Solutions for Math 411 Assignment #8 1

Transcription

1 Solutios for Math Assigmet #8 A8. Fid the Lauret series of f() i (a) { < }; (b) { < < }; (c) { < 2 < 3}; (d) {0 < + < 2}. Solutio. We write f() as a sum of partial fractios: f() ( ) 2 ( + ) c + c 2 ( ) + c ( ) + 2( ) 2 ( + ) where c i are computed by c ( ( ) 2 f() ) c 2 ( ( ) 2 f() ) 2 For <, f() c 3 (( + )f()) ( ) + 2( ) 2 ( + ) ( ) + 2 ( ) 2 ( + ) + + ( ) ( ) xiche/math7f/hw8sol.pdf

2 2 Here we apply the geeralied biomial theorem ( + ) r r(r )...(r + ) r! 0 0 r( r + )...( r + ) ( )! to obtai ( ) ( ) ( r + ) 2 + ( ) ( ) + 0 for <. For < <, f() ( ) + 2( ) 2 ( + ) + 2 ( 2 ) 2 ( + ) + + ( ) ( ) + ( ) 2 2 (2 + ( ) ).

3 3 For < 2 < 3, f() ( ) + 2( ) 2 ( + ) ( 2) (( 2) + ) 2 ( 2) + 3 ( ) + ( 2) + ( 2) 2( 2) ( + ( 2 2) ) 2 2 ( + 3 ( 2)) ( ) ( 2) + + ( ) ( 2) ( 2) 2( 2) ( ) 3 ( 2) 2 0 ( ) ( 2) + ( ) ( )( 2) 2 2 ( ) 3 ( 2) 2 (2 3) ( ) ( 2) 2 ( ) 3 ( 2).

4 For 0 < + < 2, f() ( ) + 2( ) 2 ( + ) ( + ) (( + ) 2) 2 ( + ) ( 2 ( + ) ) ( 2 ( + )) 2 2 ( + ) ( + ) ( + ) ( + ) 0 ( + ) ( + ) A8.2 For each of the followig aalytic fuctios o {0 < < }, determie the type of its sigularity ad pricipal part at 0: cos (a) 2 ( ; ) + (b) cos. Solutio. (a) Sice cos 2 cos ( ) cos cos 0 cos ( ) ( ) 2 (2)! ( ) 2 (2)! where cos /( ) is aalytic at 0, we coclude that it has a pole of order at 0 with pricipal part /.

5 (b) Sice ( + cos cos + ) cos() cos si() si ( ) 2 ( ) 2 cos() si() (2)! (2 + )! 0 0 ( ) 2 ( ) 2 cos() + cos() si() (2)! (2 + )! we coclude that it has a essetial sigularity at 0 with pricipal part ( ) 2 ( ) 2 cos() si(). (2)! (2 + )! Alteratively, usig Euler formula, we obtai ( + cos cos + ) ( ( exp i + i ) ( + exp i i )) 2 ( ei i 2 exp + e i 2 exp i ) ei i + e i ( i) 2! 2! 0 0 i e i + ( i) e i ei + e i i e i + ( i) e i + 2(!) 2 2(!) So it has a essetial sigularity at 0 with pricipal part i e i + ( i) e i. 2(!) 5

6 6 Note that ( i e i + ( i) e i ) 2(!) i 2m e i + ( i) 2m e i 2m + 2((2m)!) 2((2m + )!) m m0 e ( ) m i + e i ( ie 2m + ( ) m i ie i 2((2m)!) 2((2m + )!) m m0 ( ) 2 ( ) 2 cos() si(). (2)! (2 + )! 0 0 i 2m+ e i + ( i) 2m+ e i 2m ) 2m A8.3 Let f() ad g() be two aalytic fuctios o {0 < < }. For each of the followig statemets, determie whether it is true or false. If it is true, prove it. If it is false, give a couterexample. (a) If both f() ad g() have poles at 0, the f()g() has a pole at 0. (b) If both f() ad g() have essetial sigularities at 0, the f()g() has a essetial sigularity at 0. (c) If f()g() has a pole at 0, the at least oe of f() ad g() has a pole at 0. (d) If f()g() has a essetial sigularity at 0, the at least oe of f() ad g() has a essetial sigularity at 0. Proof. (a) True. Suppose that f() m p() ad g() q() for some positive itegers m ad ad aalytic fuctios p() ad q() o { < } satisfyig that p(0) 0 ad q(0) 0. The f()g() p()q() m+ has a pole of order m + at 0 sice p()q() is aalytic o { < } ad p(0)q(0) 0. (b) False.

7 Let f() exp( ) ad g() exp( ). The ( ) f() (!) + g() 0 0 (!) i C ad hece both f() ad g() have essetial sigularities at 0. But f()g() has a pole at 0. (c) False. Use the same couter-example for (b). (d) True. Suppose that both f() ad g() have removable sigularities or poles at 0. The f() m p() ad g() q() for some o-egative itegers m ad ad aalytic fuctios p() ad q() o { < }. Thus f()g() p()q() m+ has a removable sigularity or pole at 0 sice p()q() is aalytic o { < }. Cotradictio. So at least oe of f() ad g() has a essetial sigularity at 0. A8. Louville s theorem says that there are o ocostat bouded etire fuctios. Prove the same for aalytic fuctios o C : If f() is aalytic o C ad f() M for some M 0 ad all 0, the f() is costat. Proof. Sice f() M for 0 < <, f() has a removable sigularity at 0 by Riema Extesio Theorem. That is, f() ca be exteded to a aalytic fuctio o C, i.e., a etire fuctio. By cotiuity, f() M for all C. So f() is a bouded etire fuctio. The f() is costat by Louville s theorem. A8.5 Let D { < }, D {0 < < } ad f : D D be a biholomorphic map. Do the followig: (a) Show that f exteds to a aalytic fuctio o D ad D f(d) D. (b) (Bous +0 poits) Show that f(0) 0. (c) (Bous +0 poits) Show that f() c for some costat c satisfyig c. 7

8 8 Proof. Sice f(d ) D, f() < for all 0 < <. By Riema Extesio Theorem, f() has a removable sigularity at 0 ad f ca be exteded to a aalytic fuctio o D. By cotiuity, f(0) D { }. We claim that f(0) <. Otherwise, f(0) ad f(d) f(d ) {f(0)} D {f(0)} fails to be ope sice { f(0) < r} D {f(0)} for all r > 0; this cotradicts Ope Mappig Theorem. Therefore, f(0) < ad f(d) f(d ) {f(0)} D {f(0)} D O the other had, sice f : D D is a biholomorphic map, D f(d ) f(d). Therefore, D f(d) D. Let g f be the iverse of f : D D. Applyig the same argumet to g, we coclude that g exteds to a aalytic fuctio o D satisfyig that g(d) D. Let us cosider f g : D D. It is a aalytic fuctio o D satisfyig that f g() f f () for all 0. By cotiuity, f g() for all D. Similarly, we ca prove that g f() for all D. That is, f ad g are iverse to each other o D. I particular, f : D D is biholomorphic. The f(0) 0 sice f(d ) D ad f is -. Sice f : D D is biholomorphic ad f(0) 0, f() c for some c by Schwart Lemma.