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1 Complex Analysis II MT433P Problems/Homework Recommended Reading: Conway: Functions of One Complex Variable, Springer Ahlfors: Complex Analysis, McGraw-Hill Jaenich: Funktionentheorie, Springer Arnold: Complex Analysis, Freitag, Busam: Complex Analysis, Springer Current Homework is on the last page. Compute the Jacobi matrix of the map f : R 3 R with fx, y, x + y + x, e x xy.. Compute d x,y, f + x x ye x xy xe x xy 0 d t dt x 34 + x dx. t t 0 3. Compute the its x x3 e x, x logx. x 0 Both its are Express the following in the form a + ib with a, b R: 3 + 7i3 + 3i, 3 + i 7i +, cosi i3 + 3i 9 + i i, 3 + i 7i i 7i 7i + 7i 3 + i 7i 0 30i Compute modulus and an argument for cosi eii + e ii e + e cosh + i, + 3i, 3 i. + i exp i π 4, + 3i 3 3 exp i arctan, 3 i exp i π 3 6. Write the polynomial px x R[x] as a product of quadratic/linear real polynomials. Abbreviating j : exp iπ 4, hence j i we compute x x ix + i x + jx jx + ijx ij x + jx ij x + ijx j x + x + x x + please hand up -6 Monday, 7/09 in class

2 7. Compute d where : [0, 3] C is the curve with t t + it for all t [0, 3]. Since the integrand is holomorphic on all of C, the integral depends on only through the endpoints 0 0 and i. Since d d we have d [ ] 3+9i i 0 8. Find a C -map φ: [0, ] [0, ] U : {x + iy x [, or y [, } such that φt, 0 + t and φt, i + it for all t [0, ]. For s [0, ] we have cs : + i + e iπs iπ/4 U since and Hence we have a smooth curve Rcs + cosπs π/4 for 0 s Ics + sinπs π/4 for s c: [0, ] U, s cs : + i + e iπs iπ/4 with c0 and c i. The domain U has the property that λ U if U, λ [,. Hence we can define a map φ: [0, ] [0, ] U : {x + iy x [, or y [, } by φt, s : cs + t. 9. Let f : U : {x + iy y 0 or y 0 and x > 0} C be holomorphic. Let, µ: [0, ] C \ R 0 be C -curves with 0 µ0 and µ. Prove that fd fd. On U the argument defines a smooth function arg: U π, π so that e i arg for all U. Thus for all t [0, ] t t e i argt µt µt e i argµt and φt, s iargµt+ s argt sµt + s t e defines a C map φ: [0, ] [0, ] U so that for all t [0, ] we have φt, 0 t, φt, µt and for all s [0, ] we have φ0, s 0 µ0 and φ, s µ. Thus and µ are homotopic relative endpoint and therefore the claim follows from the Cauchy Integral Theorem. please hand up 7-9 Thursday, 7/0 in class 0. For 0 < r πz compute the integral r µ d sin. Hint: Recall that 0 r fd is short for fd where : [0, π] C is the curve with t 0 + re it. Use your freedom to cut and deform the curve! Since the integrand is holomorphic in C\πZ, by Cauchy s Integral Theorem we can replace the curve by a family of cirles around the eros of sin within the circle { r}. Thus d r sin resp. if kπ < r < k+π with k even respectively odd.. Compute the integral d The eros of the polynomial all lie in B , hence the integrand is holomorphic outside this ball. By Cauchy s Integral Theorem, d R d

3 for any R > We estimate d π ire it Re it Re it + dt R 0 π 0 R R 9 30R dt πr R R R. What are the winding numbers w, i of the curve around the i. 3,, Compute the residues a cos, π/ cos, π/ π/ sin π/, π/ h π/, π/ since h sin π/ has a removable singularity at π/ and its extension evaluates to h π/. b sin, π sin, π sin, 0 since sin + π sin, hence sin + π sin for all. Now sin sin and therefore all monomials in the Laurant series of sin. Hence c sin 3, π the function is holomorphic at π e d 3, 0 e 3, 0 please hand up 0-3 Thursday, /0 in class sin, π of odd degree vanish, in particular the coefficient with 0. sin 3, π 4. Let S πz {i, i} and consider the function f : C \ S C, 0, k0 k k! k k k! 3, 0 3, 0 k k 3, 0 k! f e + sin for all C \ S. For all 0 S determine the type removable, pole, essential of the singularity. For poles give their order. Recall that a singularity 0 of f is removable if 0 f is finite, a pole if 0 f and

4 essential if the 0 f does not exist. Thus for the function given in : 0 is removable since 0 e + sin exists. πz \ {0} are poles of order since for p 0 sin e + e 0 kπ p f kπp e + sin + kπp e sin kπ and therefore the it kπ p e + kπ kπp f kπ sin kπ u 0 u p u + kπ e sinu +u+kπ { u p u 0 sinu u + kπ e +u+kπ kπ e +kπ : p u 0 : p < exitst for p but for no p <. ±i are essential singularities: We prove that i is an essential singularity. Treating i only requires a few sign changes. We have hence, with h : sin and g e i+i, f he + he + i i + i i i e i+i hge i i Also note that Hence whereas hi i, i ir 0 and gi e 4 0. sini + f higi e i i i, i ir + higi e λ higi e λ 0 λ 0,λ R + λ,λ R + i, i ir f higi e i i i, i ir 5. Prove that for all C, we have e e R. higi e λ higi eλ λ 0,λ R λ,λ R + e e R e ii, hence e e R e ii e R since e ii. 6. Find all entire functions f functions holomorphic on C such that f e R for all C. Hint: R is the real part of, Ru + iv u for u, v R. You might use Liouvile s Theorem: A bounded entire i.e. holomorphic on all of C function is constant. Look at the function f/e! If f is such a function then f e R e, hence f e is bounded, and holomorphic since e 0 for all. By Liouvilles Theorem f e c is constant. Because of the estimate, c. So the entire functions satisfying the above estimate are precisly the functions f of the form f ce with c. 7. Let f be an entire function, i.e. a holomorphic function f : C C. Assume that f preserves the real line, i.e. fr R. Show that f then commutes with complex conjugation, f f for all C. Hint: One way of proving this is to show that the function f is holomorphic. A second method of proof uses the representation of an entire function as an everywhere convergent power series. Show that the Taylor coefficients of such a function must be real!

5 First method: If f is holomorphic at p then f fp p p exists. We write C for the complex conjugate of a complex number here. Observe that this defines a continuous map C C. We want to show that C fc is holomorphic, i.e. that for all p C, exists. But by continuity of C C fc C fc p C p p p C fc C fc p p p fc fc p C C p C p fc fc p C C C p Cp f fc p C p Now this it exists since f is holomorphic at C p. Since f and the function C fc are both holomorphic on all of C and coincide on R they must coincide on all of C by the Uniqueness Theorem. Second solution: Since f is entire, there are a k such that f k0 a k k for all C. By the assumption, the Taylor coefficients a k dfx k!dx R. x0 Hence a k a k and 8. Compute the integral d where is the curve shown in the picture. f a k k a k k f. k0 k0 0 The winding numbers are w, 0, w, 0, w, and the residues f, 0, f, 0, f, 7. By the idue Theorem, 3 + d πi πi. + + please hand up 4-8 Thursday, 4/ in class 9. Compute the residues 3 e sincos, 0, ,. Both singularities are simple poles. The residues are therefore 3 e sincos, 0 4 e sincos e , sin + 4! 4 ±. 0 6, Compute the integral For R we have + x 4 + 5x + 4 dx. R R R R 4 5R 0. 4

6 Thus the integrand decays sufficiently rapidly to close the path of integration with a negligible arc in the upper half plane at infinity. The integral is therefore given by the sum of the residues in the upper half plane, + x 4 + 5x + 4 dx πi , p. p H Since i i + i i the singularities are simple poles at ±i, ±i. The residues in the upper half plane are , i + i i + i i, i i + i i + i i i i 3i i 6i, , i + i i + i i, i i + i i + i i i 3i i 4i i. For the integral we get from the idue Theorem + x 4 + 5x dx πi + 4 6i + π i 6. please hand up 9-0 Tuesday, 9/ in class. Find the number of solutions of each of the following equations in the given domains: a in { C > }, For we have 5 > 4 +. Hence we have as many eros in the disk as the function 5 and therefore 3 eros in the given domain. b i 0 in { C < }, As before, 5 4 dominates for. There are 4 eros in the domain. c 5 + i i 0 in { C < < }. For, the linear term 4 dominates and we have ero in B 0. For the leading term dominates, we have all 5 eros in B 0. Thus there are 4 eros in the domain.. Prove that for all C, coshi : ei + e I cos sinh I : e I e I. If u + iv, u, v R, we have cos e i e i e v e iu e v e iu e v e iu e v e iu e v e v e v e v and cos e i e i e v e iu e v e iu e v e iu + e v e iu e v + e v 3. How many eros has the function f 4 +cos in B π/ 0 when counting with multiplicities and when disregarding multiplicities? From we estimate for π/ that 4 π > 9 and cos ei + e I eπ/ By Rouché s Theorem 4 has as many eros in B π/ as f 4 + cos, i.e.. Now since f is real, we have for R, π/ π/ that f R and f > 0. Therefore we must have two different complex conjugate eros. please hand up -3 Thursday, 5/ in class 4. Prove that for any r > 0 and any piecewise C -curve : [0, ] C \ B r 0 we have Hint: L t dt 0 πr w, 0 r d L πr w, r t t dt t r dt L. r 0

7 5. Is there a curve : [0, ] C such that the map C \ [0, ] Z, p w, p, has infinite image? Hint: It suffices to sketch the image of. 4 is relevant. 6. Recall that a rational function f : C C is one of the form f p q where p, q C[] and q is not the ero polynomial. Express the sum ordf : ordf, x x C in terms of the degrees of p and q. If f, g are two rational functions with given ordf, ordg, what is ordfg? For any meromorphic f, g we have ordfg, p ordf, p + ordg, p, hence ordfg ordf + ordg. Now any non ero polynomial p C[] can be written in the form p c s ordp,s and s p 0 s p 0 ordp, s degp hence ord p degp degq. q 7. The order at infinity, ordf,, of a meromorphic function f : C C is defined by inserting, i.e. ordf, ordf/, 0. Prove that a meromorphic function whose order at infinity is finite must be rational. Let f be a meromorphic with ordf, ordf/, 0 k Z. Then f/ k is holomorphic near 0. We may assume k 0. Then there are M, r > 0 such that for all B r 0 we have f/ M. Hence for > /r we have f M, f is bounded outside B 0. Since B 0 is compact, f has only finitely many poles r r p,..., p r of order m,..., m r, all within B 0. Let q r r i p i mi. Then fq is an entire function in and fq M q whenever r <. The estimate for the Taylor coefficients a k of fq k0 a k k, a k max { fwqw w R} M max { qw w R} πr k πr k, we get that a k 0 for all k > degq and fq is a polynomial. please hand up -7 Monday, / in class 8. Show that the maximum principle holds with φ: C R, φ R 4 + I 4, in place of the modulus: If U C is open and connected, f : U C is holomorphic and p U is such that φfp Rfp 4 + Ifp 4 φf Rf 4 + If 4 then f is constant. Can one also replace the modulus by the sine of the modulus, i.e. the function φ with φ sin? Let M : φp and Z : { C φ M}. We have fu Z and fp Z, hence fu can not contain a neighbourhood of fp. By the Open Mapping Theorem, f is constant. The point here is that φ: C R is open. The function φ with φ sin is not open and one actually does not have a maximum principle for this function. Take the function f with f. We then have max φf max sin which is assumed at for instance p π/. 9. Find a biholomorphic map h: { re it r, t R, 0 < r, 0 < t < π/8 } B 0. Hint: Compose h from maps between a half plane, the sector and B 0. Try Moebius transformations a+b c+d and power functions k. h 8 i 8 +i does the job. 30. Let U C be open, p U and f : U \ {p} C be holomorphic. Let f k c k p k be the Laurent series of f near p. How can one compute the residue from c, c, c 0, c? f, p f, p

8 We may assume p 0. Then f c near 0 with a holomorphic function g and f c + c with holomorphic g. The residue therefore is + c + c 0 + c + g + c 0 + c + c g c c + c c 0 + g f, 0 c c + c c 0. please hand up 8-30 Thursday, 9/ in class 3. Let p C and r R so that 0 < r < p. Find P p, r C and Rp, r R + so that { } B r p : p < r { P p, r < Rp, r} : B Rp,r P p, r P p, r Rp, r p pp r and r pp r. 3. For each of the following domains U C and holomrphic functions f : U C describe the image fu by equations or inequalities for modulus, real or imaginary part. a U B 0, f : / +, b U {x + iy x, y R, x < 0}, f :, c U {x + iy x, y R, < x < }, f : /, Hint: As an example, if U { < < } and f : / + 3, then { } { } { fu + 3 < < < 3 < 4 < 3 8 and 3 5 < } 5. You can use 3! 33. Let p E B 0 and let w p : E C, w p p p a Prove that w p E E. p + pp p p < p + pp p p because for any real numbers a, b pp, 0 a, b < we have b Prove that w p : E E is biholomorphic. The inverse of w p is w p. 34. Let h: Ω Ω be biholomorphic where Ω C is the domain + ab a b a b > 0. Ω : { x + iy x, y R, y < x + }. Assume that h 0, h0. What is h h h? Prove your statement! Hint: Use 33, the Schwar Lemma and the Riemann Mapping Theorem! The domain Ω is simply connected and different from C. Hence by the Riemann Mapping Theorem there is a biholomorphic map φ: Ω E. By 33 the composition ψ w φ0 φ: C E is biholomorphic and takes 0 0. Now h 0 0 and h, hence ψ h ψ : ĥ: E E is biholomorphic and ĥ0 0. By the Schwar lemma, ĥ is a rotation. Finally, ĥψ ψh ψ 0 ψ0 hence ĥ fixes a point other than 0, so ĥ id E. It follows that h ψ ĥ ψ id Ω. please hand up 3-34 Tuesday, 4/ in class

9 Some Problems for the Study Week 35. For each of the following pairs of domains U i, V i in C find a biholomorphic map h i : U i V i or h i : V i U i a U a {x + iy x, y R, y > 0}, V a {x + iy x, y R, π > y > 0} Hint: Think of polar coordinates! The exponential function exp: e is πi-periodic and takes S {x + iy x, y R, y 0, π} biholomorphically to almost all of C C \ {0}. Now rescale! h a e / b U b { x + iy x, y R, y > 0, x + y < }, V b {x + iy x, y R, y > 0} We map V b {x + iy x, y R, x, y > 0} {x + iy x, y R, x > 0, y > } {x + iy x, y R, x > 0, y > } B i/ {x + iy x, y R, x > 0} B {x + iy x, y R, y > 0} B 0 {x + iy x, y R, y > 0} U b using square root, shifts +q, rotations a, a, and the inversion /. Thus the map h b defined by h b i i/ + / + i + i + i does the job. 36. For the curve as in the picture compute e a d and sin b + πe sin d The singularities of the integrand in 36a surrounded by are a simple pole at π and a removable singularity at 0. The winding numbers and the residues for the first integrand are Thus the integral in 36a is π 0 w, π, w, 0 3 e e sin, π e π e π, sin, 0 0 e sin d πi e π πi e π The integrand in 36b has a double pole at π and a pole of order 3 at 0. To compute the residues of these higher order pole we demonstrate the two basic techniques: first the differentiation formula and second the Laurent series. Both are always valid methods, but the second often requires less computation. Take care to expand everything to sufficient high order! + πe sin, π + πe sin + π, π d d + π π + πe sin + π d + π d π e sin + π [ e sin + π + + πe sin + π + e sin + π + e ] cos + π e sin + π π [ e + e sin + π + e sin + π + e ] cos + π e sin + π π [ e sin + π + e ] sin + π e sin + π [ e + e ] e π e π πe π π e π π eπ e π πe π π e π You can also compute this from the Laurent series. We first substitue u + π and replace each factor by the first part of its Laurent series. + πe sin, π u πue u π sinu, u 0 π u π u e π e π eπ u e π u u 6, u 0

10 Similiarly, + πe sin, 0 e π π e π e π π e π πeπ e π e π π e π π π + π 3 + +, 0 6 π + π + π π + 6π + + π π 3 3π + 6π + π 3 π + π + 4 4π 3 Thus the integral in 36b is + πe d πi πeπ e π e π sin π e π + 3 π + π + 4 4π Calculate the improper integral 0 x x + x + 4x + 9 dx. The integrand is even, so the integral is half the integral over the full real line. The integrand decays rapidly enough for x so that we can close the path of integration in the upper half plane. The singularities of the integrand in the upper half plane are simple poles at x i, i, 3i. The residues there are x x + x + 4x + 9, x i x x i x x + x + 4x + 9 xi x + ix + 4x + 9 xi Hence πi p C,Ip>0 i i + ii + 4i i x x + x + 4x + 9, x i x x + x + 4x + 9, x 3i x 5i 3 80i 0 x + x + 4x + 9 dx x + x + 4x + 9 dx x x + x + 4x + 9, x p πi 48i + 5i π π 80i 40 0 x 38. Prove that there is a biholomorphic map W : {x + iy x, y R, x 0 or y > or y < 0} E : B 0 \ {0}. Hint: If a: C \ {0} C \ {0} is the map with a /, what is aw? a: W aw {x + iy x, y R, x 0 or y > 0 or < y < 0} is biholomorphic. By the Riemann Mapping Theorem there is a biholomorphic map h: aw {0} E. If h0 0 then recall from problem 3 that there is a biholomorphic map w h 0: E E taking h0 to 0. We then continue to work with w h0 h in place of h. For this one would simply say without loss of generality we may assume h0 0. Thus we may assume h0 0. tricting h we get a biholomorphic map k : aw B 0 \ {0}. 39. Let U C, U C, be a simply connected open subset and h: U U a biholomorphic map. Assume that h fixes two points of U. Prove that h is the identity. By the Riemann Mapping Theorem, we can assume that U B 0 and, since the biholomorphic maps of the disc act transitively, we may also assume that one of the points fixed by h is 0. By the Schwar Lemma, h must be a rotation and the claim follows. 40. For each of the following functions f i : U i C find the number of eroes counted with multiplicities i.e. p U i ordf i, p and disregarding multiplicities i.e. #f i 0 a U a B 0, f a For B 0 U, i.e., we have 4 4 > 8 +. Hence f a has as many eros in U a as that is, when counting multiplicities. Thus f a either has two different eros or one double ero in U a. Since f is even, i.e. f a f a, for a double ero p U a we would have f a p 0 f a p, p U a, hence p p 0. But f a 0 0. Hence we have two simple eros in U.

11 b U b B 0 \ B / 0, f b As in 40a, we have 6 eros in B 0 for, 5 6 > and none of these lie in B / 0 for /, 3 + /8 > > Hence f b has as many eros in B / 0 as 3 +, that is none. Let ɛ e πi/3, ɛ 3 ɛ. Obviously f b ɛ f b ɛ f b for all C. In particular, the eros of f b come in triples {ɛ, ɛ, }. Also if is a ero then so is its complex conjugate. Since f b has no real eros in B 0 there are no multiple eros in U b, hence we have 6 simple eros in U b. 4. Let B 0 U C be open and f : U B 0 be holomorphic. How many fixed points has f? Prove your answer! Hint: A fixed point of f is a point with f. B 0 { C }. If then > f. Hence, by Rouché s Theorem, the function f has as many eros in B 0 as the function, i.e.. Thus we have fixed point. 4. Let h: E : B 0 H { C I > 0} be a biholomorphic map with h0 i and Ih/. Find all possible values for h + i/. The inverse of the Cayley transform, c: H E, c i +i, is biholomorphic and φ c : E H, i + i φ0 i, φ 3i. By the Schwar Lemma, we must have h φ A with a suitable rotation A: E E, α, for some α S { C }. We find α by solving the equation Ih/ Iφα/ : α Iφ I i + i α α i + iα i + α α αα + α α 3 I I Ii4 α 4 + αα α + α 4 + αα α + α 5 α + α wich gives α + α, hence α ± i 3 With these two values we get + i α + i h φ 4i + i ± i 3 + i 4 ± i 3 + i 5 3i ± 3 3 ± 3 i ± For i a, b, c let U i C and f i : U i C be as below. For each i determine the number of eros of f i on U i, counted with multiplicities. a U a { C < < 3}, f a for all U a, The function has 9 eros, all in B 3 0 since for 3 we have 9 > 3 and therefore 9 and have the same number of eros in B 3 0. For we have 3 > 9 + which gives one ero in B 0. Hence there are 8 eros in U a. b U b {x + iy C x, y R, x < 0}, f b e for all U b, The function has no eros with large modulus. We can therefore assume U b { C R < R < 0} with some R R sufficiently large. On the boundary of this U b we have e < + 6 and therefore f b has as many eros there as + 6, that is 8. c U c { C > }, fc for all U c. If / then 8 3 > and f c has as many eros in B / 0 as 3, that is 3. Overall f c has 9 eros. Thus f c has 6 eros in U c 44. Let be the simple curve as shown in the picture. Compute the integrals d and + d + 4 C ir i i 0 i R i

12 The singularities of the integrands are at i, i, 0, i, i. The winding numbers of are w, i 0, w, i, w, 0, w, i 0, w, i. The first integrand has only simple poles at i, 0, i with residues + i i, i + i i, 0 + i i, +i i i i i ii For the first integral we thus get πi + 5πi. The relevant residues of the second integrand are + 4, 0 0, since the function is even The second integral therefore is 8. i + i, i i4i a Compute the integral + + x + x 4 dx + + x + x 4 dx πi p,ip> , p The nonvanishing residues are at simple poles at ɛ and ɛ 3 where ɛ e πi/4 +i , ɛ + ɛ ɛ ɛ 3 ɛ ɛ 5 ɛ ɛ 7 + i + i ɛ i 4iɛ + + 4, + ɛ 6 ɛ3 ɛ 3 ɛɛ 3 ɛ 5 ɛ 3 ɛ 7 i i ɛ i 3 4iɛ 3 + i 4iɛ For the integral this yields πi + i 4iɛ π. b Let f : C C be a holomorphic function, not identically ero. Prove that f 3, 0 0. f has a ero of finite possibly 0 order k at 0. Hence f has an isolated possibly removable singularity at 0. For sufficiently small we have the Laurent series f n k a k k.. and therefore f 3 n k a k 3k b k k n k with b k a k/3 if k is divisible by 3 and b k 0 else. In particular f 3, 0 b 0.