Solutions to Exercises 4.1

Transcription

1 Sectio 4. Sequeces ad Series of Complex Numbers 73 Solutios to Exercises 4.. We have a e i π as. So a by the squeee theorem or by applyig, Sec We have cosh i cos see 5, Sec..6. Hece a cosh i cos as. So a by the squeee theorem. 9. a Suppose that {a } is a coverget sequece ad let b a +. We wat to show that {b } is a coverget sequece ad that lim a lim b. Let L deote the limit of the sequece {a }. By defiitio of covergece, give ɛ>, we ca fid N, such that for all N, a L <ɛ. But N implies that + N. So for all N, a + L <ɛ. But a + b, so for all N, b L <ɛ, which implies that {b } is a coverget sequece with same limit as {a }. b Defie a i ad a + 3 +a. Give that {a } is coverget, the by part a, we have lim a lim a +. Let L lim a. The Solvig for L, we fid L lim a lim a 3 + lim 3 +a +L. L 3 +L L + L 3; L +L 3, L + 3L,L 3 orl. Because the limit of a coverget sequece is uique, we have to decide whether L orl 3. Note that a. If we ca show that Re a, the Re L lim Re a, ad this would elimiate the value L 3. Let s prove by iductio that Re a. The statemet is clearly true if, because a i ad so Re i. Now suppose that Re a ad let s prove that Re a +. For this purpose, we compute where α so L. Re a + 3 +a Re 3Re +a + a + a 3 + a + a Re + Re a i Im a 3α + Re a, +a +a +a. This completes the proof by iductio ad shows that L 3,

2 74 Chapter 4 Power Series ad Lauret Series 3 i 3. The series is a costat multiple of a coverget geometric series ad so it is + i 3 coverget. To fid its sum proceed as follows: 3 i + i 3 i + i i i + + i 3 +i + + i + 3 i + i 3 where r +i, r <. So the sum is 7. The series S 3 i + i 3 +i + i 3 i + i 3 r, 3 i 3i + i i + i 3 + i. + i + i is absolutely coverget by compariso to the series : + i + i + i + i if ; so + i + i. To sum the series, use partial fractios ad the terms will telescope, as follows: So the Nth partial sum is + i + i + i + i. s N N + i + + i + i +i i 3+i + + N + i N + i +i N + i +i, as N. So the sum is S +i i. +3i. The series is a geometric series with +3i 4 4. Sice 4 <, the series coverges to +3i 4 3 3i 4 3 i + i Write +i +i.

3 Sectio 4. Sequeces ad Series of Complex Numbers 75 It is ow clear that the series is a geometric series with +i. Sice Im, it follows that the series diverges. 9. For, we have cosi e 3 cosh e 3 e + e e 3 3 e + + e 3 e 3 /. The series is absolutely coverget by compariso with the coverget series 33. The series coverges as log as < <. 37. The series coverges as log as < < < 5 < 5. e 3 /. Thus the series coverges outside the closed disk of radius ad ceter at 5. The limit i this regio is use the geometric series mius the first term, which is. 4. If the th partial sum of the series is s i, the the series coverges ad its limit is lim s i lim. A example of such a series is + i + i. 45. The test of covergece i Theorem is really about series with positive real terms. For a proof, see ay calculus book ad use the same proof for real series. 49. To establish the additio formula for cosies, we will maipulate the series for cosie, usig several properties of absolutely coverget series ad Cauchy products. We have cos! ; cos! ; cos cos k k k k! k k! k!! k! k! k k k k k! k. k

4 76 Chapter 4 Power Series ad Lauret Series Similarly, si si si + +! ; si + +! ; k k+ k+ k +! k k +! k +! +! k! + k +! k+ k + k+ +! k + + k+. k + k+ Hece cos cos si si +! k k +!!! k k k k k + k + k k + k k k+ + k+ k+ k+, where i the first series o the right, we wrote the first term separately, ad i the secod series o the right, we shifted the idex by chagig to. Combiig the two series, we see that the last displayed sum equals +! { k k k k + k k + } k+ k+. But the two ier sums i k add up to k k k k +, because the first sum adds the eve-idexed terms ad the secod sum adds the odd-idexed terms. Hece cos cos si si +! +! + cos +, as desired. Note: While this was a good exercise i Cauchy products, it is ot the most efficiet way to prove the additio formula for the cosie. For a more elegat proof, see Example b, Sec. 4.6.

5 Sectio 4. Sequeces ad Series of Fuctios 77 Solutios to Exercises 4. si x. The sequece of fuctios f x coverges uiformly o the iterval x π. To see this, let M max f x max f x, where the maximum is take over all x i [, π]. The M. Sice M, as, it follows that f coverges uiformly to f o[,π]. I fact, we fact uiform covergece o the etire real lie. 5. a ad b First, let us determie the poitwise limit of the sequece of fuctios f x x x.forx i the iterval x, we have x + x f x x x + x x x x + x x +, as. Does the sequece coverge to uiformly for all x i [, ]? To aswer this questio we estimate the maximum possible differece betwee ad f x, as x varies i [, ]. For this purpose, we compute M max f x for x i [, ]. We have f x 3 x x + x ; f x 3 x x ; f > M >. Sice M does ot coverge to, we coclude that the sequece does ot coverge uiformly to o [, ]. c The sequece does coverge uiformly o ay iterval of the form [a, b], where <a<b. To see this, pick so that < <a. The, f x < for all a<xcheck the sig of f x if <x. Hece f x is decreasig o the iterval [a, b]. So, if M max f x for x i [a, b], the M f a. But f a, by part a, so thus M, ad so f x coverges uiformly o [a, b] The sequece f coverges to for all, as we ow show: , as. + b Let M max f for, To prove that f coverges uiformly, we must show that M, as. We have f This shows that M is smaller tha the last displayed expressio, which teds to as. Hece f does coverge to uiformly for all. 3. If, the + +. Apply the Weierstrass M-test with M +. Sice M is coverget, it follows that the series, coverges uiformly for all If, the

6 78 Chapter 4 Power Series ad Lauret Series Apply the Weierstrass M-test with M 4. 5 Sice M is coverget a geometric series with + r<, it follows that the series coverges uiformly for all. 5. If..9, the.9 3 A, 3 ad B.. Apply the Weierstrass M-test with M A + B. Sice M is coverget two geometric { } series with ratios <, it follows that the series + coverges uiformly i 3 the aular regio a If < 6 the + + < <. So the series coverge uiformly o < 6 by the Weierstrass M-test with M 3. b If <, the ca get very close to the value, where the series does ot coverge. So the iitial guess is that the series is ot uiformly coverget i the regio <. To prove this assertio, you ca repeat the proof give i Example 4; i particular, the iequalities i 3 ad the argumet that follows them still hold. 9. a Let δ>be a positive real umber. To show that the series ζ pricipal brach of coverges uiformly o the half-plae H δ { : Re δ > }, we will apply the Weierstrass M-test. For all H δ, we have e x+iyl e x l x > δ. So M. δ Sice M is a coverget series because δ>, it follows from the Weierstrass M-test δ that that coverges uiformly i H δ. b Each term of the series is aalytic i H { : Re > } i fact, each term is etire. To coclude that the series is aalytic i H, it is eough by Corollary to show that the series coverges uiformly o ay closed disk cotaied i H. IfS is a closed disk cotaied i H, S is clearly disjoit from the imagiary axis. Let H δ δ> be a half-plae cotaiig S. By part a, the series coverges uiformly o H δ, cosequetly, the series coverges uiformly o S. By Corollary, the series is aalytic i H. Note the subtilty i the proof. We did ot show that the series coverges uiformly o H. I fact, the series does ot coverge uiformly i H. c To compute ζ, accordig to Corollary, we ca differetiate the series term-by-term. Write e l e l.

7 Sectio 4. Sequeces ad Series of Fuctios 79 Usig properties of the expoetial fuctio, we have So, for all H, d d d d e l l e l l. ζ l. 33. To show that f {f } coverges uiformly o Ω, it is eough to show that max Ω f f m ca be made arbitrarily small by choosig m ad large. I other words, give ɛ>, we must show that there is a positive iteger N such that if m, N, the max f f m <ɛ. Ω This will show that the sequece {f } is uiformly Cauchy, ad hece it is uiformly coverget by Exercise 3. Sice each f is aalytic iside C ad cotiuous o C, it follows that f f m is also aalytic iside C ad cotiuous o C. Sice C is a simple closed path, the regio iterior to C is a bouded regio. By the maximum priciple, Corollary, Sec. 3.7, the maximum value of f f m occurs o C. But o C the sequece {f } is a Cauchy sequece, so there is a positive iteger N such that if m, N, the max f f m <ɛ. C Hece max f f m max f f m <ɛ, Ω C which is what we wat to prove. The key idea i this exercise is that the maximum value of a aalytic fuctio occurs o the boudary. So the uiform covergece of a sequece iside a bouded regio ca be deduced from the uiform covergece of the sequece o the boudary of the regio.

8 8 Chapter 4 Power Series ad Lauret Series Solutios to Exercises 4.3. Apply the ratio test: For, ρ lim a + a lim lim lim Thus the series coverges absolutely if < ad diverges if >. The radius of covergece is, the disk of covergece is <, the circle of covergece is. 5. Apply the ratio test: For 4i, 4i. ρ lim a + a 4i. Thus the series coverges absolutely if lim 4i + + 4i 4i < 4i 4i < i < 4 i <. The series diverges if i >. The radius of covergece is ; the disk of covergece is i < ; ad the circle of covergece is i. 9. We compute the radius of covergece by usig the Cauchy-Hadamard formula R limsup e i π 4 lim sup e i π 4. To uderstad why the limsup is equal to, recall that the limsup is the limit of the sup of the tail of the sequece { e i π 4 } N,asN teds to. The terms e i π 4 take values from the set {± ±, ±i, ±}. So the largest value of e i π 4 is, ad this value repeats ifiitely ofte, which explais the value of the limsup. Thus, R. 3. Start with the geometric series, <. Differetiate term-by-term: Multiply both sides by :, <., <.

9 Sectio 4.3 Power Series 8 7. We have 3 i 3 3 i 3 3 i 3 w w i 3 w i i 3, i 3 < which is valid for i 3 <.. This exercise is a simple applicatio of Theorem 5, Sec. 4., ad basic properties of power series. If a has radius of covergece R > ad b has radius of covergece R >, the these series coverge absolutely withi their radii of covergece. Apply Theorem 5, Sec. 4.: If is iside the disk of covergece of both series; that is, if <R, where R is the smallest of R ad R, the a b c, where c a k k b k k k a k b k k a b + a b + + a b + a b. Thus, for <R, a b k a k b k a b + a b + + a b + a b. Cauchy products work icely with power series. The Cauchy products of two power series, a ad b, cetered at is aother power series cetered at, c. Moreover, the coefficiet of i the product series, c, is the th term i the Cauchy product of the series a ad b. 5. a I the formula, take, the [Γ ] Γ π so Γ π. b I 9, let u t, udu dt, the c Usig b Γ t e t dt Γ Γ 4 cos θ si θdθ π u e u udu u e u du dθ π π, v e v dv e u +v u v du dv. u e u du.

10 8 Chapter 4 Power Series ad Lauret Series d Switchig to polar coordiates: u r cos θ, v r si θ, u + v r, dudv rdrdθ; for u, v varyig i the first quadrat u< ad v<, we have θ π,ad r<, ad the double itegral i c becomes implyig d. Γ Γ 4 π π Γ + e r r cos θ r si θ rdrdθ Γ + { }}{ cos θ si θ dθ r + e r dr use b with + i place of π cos θ si θ dθ,

11 Sectio 4.4 Taylor Series 83 Solutios to Exercises 4.4. Accordig to Theorem, the Taylor series aroud coverges i the largest disk, cetered at, i which the fuctio is aalytic. Clearly, e is etire, so radius of covergece is R. 5. Accordig to Theorem, the Taylor series aroud coverges i the largest disk, cetered at +i, i which the fuctio is aalytic. The fuctio f + is aalytic for all i. So i the largest disk aroud o which f is aalytic has radius R i +i i. 9. Accordig to Theorem, the Taylor series aroud coverges i the largest disk, cetered at, i which the fuctio is aalytic. The fuctio f ta is aalytic for all π +kπ. So the largest disk aroud o which f is aalytic has radius equal to the distace from to the earest poit where f fails to be aalytic. Clearly, the R π π. 3. Arguig as we did i Exercises -9, we fid that the Taylor series of f aroud has radius of covergece equal to the distace from to the earest poit where f fails to be aalytic. Thus R. This will also come out of the computatio of the Taylor series. Now, for <, the geometric series tells us that. Multiplyig both sides by, we get, for <, Because the fuctio is etire, the Taylor series will have a ifiite radius of covergece. Note that the series expasio aroud is easy to obtai: e! e! +.! But how do we get the series expasio aroud? I the previous expasio, replacig by, we get e +.! The expasio o the right is a Taylor series cetered at, but the fuctio o the left is ot quite the fuctio that we wat. Let f e. We have e e e e e f e. So f e [ e + e ]. Usig the expasio of e ad the expasio of e,wefid! f e [ + ] [ + e!!! + ]! e [! +++ ]! e [ +! + ] [ e +!! + ]! e [ + + ].!

12 84 Chapter 4 Power Series ad Lauret Series. a To obtai the partial fractios decompositio, we proceed i the usual way: A + B A +B ; A +B Take B, B. Take A. Thus we obtai the desired partial fractios decompositio. Expadig each term i the partial fractios decompositio aroud, we obtai, < ; So, for <,, <, or <. +. b We a derive the series i a by cosiderig the Cauchy products of the series expasios of ad, as follows. From a, we have + c, where c is obtaied from the Cauchy product formula see Exercise, Sec. 4.3: c a k b k, k k a k, b k, k+ c k+ + k. c To show that the Cauchy product is the same as the series that we foud i a, we must prove that + k. + k But this is clear sice k , k k

13 Sectio 4.4 Taylor Series 85 ad so + k + + k +. The radius of the Maclauri series is. This follows from our argumet i a or from Theorem, sice the fuctio has a problem at. 5. The easiest way to do this problem is to start with the series expasio of f i ad the differetiate it term-by-term, twice. Let s see: i i i i i i i i i i i, which is valid for i < or <. Withi the radius of covergece, the series ca be differetiated term by term as ofte as we wish. Differetiatig oce, we obtai Differetiatig a secod time, we obtai i 3 i i i i. i i 3 i 4 i. All series are valid i <. If we shift the idex of summatio of the series chage to +, we obtai the series i 4 9. For all, we have i i i e!! + + 4! +. Hece, for all, If, we have e +! +. e + 4! + 6 3! + +! + 4 3! +, so, if, e +! + 4 3! + +! + 4 3! +, where the series coverges for all. But a power series is aalytic i the disk where it coverges. So the series +! + 4 3! + is etire. Call g +!. We just proved that +! for,g e f. Now for, we have g +! + 4 3! + f. Thus f g for all ad sice g is etire, it follows that f is etire ad its Maclauri series is g.

14 86 Chapter 4 Power Series ad Lauret Series 33. a The sequece of itegers {l } satisfies the recurrece relatio l l + l for, with l adl 3. As suggested, suppose that l occur as the Maclauri coefficiet of some aalytic fuctio f l, <R. To derive the give idetity for f, multiply the series by ad, ad the use the recurrece relatio for the coefficiets. Usig l ad l 3,we obtai f l +3 + l ; f l + l l + l ; f + l ; f l + l. Usig the recurrece relatio ad the precedig idetities, we obtai f +3 + l l + l f {}}{ l + f {}}{ l +3 + f + f + + f+ f. Solvig for f, we obtai f +. b To compute the Maclauri series of f, we will use the result of Exercise : + + +, <,,. To derive this idetity, start with the partial fractios decompositio [ ] [ ]. Apply a geometric series expasio ad simplify: [ ]

15 Sectio 4.4 Taylor Series 87 Now, cosider the fuctio + where ad are the roots of + : + 5, ad 5, arraged so that <. These roots satisfy kow relatioships determied by the coefficiets of the polyomial +. We will eed the followig easily verified idetities: 5 ad. We will also eed the followig idetities: Similarly, We are ow ready to derive the desired Maclauri series. We have {}}{ ;

16 88 Chapter 4 Power Series ad Lauret Series So Thus + f l , We use the biomial series expasio from Exercise 36, with α. Accordigly, for <, +, where, for, +! ! !! 4!!!!!!!!!!!!!!!. Thus +, <. 4. a ad b There are several possible ways to derive the Taylor series expasio of f Log about the poit +i. Here is oe way. Let +i, so. The fuctio Log is aalytic except o the egative real axis ad. So it is guarateed by Theorem to have a series expasio i the largest disk aroud that does ot itersect the egative real axis. Such a disc, as you ca easily verify, has radius Im. However, as you will see shortly, the series that we obtai has a larger radius of covergece, amely of course, this is ot a cotradictio to Theorem.

17 Sectio 4.4 Taylor Series 89 Cosider the fuctio Log i B, where it is aalytic. The disk of radius, cetered at is cotaied i the upper half-plae. For B, we have d d Log. Istead of computig the Taylor series of Log directly, we will first compute the Taylor series of, ad the itegrate term-by-term withi the radius of covergece of the series Theorem 3, Sec Gettig ready to apply the geometric series result, we write, where the series expasio holds for < <. Thus the series represetatio holds i a disk of radius, aroud. Withi this disk, we ca itegrate the series term-by-term ad get ζ dζ ζ dζ Reidexig the series by chagig +to, we obtai ζ dζ + + <. +. Now we have to decide what to write o the left side. The fuctio Log is a atiderivative of i the disk of radius, cetered at. Remember that Log is ot aalytic o the egative real axis, so we caot take a larger disk. So, for <, we have ζ dζ Log ζ Log Log. Thus, for <, we have Log Log + +, eve though the series o the right coverges i the larger disk <.

18 9 Chapter 4 Power Series ad Lauret Series Solutios to Exercises 4.5. Apply 5 with w ad get equivaletly, +, < ; + +, <. It is worth doig this problem without appealig to formula 5. Sice <, we ca use a geometric series i as follows. We have Expad usig a geometric series. + {}}{, + where i the last series we shifted the idex of summatio by. Note that +, ad so the two series that we derived are the same. 5. We have Log + w w, w <. Put w, the equivaletly Log +, < ; Log +, 9. Simplify the fuctio by usig si a si a cos a, ad get si cos Start with the Maclauri series for si w: for all w, si w For, put w, si +! si si. +! w+. + <. + +! +.

19 Sectio 4.5 Lauret Series 9 To get the desired expasio, multiply by : for, si + +! + +! Applyig a partial fractios decompositio, we fid that I the aulus < < 3, we have < ad 3 <. So to expad 3 +3, factor the 3 i the deomiator ad you ll get , where 3 <. So we ca apply a geometric series expasio ad obtai: This series expasio is valid for < 3; so, i particular, it is valid i the aulus < < 3. To expad +, i the aulus < < 3, because <, we divide the deomiator by ad get + +. Expad, usig a geometric series, which is valid for / < or<, ad get +, + + which is valid for <. Hece, for < < 3, First, derive the partial fractios decompositio + i + i + + i +. The first step should be to reduce the degree of the umerator by dividig it by the deomiator. As i Exercise 3, we hadle each term separately, the costat term is to be left aloe for ow.

20 9 Chapter 4 Power Series ad Lauret Series I the aulus < <, we have < ad <. So to expad i, factor the i the deomiator ad you ll get i i i, where i < or<. Apply a geometric series expasio: for <, i i i i + i. To expad +, i the aulus < <, because <, we divide the deomiator by ad get + +. Expad, usig a geometric series, which is valid for <, ad get Hece, for < <, +. + i + i +.. The fuctio f has isolated sigularities at ad i. If we + i start at the ceter, the closest sigularity is i ad its distace to is. Thus f is aalytic i the disk of radius ad ceter at, which is the aulus + <. Movig outside this disk, we ecouter the secod sigularity at. Thus f is aalytic i the aulus < + <, ad has a Lauret series represetatio there. Fially, the fuctio is aalytic i the aulus < + ad so has a Lauret expasio there. We ow derive the three series expasios. Usig a partial fractios decompositio, we have f + i A A + i, where A i i. We have, for + <, For + <, we have + i + i <, ad so i + + i i +. i + + i

21 Sectio 4.5 Lauret Series 93 Thus, for + <, we have For < +, we have f + i A A + i A + + A + i i i i i + <, ad so + i So, if < + <, the i i i + i +. + i + i + + f Fially, for < +, we have + i A + i 4 i A + i i So, if < +, the f + i A i i i + A + i i +. i I this problem, the idea is to evaluate the itegral by itegratig a Lauret series termby-term. This process is justified by Theorem, which asserts that the Lauret series coverges absolutely ad uiformly o ay closed ad bouded subset of its domai of covergece. Sice a path is closed ad bouded, if the path lies i the domai of covergece of the Lauret series, the the series coverges uiformly o the path. Hece, by Corollary, Sec. 4., the series ca be

22 94 Chapter 4 Power Series ad Lauret Series differetiated term-by-term. We ow preset the details of the solutio. Usig the Maclauri series of si, we have for all, si +! +. Thus si C d C +! + d + d. +! C We ow recall the importat itegral formula: for ay iteger m: { m πi if m, d otherwise, C where C is ay positively orieted simple closed path cotaiig see Example 4, Sec Thus, { + πi if, d otherwise. C Hece all the terms i the series + d +! C are, except the term that correspods to, which is equal to πi. So si d πi. C 9. We follow the same strategy as i Exercise 5 ad use the series expasio from Exercise 5. We have Log + d d C 4 C 4 C 4 d πi, where we have used the fact that C 4 d πi if ad otherwise. 33. a Cosider the power series i, where the coefficiets are give by. Sice f ad are aalytic i the aulus A + R,R, the path i the itegral defiig the coefficiets i, C R, ca be replaced by ay positively orieted path C r, where R <r<r. The reaso for this is that C r ad C R are mutually deformable i A R,R. Fix, r ad r, such that A R,R is arbitrary but fixed, R <r < <r <R. Let M be the maximum value of fζ for ζ i the closed aular regio r ζ r. Note that because f is aalytic, hece cotiuous, i this aular regio, M is fiite. Use the circle of radius r to evaluate the coefficiets i ad estimate as follows: a πi π πr M C r r + fζ ζ M r. + dζ

23 Sectio 4.5 Lauret Series 95 For r < <r, let ρ r, the ρ<. Thus a M r Mρ, ad so a coverges absolutely, by compariso with the series M ρ. Sice is arbitrary i A R,R, we coclude from Lemma, Sec. 4.3, that the series a coverges absolutely for all <R ad uiformly o ay subdisk r <R. Note that this proof shows that the series i defies a aalytic fuctio i the disk B R. Were it ot for the other series with egative powers i, the fuctio f would be aalytic i B R. b The proof i this part is very similar to the proof i part a. I fact, we will use the otatio from a. Fix, r ad r, such that A R,R is arbitrary but fixed, R <r < <r < R. Let M be the maximum value of fζ for ζ i the closed aular regio r ζ r. Use the circle of radius r to evaluate the coefficiets i ad estimate as follows: for <, fζ a dζ πi C r ζ + π πr M M r Mr. r + Equivaletly, for, a Mr., For r < <r, let ρ r the ρ<. Thus, for, ad so a a Mr Mρ, coverges absolutely, by compariso with the series M ρ. Sice is arbitrary i A R,R, we coclude that the series a coverges absolutely for all A R,R. To show that the series coverges uiformly o closed subsets of A R,R, it suffices to show that it coverges uiformly o ay closed subaulus A r,r, where R <r <r <R. For all i a A r,r, we have a. From the previous part, we kow that the series a r r coverges absolutely, sice the poit r is i the aulus A R,R to see this, compute r r, ad R <r <R. Thus, if we take M a, we ca apply the Weierstrass r M-test ad ifer that the series coverges uiformly o A r,r, as desired. a

24 96 Chapter 4 Power Series ad Lauret Series Solutios to Exercises 4.6. The fuctio f si has eros at ± ad kπ, where k is a iteger. All these eros are simple eros. For, write f + si λ. Sice λ,we coclude that is a simple ero. A similar argumet applies to. For the remaiig eros, which are the eros of si, see Example. 5. We will use the fact that the eros of si are all isolated simple eros. At, we have si λ, where λ. So, f si7 4 3 λ 7 λ 7. This shows that if we defie f, the f has a ero of order 3 at. All other eros of f occur at the eros of si, kπ, k, ad they are of order We have cos! + 4 4! 6 6! + So f cos 4 4! + 6 6! + 4 4! + 6! +. Note that λ 4! + 6! + is a power series that coverges for all. Thus λ is etire. Moreover, λ 4. Thus f λ, where λ is aalytic ad λ, implyig that f has ero of order at. 3. Clearly, the fuctio f si + + has isolated sigularities at ad kπ, where k is a iteger. These sigularities are all simple poles. To prove the last assertio, it is easier to work with each part of the fuctio separately. First, show that has a simple pole at the eros of si, which follows immediately from the si fact that the eros of si are simple eros. Secod, show that has a simple pole at, + which follows immediately from the fact that is a simple ero of +. Now to put the two terms together, you ca use the followig fact: If f has a pole of order m at ad g is aalytic at, the f +g has a pole of order m at. This result is easy to prove usig, for example, Theorem Write ta si cos. The problem poits of this fuctio are at ad at the eros of the equatio cos. Solvig, we fid π + kπ k,ka iteger. πk + Sice, as k, k, the fuctio f is ot aalytic i ay puctured disk of the form <. Thus is ot a isolated sigularity. At all the other poits k, the sigularity is isolated ad the order of the sigularity is equal to the order of the ero of cos at k. Sice the eros of cos are all simple this is very similar to Example, we coclude that f has simple poles at k.. We have a isolated sigularity at, which is clearly a essetial sigularity. To see this, we cosider the Lauret series expasio of f: si 3! 3 + 5! 5 3! 3 5! 5 +.

25 Sectio 4.6 Zeros ad Sigularities Determiig the type of sigularity of f + at is equivalet to determiig the type of sigularity of f + + at. Sice f has a removable sigularity at, we coclude that f has a removable sigularity at. Note that this is cosistet with our characteriatio of sigularities accordig to the behavior of the fuctio at the poit. Sice f as, we coclude that f has a removable sigularity ad may b redefied to have a ero at. 9. Arguig as i Exercise 5, it follows that si has a removable sigularity at ad may be redefied to have a ero at. 33. a Suppose that f is etire ad bouded. Cosider g f. The g is aalytic at all. So is a isolated sigularity of g. For all, we have g f M<, where M is a boud for f, which is supposed to exist. Cosequetly, g is bouded aroud ad so is a removable sigularity of g. b Sice f is etire, it has a Maclauri series that coverges for all. Thus, for all, f sum a. I articular, we ca evaluate this series at ad get, for, g f By the uiqueess of Lauret series expasio, it follows that this series is the Lauret series of g. But g has a removable sigularity at. So all the terms with egative powers of must be ero, implyig that g a ad hece f a is a costat. 37. a If f has a pole of order m at, the f a. m φ, where φ is aalytic at ad φ. See 6, Sec So if is a positive iteger, the [f] m φ ψ, m where ψ is aalytic at ad ψ. Thus [f] has a pole at of order m if >. If <, the [f] m φ m ψ, where ψ is aalytic at ad ψ. Thus [f] has a ero at of order m if <. b If f has a essetial sigularity at the f is either bouded or teds to ifiity at. Clearly, the same holds for [f] f : It is either bouded or teds to ear. Thus [f] has a essetial sigularity at. 4. Suppose that f ad g are aalytic i a regio Ω ad fg is idetically ero i Ω. Suppose that g is ot idetically i Ω ad let be such that g. Because g is cotiuous ad g, we ca fid a eighborhood of, B r, such that g for all B r. Sice fg for all Ω, it follows that f for all B r. Thus, the eros of f are ot isolated, ad so, by Theorem 3, f is idetically. 45. Suppose that p is a polyomial such that p for all. Sice p for all, the polyomial has o eros o the uit circle. Let m be the degree of the polyomial ad let a, a,...,a deote the eros of p iside the uit disk, couted accordig to multiplicity.

26 98 Chapter 4 Power Series ad Lauret Series The m. For each j,...,,, cosider the liear fractioal trasformatios φ aj aj a. j We kow form Example 3, Sec. 3.7, that φ j for all. Let φ Π j φ j deote the product of the φ j, where each φ j is repeated accordig to the multiplicity of the ero. We have φ for all. Moreover, φ has exactly eros at the a j. If some of the a j s are repeated, the the order of the ero at a j is equal to the umber of times we repeat a j. Cosider g p φ a j pπ j a j. Factorig out the eros i p ad cacelig the factors a j i the umerator with the same i the deomiator i g, we see that g has removable sigularities at a j, hece it ca be redefied to be aalytic at these poits, ad so we will cosider g to be aalytic i B. Moreover, g has o eros i B, ad for, we have g p φ. By the maximum modulus priciple Corollary 3ii, Sec. 3.7, it follows that g A is costat i B, ad sice g for, it follows that A. As a cosequece, we have p AΠ a j j a j. Thus pπ j a j AΠ ja j. O the right side we have a polyomial of degree. O the left side, we have a polyomial p of degree m, multiplied by a polyomial of degree equal to the umber of oero a j s. It is clear that such a equality is impossible uless m ad all a j ; otherwise the degree of the polyomial o the left becomes strictly greater tha the degree of the polyomial o the right. Cosequetly, p A B, where B.

27 Sectio 4.7 Harmoic Fuctios ad Fourier Series 99 Solutios to Exercises 4.7. a The graph of the boudary fuctio { π + θ if π θ, fθ π θ if <θ<π, is a triagular wave that repeats every π-uits. See Figure 8, Sec. 7.. b The solutio of the Dirichlet problem with boudary fuctio fθ is give by ur, θ a + r a cos θ + b si θ r<, where a ad b are the Fourier coefficiets of f, give by formulas 6 8. However, sice the formula for f is give o the iterval [ π, π], it is more coveiet to use equivalet formulas that are obtaied by chagig the iterval of itegratio i 6 8 from [, π]to[ π, π]. More precisely, we ca use a π fθ dθ, π a π b π π π π π π fθ cos θ dθ, fθ si θ dθ. The reaso is that the precedig itegrads are π-periodic, ad so the values of the itegrals are the same over ay iterval of legth π. See Theorem, Sec. 7. c Sice f is eve, we have a π π π Similarly, sice fθ cos θ is eve, π fθ dθ π π a fθ cos θ dθ π π π cos θ π θ π Now usig the fact that fθ si θ is odd, we obtai odd π fθ dθ π π si θ π π θ dθ π. π θ cos θ dθ π { } cos π π. b fθ si θ dθ. π π d Pluggig the coefficiets ito the series, we obtai the solutio for <r<, ur, θ π + 4 π r cos θ π + 4 r k+ cosk +θ. π k + e We kow that the solutio of the Dirichlet problem iside the uit disk approaches the boudary fuctio as r. Hece lim r ur, θ fθ. Now assumig that lim r ur, θ u, θ, we obtai u, θ fθ, or fθ π + odd 4 π cos θ π + 4 π k k cosk +θ. k +

28 Chapter 4 Power Series ad Lauret Series 5. a Startig with the solutio that we derived i Example ad usig the series i Exercise 4, we obtai ur, θ 5+ π k where ρ r, ρ<. So by Exercise 5, ur, θ 5+ π 5+ π 5+ π r k sik +θ 5+ k + π ta ta ta ρ si θ ρ cos θ r si θ r cos θ y x k + ta ρ si θ +ρcos θ + ta r si θ +rcos θ + ta y +x k + ρk sik +θ, where we have used the relatios x r cos θ ad y r si θ. b Let <T < be a give temperature. Suppose further that T 5. Let x, y be a poit iside the disk x + y 4 such that ux, y T. The T 5+ π ta y x + ta y +x, π T 5 y ta + ta y. x +x Apply the taget to both sides ad use the idetity taa + b write A ta π T 5. The π ta T 5 y x + y +x y +x y x 4y A x + y +4, x + y +4 4Ay, x + y + 4 A y 4, x + y + 4 A A 4. This is the equatio of a circle cetered at,y, where ad radius, y A π ta π cot T 5 T 5, 4 π R A 4 A cot T 5 π π csc T 5 csc T 5. ta a+ta b ta a ta b. To simplify otatio, Note that the cosecat is positive for π T 5, with <T <. So there is o eed to use absolute values whe evaluatig the square root.,