Math 220A Homework 2 Solutions

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1 Mth 22A Homework 2 Solutions Jim Agler. Let G be n open set in C. ()Show tht the product rule for nd holds for products of C z z functions on G. (b) Show tht if f is nlytic on G, then 2 z z f(z) 2 f (z) 2 for ll z G. (c) Show tht if f nd f 2 re nlytic on G nd f (z) 2 + f 2 (z) 2 for ll z G, then f nd f 2 re constnt. Solution. (): A rel liner opertor D defined on n lgebr of rel vlued functions is rel derivtion if it stisfies the product rule, i.e., D(uv) D(u)v + ud(v). Likewise, complex liner opertor L defined on n lgebr of complex vlued functions is complex derivtion if Recll tht we defined nd x y the web pge) by setting (u + iv) x x u + i x v Therefore, tht known fct tht following generl fct. nd x y nd x y L(fg) L(f)g + fl(g). in clss (cf. notes on del z nd del z-br on nd y (u + iv) y u + i y v re complex derivtions follows from the well re rel derivtions (first yer clculus) nd the Fct.. If D is rel derivtion on rel lgebr A, then L defined on A + ia by L(u + iv) D(u) + id(v) is complex derivtion on A + ia.

2 Proof. Let f u + iv nd g s + it. Then L(fg) L((u + iv)(s + it)) L((us vt) + i(ut + vs)) D(us vt) + id(ut + vs) D(u)s + ud(s) (D(v)t + vd(t)) + i(d(u)t + ud(t)) + i(d(v)s + vd(s)) [ D(u)s D(v)t + i(d(u)t + D(v)s) ] + [ ud(s) vd(t) + i(ud(t) + vd(s)) ] (D(u) + id(v))(s + it) + (u + iv)(d(s) + id(t)) L(f)g + fl(g) Since from elementry clculus we know tht D nd D re x y rel derivtions, it follows tht nd cting on complex vlued functions x y re derivtions s well. Once it is known tht re complex derivtions, then the fcts tht nd x y z 2 ( x i y ) nd z 2 ( x + i y ) stisfy the product rule follows from the following fct. Fct.2. If, b C nd L nd M re complex derivtions then L + bm is complex derivtion. Proof. (L + bm)(fg) L(fg) + bm(fg) (L(f)g + fl(g)) + b(m(f)g + fm(g)) (L(f) + bm(f))g + f(l(g) + bm(g)) (L + bm)(f)g + f(l + bm)(g). (b): Note tht if u is rel vlued, then z u 2 ( x u + i y u) 2 ( x u i y u) z u. 2

3 Hence, if f u + iv, z f z u + i z v z u + i z v z u i z v z f Thus since f is nlytic implies tht f nd f f, we see tht if f z z is nlytic, then z f nd z f f. Using these fcts nd the product rule, we hve tht 2 z z f 2 z z f 2 z z (ff) z (( z f)f + f z f) z (ff ) ( z f)f + f z f f f f 2. (c): By prt (b) bove, if f (z) 2 + f 2 (z) 2 for ll z G, then 2 z z ( f (z) 2 + f 2 (z) 2 ) f (z) 2 + f 2(z) 2 3

4 for ll z G. Therefore both f nd f 2 re identiclly on G. This implies by Proposition 2. (Conwy pg. 37) tht f nd f 2 re constnt. Note. Something is not quite right here. Wht is it nd how cn it be fixed?. (#8 pg. 55 Conwy) If T z z+b cz+d show tht T (R ) R iff we cn chose, b, c, d to be rel numbers. Solution. We ssume tht T is Moebius trnsformtion, i.e., d bc. If, b, c, d to be rel numbers, then it is esy to see tht T : R R nd lso tht d bc implies tht T is surjective. To prove the converse, first consider the cse c, so tht T Az + B where A /d nd B b/d. Since, T () nd T () re both rel, it follows tht A nd B re rel. Hence, the converse holds when c. Now ssume tht c so tht T z Az+B where A /c, B b/c, nd z+d D d/c. The ssumption tht T x R for ll x R gives tht or equivlently tht Hence, x R Ax + B x + D Ax + B x + D x R Ax 2 + (B + AD)x + BD Ax 2 + (B + AD)x + BD. A A, (.3) B + AD B + AD, nd (.4) BD BD. (.5) Clerly, (.3) implies tht A R. There remins to prove tht B R nd D R. The proof splits into the two cses A nd A. If A, then (.4) implies tht B R. Furthermore, s T is onto, B. Hence, (.5) implies tht D R s well. Now ssume tht A. (.4) implies tht Im B A Im D (.6) 4

5 nd tking the imginry prt of (.5) gives tht These equtions imply tht Re B Im D Im B Re D. Im D (Re B A Re D). (.7) Note tht if Re B A Re D then s we hve tht (.6) holds s well, B A D. This would imply tht T is constnt, contrdiction. Hence, Re B A Re D. But then (.7) implies tht Im D. Since Im D, by (.6) nd the fct tht A is rel, so lso Im B. Since Im B Im D, B, D R, s ws to be shown. Note. This is very plodding solution to the problem. By using ides in the two solutions given to the following problem one cn come up with vrious shorter solutions. 2. (#9 pg. 55 Conwy) If T z z+b, find necessry nd sufficient conditions tht T (Γ) Γ where Γ is the unit circle {z C z cz+d }. Solution. First observe tht since T is Moebius trnsformtion, either T (D) D or T (D) C \ (D Γ). Also notice tht T (D) C \ (D Γ) if nd only if S defined by Sz (T z) hs the property tht S(D) D. Therefore, it suffices to solve the problem under the ssumption tht T (D) D. (.8) Now, (.8) implies tht there exists β D such tht T β. Hence, since T is not constnt, there exist C, D C such tht T is given by T z z β Cz + D. (.9) T (Γ) Γ implies tht z β 2 Cz + D 2 for ll z Γ. Hence, for ll z Γ This implies the two equtions + β 2 2Re βz C 2 + D 2 + 2Re DCz. + β 2 C 2 + D 2 nd β DC, (.) 5

6 which in turn imply tht ( C 2 )( D 2 ). Since C T ( ) Γ, C. Hence, D. Using (.9) nd (.) we my rewrite the formul for T in the form, T z D z β βz. (.) Summrizing, we hve shown tht if T is Moebius trnsformtion stisfying T (Γ) Γ nd T (D) D, then there exist β D nd τ Γ such tht T z τ z β βz. The converse of this ssertion is true s well. If T is defined s bove, then T (Γ) Γ by strightforwrd clcultion. Since T is Moebius trnsformtion, either T (D) D or T (D) C \ (D Γ). Hence, since T (β), T (D) D. Solution 2. Let T be s in (.9) nd define S by S(z) T ( z ). Noting tht w Γ if nd only if w, it follows tht S(z) T (z) for w ll z Γ. Since S nd T re Moebius trnsformtions, it follows from Proposition 3.9 (pg. 48 Conwy) tht S T. Hence, T (z) S(z) T ( ) z /z β C/z+D C/z D /z β Dz + C βz D z + C/D βz 6

7 Since, T (β), it follows tht C/D β nd which is (.). T (z) D z β βz 3. (#7 pg. 55 Conwy) Let G be region nd suppose tht f : G C is nlytic such tht f(g) is subset of circle. Show tht f is constnt. Solution. I clim tht f (z) for ll z G. For if G nd f (), then f is conforml t (Theorem 3.4 pg. 46). But, s f(g) is subset of circle, f cnnot be conforml t. Therefore, f (z) for ll z G. As G is region, in prticulr G connected (cf. Conwy pg. 4). Therefore, the ssertion follows from Proposition 2. (pg. 37) which sserts tht n nlytic function defined on connected open set whose derivtive vnishes identiclly must be constnt. Solution 2. Suppose Γ is subrc of of circle in C nd f : G Γ is nlytic. There exists Moebius trnsformtion T (z) z+b such tht T cz+d mps Γ to subintervl of R. Hence, T f is rel vlued nlytic function on G. But then (Exercise #4 Conwy pg. 44 or HW #5), T f is constnt function on G. Therefore, f is constnt on G. Solution 3. Let f : G Γ, where Γ circle. Let denote the center of Γ, let r denote the rdius of Γ nd let g be defined by g(z) f(z). Evidently, g(z) 2 r 2 for ll z G. Hence, by Exercise (b), we hve tht g (z) 2 2 z z g(z) 2 for ll z G. Hence, g is constnt on G. 4. (#4 pg. 67 Conwy) Prove Proposition.8 (pg. 6) (use induction). Solution. The result follows by induction from the following clim. 7

8 Clim.2. Let γ : [, b] C be of bounded vrition, let f : [, b] C be continuous nd ssume tht c [, b]. Then f dγ c f dγ + c f dγ. Proof. Fix ɛ >. By the definition of f dγ, there exists δ > such tht prtitions P of [,b] smple choices τ for P f dγ S γ (f, P, τ) < ɛ 3. (.3) By the definitions of c f dγ nd f dγ there exist prtitions P of [, c] c nd P + of [c, b] nd smple choices τ for P nd τ + for P + such tht c P < δ nd P + < δ, (.4) f dγ S γ (f, P, τ ) < ɛ, nd (.5) 3 c f dγ S γ (f, P +, τ + ) < ɛ 3. (.6) We define prtition P of [, b] by setting P P P + nd select smple points τ for P by letting τ τ τ +. Note tht with this construction, S γ (f, P, τ) S γ (f, P, τ ) + S γ (f, P +, τ + ). (.7) Also, s (.4) holds, P < δ so (.3) implies tht Using (.5), (.6), (.7), nd (.8) we see tht f dγ ( c f dγ + c f dγ S γ (f, P, τ) < ɛ 3. (.8) f dγ) ( f dγ S γ (f, P, τ) ) + ( S γ (f, P, τ ) f dγ S γ (f, P, τ) + <ɛ/3 + ɛ/3 + ɛ/3 ɛ. S γ (f, P, τ ) c c f dγ ) + ( S γ (f, P +, τ + ) f dγ + Clim.2 follows from this inequlity by letting ɛ + 8 S γ (f, P +, τ + ) c c f dγ ) f dγ

9 5. (#5 pg. 67 Conwy) Let γ(t) exp (( + i)t ) for < t nd γ(). Show tht γ is rectifible pth nd give rough sketch of the trce of γ. Solution. The proof tht γ is rectifible, i.e., hs bounded vrition, will be bsed on the following simple fct. Clim.9. Suppose tht γ : [, b] C is continuous t nd for ech s (, b] let γ s γ [s, b]. γ hs bounded vrition iff γ s hs bounded vrition for ech s (, b] nd sup s (s,b] V (γ s ) <. Furthermore in this cse, lim V (γ s) V (γ). (.2) s + Proof. If γ hs bounded vrition, then γ s hs bounded vrition nd V (γ s ) V ( γ [, s] ) + V (γ s ) V (γ) so tht sup s (s,b] V (γ s ) < V (γ). To see tht (.2) holds note tht V (γ s ) is monotone in s so tht it suffices to prove the for ech ɛ >, there exists t > such tht V (γ) ɛ < V (γ t ). To prove this, fix ɛ > nd choose δ > such tht t [, δ) γ(t) γ() < ɛ/2. Choose prtition P of [, ] stisfying Then P < δ nd V (γ) ɛ/2 < v(p ; γ). V (γ) ɛ < v(p ; γ) ɛ/2 ( γ(t ) γ() ɛ/2) + < V (γ t ). m γ(t k ) γ(t k ) Now ssume tht γ s hs bounded vrition for ech s (s, b] nd L sup s (s,b] V (γ s ) <. Since γ is continuous t, there exists δ > such tht γ is bounded on [, δ). As γ δ hs bounded vrition, it is lso the cse tht k2 9

10 γ is bounded on [δ, ]. Hence, γ is bounded. If sup t [,] γ(t) B, it follows tht if P is prtition of [, ], then v(γ; P ) m γ(t k ) γ(t k ) k γ(t ) γ(t ) + 2B + V (γ t ) 2B + L <. This proves tht γ hs bounded vrition m γ(t k ) γ(t k ) k2 If γ is s defined in the exercise, then s lim γ(t) t + e t, it is cler tht γ is continuous t. Therefore, by Clim.9 we will be done if we cn show tht sup s (,] V (γ s ) <. But for s (, ], γ s γ [s, ] is smooth pth. Therefore, by Proposition.3 (Conwy pg. 58), V (γ s ) s s γ (t) dt i t 2 2 s 2e t t 2 e s e ( +i)t dt t 2(e e s ) Since e s s s + Clim.9 implies tht γ hs bounded vrition nd tht V (γ) 2/e. dt 6. (#6 pg. 67 Conwy) Show tht if γ : [, b] C is Lipschitz function then γ is of bounded vrition.

11 Solution. As γ is ssumed Lipschitz, there exists constnt M so tht γ(t) γ(s) M s t for ll s, t [, b]. If P { t < t <... < t m b} is prtition of [, b], then m v(γ; P ) γ(t k ) γ(t k ) k m M(t k ) t k ) M(b ). Hence, sup{v(γ; P ) P is prtition of [, b]} M(b ). Therefore, γ hs bounded vrition. k 7. (# 7 pg. 67 Conwy) Show tht γ : [, ] C, defined by γ(t) t + it sin for t nd γ(), is pth but is not rectifible. Sketch this t pth. Solution. Define sequence of points x, x 2, x 3... in [, ] by Evidently, x n γ(x n ) 2 (2n + )π, n. 2 (2n + )π ( + ( )n i). Hence, for ech n, γ(xn+ ) γ(x n ) Im (γ(xn+ ) γ(x n )) 2 2 (2n + 3)π ( )n+ (2n + )π ( )n 4 (2n + 3)π.

12 It follows tht if for ech positive integer N we define prtition P N of [, ] by P N {, x N, x N,..., x 2, x, }, then v(γ, P N ) γ(x N ) γ() + N n n N n γ(x n ) γ(x n+ ) N 4 (2n + 3)π. As n /(2n + 3) is divergent series, it follows tht γ(x n ) γ(x n+ ) + γ() γ(x ) lim v(γ; P N). N Therefore, γ does not hve bounded vrition. 8. (# pg. 67 Conwy) Let γ be the closed polygon [ i, + i, + i, i, i]. Find dz. γ z Solution. We brek the polygonl contour up into 4 smooth pths γ, γ 2, γ 3, γ 4 defined by γ (t) + ti, t, γ 2 (t) t + i, t, γ 3 (t) ti, t, nd γ 4 (t) t i, t. 2

13 We hve γ z dz γ (t) dγ (t) γ (t) γ (t)dt + it idt t + i + t 2 dt t + t dt + i 2 + t dt 2 2 log( + t2 ) + i rctn t + i( π 4 π 4 ) π 2 i. In like fshion, ech of the other integrls, dz, dz, nd γ 2 z γ 2 z γ 3 z cn be shown to be equl to π i. Therefore, 2 γ z dz γ z dz + γ 2 z dz + γ 3 z dz + γ 4 z dz π 2 i + π 2 i + π 2 i + π 2 i 2πi. dz re 9. (#2 pg. 67 Conwy) Let I(r) dz where γ : [, π] C is defined γ by γ(t) re it. Show tht lim r I(r). e iz z 3

14 Solution. Noting tht I(r) i γ π e iz z dz e iγ(t) γ(t) dγ(t) π e ireit re it π e ireit ire it dt dt. it follows tht e ireit e Re (ireit ) ( ) e Re ir(cos t+isint) I(r) e r sin t, π e r sin t dt. Fix ɛ >. Since e r sin t converges uniformly to on [ɛ/3, π ɛ/3] s r, there exists R such tht r > R π ɛ/3 ɛ/3 e r sin t dt < ɛ/3. Hence, if r > R, I(r) π ɛ/3 e r sin t dt e r sin t dt + π ɛ/3 ɛ/3 e r sin t dt + π π ɛ/3 < ɛ/3 mx t ɛ/3 e r sin t sin t + ɛ/3 + ɛ/3 mx e r π ɛ/3 t π ɛ/3 + ɛ/3 + ɛ/3 ɛ. e r sin t dt 4

15 This proves tht I(r) s r. 2. (#2 pg. 68 Conwy) Let γ(t) + e it for t 2π nd find γ (z2 ) dz. Solution. (z 2 ) dz γ i i 2 2π 2π 2π ( + e it ) 2 ieit dt 2 + e + 2 it dt eit dt. Furthermore, by the Weierstrss M-test, the series, + 2 eit n ( ) n 2 n e int converges uniformly on [, 2π]. Thus, using the fct tht 2π e int { 2π if n if n we hve tht γ(z 2 ) dz i 2 i 2 i 2 2π 2π n i 2 2π πi. + 2 ( ) n e int dt 2 n n 2π eit dt ( ) n 2 n e int dt 5

16 2. (#23 pg. 68 Conwy) Let γ be closed rectifible curve in G nd G. Show tht for n 2, γ (z ) n dz. Solution. If n 2 nd G, then the function F defined on G by the formul F (z) (z ) n+ n is primitive for (z ) n. Hence, Corollry.22 (pg. 66 Conwy) implies tht γ (z ) n dz.

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