez = }. Solutio 47. We c trsform Ω to D(, ) coformlly vi z z 2 z2 i z 2 + i It is esy to verify tht this mp is coforml. We c ccordigly trsform the boudry coditios to get i( + w) ũ(w) = Im w for w D(, ). Oe c verify the tht ũ correspods to u(x, y) = y o Ω vi our coforml mp bove. Observe tht w = correspods to y = o the boudry of Ω. Of course, ll of this is too complicted. Oe could hve simply foud u(x, y) = y by observtio. (b) To do: complete this lter, follow computtio i red otepd. (c) This is essetilly the sme s (). u(x, y) = x will do. Oe c fid it either by observtio, or by mppig coformlly the strip to the puctured disc, d the boudry coditios ccordigly.

20 More solutios mul t Thks to the Willim's work! Exersize 48. Prove tht if f is cotiuous, rel-vlued fuctio o the itervl [, b], < b, d if φ is covex fuctio o R, the ( ) b φ f(x)dx b (φ f)(x)dx. Solutio 48. Let m be lier, sy m(x) = αx + β. The ( ) b m f(t)dt = α b f(t)dt + β sice from bove we hve ( m b f(t)dt b ) βdt = β = b m(f(t))dt Now, by covexity of φ, φ(x), is the supremum over ll of its supportig tget lies m evluted t x. Hece we hve ( ) ( b ) b φ f(t)dt m f(t)dt = b m(f(t))dt where m is supportig tget lie. Tkig the supremum over ll such m we obti immeditely ( ) b φ f(t)dt φ(f(t))dt