复变函数与积分变换chapter5answer

1、08?09 ? 1 11. f (z) = , KResf (z), 0 =.ez1zo#?c?K?Y08?09 ? 1 11. f (z) =) , KResf (z), 0 = .ez1z11lim f (z)z0=limzz0 e 1zz (ez 1)=limz0z(e 1)z1 ez=limz0 e 1 + e zzzez1=lim= .zzzz0 e + e + e z2 1 1?,z = 0 ?f (z) =? ?:,?dez1zResf (z), 0 = 0.o#?c?K?Y08?09 ? 1 11. f (z) =) , KResf (z), 0 = 0.ez1z11lim f

2、 (z)z0=limzz0 e 1zz (ez 1)=limz0z(e 1)z1 ez=limz0 e 1 + e zzzez1=lim= .zzzz0 e + e + e z2 1 1?,z = 0 ?f (z) =? ?:,?dez1zResf (z), 0 = 0.o#?c?K?YZz312. O?:e z dz, ? C ? ? ? |z| = 2. 1 + zCo#?c?K?YZz312. O?:e z dz, ? C ? ? ? |z| = 2. 1 + zCz31):f (z) =e3C : |z| = 2 S?k?:z = 1 ? z1+z5?:z = 0.z31Resf (z

3、), 1 =lim (1 + z)z1e z1 + z131lim z= e.e zz1o#?c?K?Y+XX z1 + z33n nn n+3=z(1) z =(1) zn=0n=0z3 z4 + z5 · · · + (1)nzn+3 + · · · .=+X1 111 11e z= 1 +z+2 z2+ · · · .nn! zn=0111Resf (z), 04! 5! + 6! · · ·= 1(1)23(1)1=e1 +1!2!3!11=e.3Zz31e z dz

4、1 + z=2i (Resf (z), 1 + Resf (z), 0)C 1211=2i e+ e= i.33o#?c?K?YZ215 + 3 cos x3. O?:dx.0o#?c?K?YZ213. O?:dx.5 + 3 cos x0)-z = eix, Kz2 + 1dzdx =, cos x =.iz2zZII I21dx5 + 3 cos x 1dz=5 + 3 z2+1iz0|z|=12z 2idz=|z|=1 3z2 + 10z + 3 2idz,=|z|=1 (3z + 1)(z + 3)o#?c?K?Y 2i13|z| = 1 S?k?:z = .f (z) =(3z+1)

5、(z+3)312i1i4Resf (z), =lim (z +)=33 (3z + 1)(z + 3)13zZ21dx5 + 3 cos x131=2iResf (z), 3 2i( 4 ) = 2 .0i=o#?c?K?Y08?09 S1. O?: I11esin dz, ?C ? ? ?|z| = 1.zzCo#?c?K?Y08?09 S1. O?: I11esin dz, ?C ? ? ?|z| = 1.zzC1):f (z) = e z sin 1 3C : |z| = 1 S?k 5?:z = 0.z+X1 111n n! znez=(1)= 1 + · · &#

6、183; ,zn=0+X 1 1sin111 1n=(1)=+ · · · ,(2n + 1)! z2n+1z3! z3zn=01 · 1 = 1,2iResf (z), 0 = 2i.Resf (z), 0=I11esin dz=zzCo#?c?K?YZ+x21 + x4 dx.2. O?:o#?c?K?YZ+x21 + x4 dx.2. O?:):1 + z4z4z=01i 2k+1e, k = 0, 1, 2, 3.4z21+z4?, f (z) =3? ?:?ii 3e4 ,e4 .o#?c?K?Y1i Resf (z), e4 =i 4e41i

7、 3Resf (z), e4 =i 34e4Z+x21 + x4 dxi i 32i(Resf (z), e 4 + Resf (z), e 4 )=22222112i( ( i) +( i) =.44=22222o#?c?K?Y09?10 ?ez1 4:, ?1. z = 0 ?f (z) =?Hz2010ez1 dz =.|z|=1 z2010o#?c?K?Y09?10 ?ez1 4:, ?1. z = 0 ?f (z) =Hz2010ez1 dz =.?|z|=1 z2010ez1 ?fez 1 ? ":,?1z2010) z = 0 ?f (z) =z2010 2010 &q

8、uot;:,?f (z) 2009 4:.ez 11zf (z)=z2010z2010 (e= 1)+X n1z=z20101n!n=11111+ · · · + 2009! z + 2010! + 2011! z + · · · .=z20091Resf (z), 0=2009!Iez 1 2i.2009!dz=2iResf (z), 0 =z2010|z|=1o#?c?K?Y09?10 ?ez11. z = 0 ?f (z) = 2009 4:, ?Hz2010ez1 dz =.?|z|=1 z2010ez1 ?fez 1 ? &

9、quot;:,?1z2010) z = 0 ?f (z) =z2010 2010 ":,?f (z) 2009 4:.ez 11zf (z)=z2010z2010 (e= 1)+X n1z=z20101n!n=11111+ · · · + 2009! z + 2010! + 2011! z + · · · .=z20091Resf (z), 0=2009!Iez 1 2i.2009!dz=2iResf (z), 0 =z2010|z|=1o#?c?K?Y09?10 ?ez11. z = 0 ?f (z) = 2009 4:,

10、 ?Hz2010ez1 dz = 2i?. 2009!|z|=1 z2010ez1 ?fez 1 ? ":,?1z2010) z = 0 ?f (z) =z2010 2010 ":,?f (z) 2009 4:.ez 11zf (z)=z2010z2010 (e= 1)+X n1z=z20101n!n=11111+ · · · + 2009! z + 2010! + 2011! z + · · · .=z20091Resf (z), 0=2009!Iez 1 2i.2009!dz=2iResf (z), 0 =z2

11、010|z|=1o#?c?K?Y2. O?: I sin z dz, ?C = C + C ,12z(z 1)2C1C1 : |z| = (_? ?), C : |z| = 4(? ?).22sin z21): f (z) =3C1 : |z| =S?k? ?2z(z1):z = 0,3C2 : |z| = 4 S?k? ?:z = 0 ? 4:z = 1.o#?c?K?YResf (z), 0=0, 1 d sin z2Resf (z), 1=lim(z 1)1! z1 dzz(z 1)2z cos z sin zlim= cos 1 sin 1.z2z1III sin zdzz(z 1)2

12、 sin zdzsin z=dz z(z 1)2z(z 1)2CCC12=2iResf (z), 0 2i(Resf (z), 0 + Resf (z), 1)2iResf (z), 1 = 2(sin 1 cos 1)i.o#?c?K?Y3. O?: I 1 1e dz, ?C ? ?|z| = 2.z1 + zC1 1 ) f (z) =3 ?*?E?k ?:? 41+z e z:z = 1, 5?:z = 0,z = ,?z = 1 ?z = 0 ?3C : |z| = 2 S?.I 1 1e dz2i(Resf (z), 1 + Resf (z), 0)2iResf (z), .=z1

13、 + zC=o#?c?K?Y11Resf (z), =Resf ( z ) · z2 , 0 1z=Rese , 0z2 + z 1z= lim ze = 1.z2 + zz0I 1 1e dz = 2i(1) = 2i.z1 + zCo#?c?K?Y4. |3?O?Z+ dx x4 + 1 .0)1 + z4z4z= 0= 1i 2k+1= e, k = 0, 1, 2, 3.4?, f (z) =143? ?:?1+zii 3e4 ,e4 .o#?c?K?Yi e4i Resf (z), e4 =124i 3ei 34Resf (z), e4 =4ZZ+ 1+ 11 + x4 d

14、x1 + x4 dx=0i i 3=i(Resf (z), e 4 + Resf (z), e 4 ) !122122i (+ i) (+ i=)42242224=.o#?c?K?Y09?10 SZez1. O?:dz, ?C : |z| = 2, ?.2z(z 1)Cez): f (z) =4:z = 1.3C : |z| = 2 S?k?4:z = 0 ? z(z1)2Zezdzz(z 1)2=2i (Resf (z), 0 + Resf (z), 1)C ! 0ze =2i1 + zz=1=2i (1 + 0)2i.o#?c?K?Y2. O?: Z (1 + z2) sin 1 dz,

15、?C ? ? ?|z| = 1.zC): f (z) = (1 + z2) sin 1 3C : |z| = 1 S?k 5?:z = 0.z+X 1sin11n(1)=(2n + 1)! z2n+1zn=011 11 1=z 3! z3 + 5! z5+ · · · .15Resf (z), 0=1 3! = 6 .Z(1 + z2) sin 1 dz5=2iResf (z), 0 =i.z3Co#?c?K?YZ+ x sin x3. |3?O?I =x2 + 1 dx.):ZZ+ x sin x+ xeixI =x2 + 1 dx = Imx2 + 1 dx.

16、z2 + 1 = 0z= ±i.3? ?:?z = i?4:?.zeiz z2+1f (z) =o#?c?K?Yzeizlim(z i)Resf (z), i=2z + 1zi12e=.Z+ xeixx2 + 1 dx=2iResf (z), ie=iI=Im i =.eeo#?c?K?Y10?11 ?sin z1. :z = 0 ?E?f (z) = 3 4:. ?×?z3sin z?fsin z 1 ":,?1z3) :z = 0 ?E?f (z) =z3sin z z3 3 ":.? :z = 0 ?E?f (z) = 3 1 = 2 4:.o#?

17、c?K?Y2. |3?O?Z+1x2 + 2x + 2 dx.)z2 + 2z + 2=0z1 ± i.Z+1 1x2 + 2x + 2 dx=2iRes, 1 + iz2 + 2z + 21=2i= .2io#?c?K?Y10?11 S1cos z 1 4:,?Res 1cos z , 0 =1 .1. z = 0 ?f (z) =z3z3 21cos z) z = 0 ?f (z) =?f1 cos z 2 ":,?1z3z31cos z 3 ":.?z = 0 ?f (z) = 3 2 = 1 4:.z31 cos zRes, 01 cos z1 cos z

18、 lim z= lim=z3z3z2z0limz0z0sin z=2z cos z1=lim=.z022o#?c?K?YI(z + 1)2 cos 1 dz.2. O?:z|z|=1(_):+X 1cos11 n=(1)(2n)! z2nzn=01 11 11 2! z2 + 4! z4+ · · · .=Res(z + 1)2 cos 1 , 00 1 + 0 = 1.=zI(z + 1)2 cos 1 dz12=2iRes(z + 1) cos , 0 = 2i.zz|z|=1(_)o#?c?K?YZ+ x sin ax+3. O?:):(a, b R ).x

19、2 + b2 dx,0ZZ+ x sin ax1+ x sin axx2 + b2 dx=x2 + b2 dx20Z+ xeiax1=Im2x2 + b2 dx.o#?c?K?YZ+ xeiaxzeiaz2iRes, bi z2 + b2x2 + b2 dx=bieab=2i2bi=eab i.1Imi2eab2eab .Z+ x sin axx2 + b2 dx=0=o#?c?K?Y11?12 ?1. O?: Z sin z dz, ?C : |z| = 2, ?.z(z 1)2C):Z sin zdzz(z 1)2 sin zsin z=2i Res, 0 + Res, 1z(z 1)2z

20、(z 1)2C ! 0sin z =2i0 + zz=12(cos 1 sin 1)i.=o#?c?K?Y2. O?: Z11esin dz, ?C ? ? ?|z| = 1.zzC):+X1 11n n! znez(1)=n=011 1=1 z+2! z2+ · · · .11+X 1sinn=(1)(2n + 1)! z2n+1zn=011 1z 3! z3+ · · · .=Z1111esin dz=2iRese z sin , 0zzzC=2i(1 · 1) = 2i.o#?c?K?Y3. |3?O?Z+ dx x4

21、 + 1 .0):ZZ+ dx x4 + 11+ dx =x4 + 1200z4 + 1 =z=i 2k+1 e, k = 0, 1, 2, 3.4?3? ? i3 iz0 = e 4 , z1 = e 4 .o#?c?K?YZ + dx 1 i 13i=2i Res, e 4 + Res, e 4 z4 + 1z4 + 1x4 + 1 !43 4iiee=2i442=.2Z+dx12=x4 + 12202=.4o#?c?K?Y11?12 SR sin z dz = 0.1. ? ?C : |z| = 1, KC ez1) duz = 0 ?sin z 1 ":,?ez 1 1 &q

22、uot;:,? sin z ? ?:,?dez1Z sin z ez 1 sin z2iRes, 0ez 12i · 0 = 0.dz=C=o#?c?K?YZez2. O?:dz, ?C :z2(z 1)|z| = 2, ?.C) Zezdzz2(z 1)ezez=2i Res, 0 + Res, 1 z2(z 1)z2(z 1)C ! 0ezez =2i+ lim(z 1) 2z 1z=0z (z 1)z1=2i(2 + e)2(e 2)i.o#?c?K?Y3. O?: Z (1 + z2) sin 1 dz, ?C ? ? ?|z| = 1.zC) f (z) = (1 + z2

23、) sin 1 3C : |z| = 1 S?k 5?:z = 0.z+X 1sin11n(1)=(2n + 1)! z2n+1zn=011 11 1=z 3! z3 + 5! z5+ · · · .15Resf (z), 0=1 3! = 6 .Z(1 + z2) sin 1 dz5=2iResf (z), 0 =i.z3Co#?c?K?Y4. |3?O?Z+ x sin xI =x2 + 1 dx.):ZZ+ x sin x+xeixx2 + 1 dx.I =x2 + 1 dx = Imz2 + 1 = 0z= ±i.3? ?:?z = i?4:?.

24、zeiz z2+1f (z) =o#?c?K?Yzeizlim(z i)Resf (z), i=2z + 1zi12e=.Z+ xeixx2 + 1 dx=2iResf (z), ie=iI=Im i =.eeo#?c?K?Y12?13 ?2 2 41. ?f (z) = (z 2) sin3: 3?z = 2 .z23)+X 2 sinz 2 2n+112n=(1)(2n + 1)! (z 2)2n+1n=0+X22 (z 2) sin 2n+1122n=(z 2)(1)z 2(2n + 1)! (z 2)2n+1n=0+X2n+1 21n(1)=(2n + 1)! (z 2)2n1n=0232511=2(z 2) +3! z