1
BQBVBZBTBRBY 2005-2006 BYBWBS 1BYBX
〈〈B9?BMA2 〉〉CZCTC5DJDJA9 (A)AFBLARA5
ATA1AAC8A8 (4 × 10 = 40B7)
1,ATB8A1B3AZAXAMCGA4CWDEBCB6AMAQBK
2pi
3 DEB3AZAXB8A1AD 1+i,B9B6B8A1DI
.
BUAUA8AXA0B2 ze2pi3 i = 1 + i,A5AWA2 z = e?2pi3 i(1 + i) =?12 +
√3
2?i
parenleftBigg
1
2 +
√3
2
parenrightBigg
(4B7).
2,BLA1 f(z) = (x2? y2 + ax + by) + i(cxy + 3x + 2y) B8B8CYCRCUAOAOC0AGA0B9
(a,b,c) =,
BU AYADu = x2?y2 + ax + by,v = cxy + 3x + 2y AOAOC7ACA0D2
ux = 2x + a,uy =?2y + b,vx = cy + 3,vy = cx + 2,
COBO C-RABBXA0BI (a,b,c) = (2,2,?3)(4 B7).
3,D8BJC DIBDAMB7BH,|z| = 12,B9
contintegraldisplay
C
ez
z(z? 1)2dz =,
BU BEC2C6AHBRB7BFDGA0B2
contintegraldisplay
C
ez
z(z? 1)2dz =
contintegraldisplay
C
ez
(z? 1)2
z dz = 2pii
bracketleftbigg ez
(z? 1)2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
= 2pii,(4B7)
4,DDC DIBDAMB7BH,|z| = 12,B9BRB7
contintegraldisplay
C
sinz
e? ez dz =,
BUAYADAABRBLA1B8 C CUAOAOC0AGA0BEC2C6AH -BHD9B0CCA0B2
contintegraldisplay
C
sinz
e? ez dz = 0,(4B7)
5,CQBVA1
∞summationdisplay
n=0
(3 + 4i)nznAXA0CEA7C1DIA2,CQBVA1
∞summationdisplay
n=0
parenleftbigg
sin1n
parenrightbigg
zn
AXA0CEA7C1DIA2,
2
BU CQBVA1
∞summationdisplay
n=0
(3 + 4i)nzn AXA0CEA7C1DI
1
5(2 B7),CQBVA1
∞summationdisplay
n=0
parenleftbigg
sin1n
parenrightbigg
zn AXA0CE
A7C1DI 1(2B7).
6,DD C DIAS y = √1?x2 AQ (1,0) AV (0,1) AXATB2B2AMD5AJA0B9
integraldisplay
C
lnzdz =
.
BU AYADAABRBLA1 f(z) = lnz B8AND7 xBIBAA7BIAXD4B5CUC0AGA0BEC2CXB4 - CA
AJCVBMBFDGA0B2
integraldisplay
C
lnzdz = [zlnz?z]
vextendsinglevextendsingle
vextendsinglevextendsingle
z=i
z=1
= ilni?i + 1,(4B7)
7,CNCBBVA1
∞summationdisplay
n=0
1
(z?2)n +
∞summationdisplay
n=0
(z? 2)n
2n AXA0CEB7BQDIA2,
BU A0CEB7BQDIA2 1 < |z? 2| < 2,(4B7)
8,DDf(z) = e
z
z2 + 1,B9Res[f(z),?i] =,
BU AYAD?iDIf(z)AXATBVBSAYA0A5AWA2
Res[f(z),?i] = limz→?i(z + i) e
z
z2 + 1 =
e?ii
2,(4B7)
9,DD z = 0 DI f(z) = z
2
sinz(1? cosz) AX m BVBSAYA0B9 m =,AH
D2Res[f(z),0] =,
BU AYAD z = 0 DI sinz(1? cosz) AXDABVCIAYA0A5AW z = 0 DIf(z) AXATBVBSAYA0
BUm = 1(2B7),AHD2
Res[f(z),0] = limz→0
bracketleftbigg
z · z
2
sinz(1? cosz)
bracketrightbigg
= 2,(2B7)
10,B0DCw = 1 + iz BYB7x2 + y2 < 4B0AKwCYCRDBAX,
BU B8B7BH x2 + y2 = 4 DBARD6DAAYA2?2,2,2i,B8B0DC w = 1 + iz AIA0A6CPAXAL
B7AGDIA2?1 + i2,1 + i2,1?i2,B4DIA0BEC2B7DGAJAPB0DCAXA9B7APA0B0DC w = 1 + iz BY
B7BHx2 + y2 = 4B0AKwCYCRDBAXB7BHA2 u2 + v2 = 12.
3
B3AYAD x2 + y2 < 4 CUAXAY z = 0 B8B0DCAIAXALDIA2 ∞,A5AWB0DC w = 1 + iz BY
B7BHx2 + y2 < 4B0AKwCYCRDBAX u2 + v2 > 12(4B7).
B5A1BWA4AICHBDA8 (4 × 7 = 28B7)
1,BWA4BRB7
contintegraldisplay
C
cosz
z(z2?a2)dz,D0BGA0|a| negationslash= 1,a negationslash= 0,C DIBDAMAS?B7BHA1
C1 AT|a| > 1DEA0AABRBLA1 f(z) =
cosz
z(z2?a2) B8AS?B7CUB2ATBCD1AY z = 0,BE
C2C6AHBRB7BFDGA0AWA2
contintegraldisplay
C
cosz
z(z2?a2)dz = 2pii
bracketleftbigg cosz
z2?a2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
=?2piia2,(3B7)
AT |a| < 1 DEA0AABRBLA1 f(z) = coszz(z2?a2) B8AS?B7CUB2DABCD1AY z = 0,?a,a.
B8C BPDAABBPAIA8BKA0BPAIAKBZAXACD5AJ C1,C2 BNC3,B7AGA8BK 0,?aBNa,BEC2
B8BOACCLB0CCBNC6AHBRB7BFDGA0AWA2
contintegraldisplay
C
cosz
z(z2?a2)dz =
contintegraldisplay
C1
cosz
z(z2?a2)dz +
contintegraldisplay
C2
cosz
z(z2?a2)dz +
contintegraldisplay
C3
cosz
z(z2?a2)dz
= 2pii
bracketleftbigg cosz
z2?a2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
+ 2pii
bracketleftbigg cosz
z(z?a)
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=?a
+ 2pii
bracketleftbigg cosz
z(z + a)
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=a
= 2pi(cosa? 1)ia2,(4B7)
2,D3BLA1f(z) = sin2zz(z + 1)2 B8C9AMB8CYCRCUAXBGCDD1AYAOAXCJA1A1
BU z = 0DIf(z)AXC7D7D1AYA0 z =?1DIf(z)AX 2BVBSAYA0 z = ∞DIf(z)AX
ABAPD1AY (1B7),A5AW
Res[f(z),0] = 0; (2B7)
Res[f(z),?1] = 11! limz→?1 ddz
bracketleftbigg
(z + 1)2 · sin2zz(z + 1)2
bracketrightbigg
= limz→?1 ddz
bracketleftbiggsin2z
z
bracketrightbigg
=?2cos2 + sin2,(2B7)
BEC2f(z)B8A5B2BDBGCDD1AYAOAXCJA1AXBNBNADCIA0AW
Res[f(z),∞] =?Res[f(z),0]? Res[f(z),?1] = 2cos2?sin2,(2B7)
4
3,BWA4BRB7
contintegraldisplay
C
1
zcoszdz,D0BGC DIBDAMB7BHA2 |z| = 2.
BU AABRBLA1
1
zcosz B8|z| < 2CUB2DABCD1AYA2 z = 0BNz = ±
pi
2,A6CPB1DIATBV
BSAY (2B7),A5AW
Res[f(z),0] = 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=0
= 1; (1B7)
Res
bracketleftBig
f(z),pi2
bracketrightBig
= 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=pi2
=?2pi; (1B7)
Res
bracketleftBig
f(z),?pi2
bracketrightBig
= 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=?pi2
=?2pi,(1B7)
A5AW
contintegraldisplay
C
1
zcoszdz = 2pii
braceleftBig
Res[f(z),0] + Res
bracketleftBig
f(z),pi2
bracketrightBig
+ Res
bracketleftBig
f(z),?pi2
bracketrightBigbracerightBig
= 2pii
parenleftbigg
1? 4pi
parenrightbigg
,(2B7)
4,BWA4BRB7
contintegraldisplay
C
z2e?1z dz,D0BGA0C DIBDAMAS?B7BHA1
C1 AABRBLA1f(z) = z2e?1z B80 < |z| < +∞CUC0AGA0A6AXCNCBBAC3DGDIA2
z2e?1z = z2
parenleftbigg
1? 1z + 12! 1z2? 13! 1z3 + ···
parenrightbigg
= z2?z + 12!? 13! 1z + ··· (5B7)
A5AW
contintegraldisplay
C
z2e?1z dz = 2pii· Res[f(z),0] = 2piic?1 =?13pii,(2B7)
DA (10B7),D3BLA1f(z) = 1z2 B8B7BQB5A2 (1) 0 < |z?i| < 1; (2)1 < |z?i| < +∞
CUAXCNCBBVA1BAC3DGA1
BU BJAXAV z = 0 DIBLA1
1
z2 AXD1AYA0BI f(z) B7AGB8BBCFBCB7BQB5CUDIAOAOC0
AGAXA0C7AWBAC3AKCNCBBVA1A1
5
(1)AT 0 < |z?i| < 1DEA0APDEB2
vextendsinglevextendsingle
vextendsinglevextendsinglez?i
i
vextendsinglevextendsingle
vextendsinglevextendsingle< 1
1
z =
1
z?i + i =
1
i ·
1
1 + z?ii
= 1i
∞summationdisplay
n=0
parenleftbigg
z?ii
parenrightbiggn
=?
∞summationdisplay
n=0
in+1(z?i)n
CFADD3AUAW
1
z2 =
∞summationdisplay
n=1
nin+1(z?i)n?1,(5B7)
(2)AT 1 < |z?i| < +∞DEA0APDEB2
vextendsinglevextendsingle
vextendsinglevextendsingle i
z?i
vextendsinglevextendsingle
vextendsinglevextendsingle< 1,
1
z =
1
z?i + i =
1
z?i ·
1
1 + iz?i
= 1z?i
∞summationdisplay
n=0
(?1)nin
(z?i)n =
∞summationdisplay
n=0
(?1)nin
(z?i)n+1
CFADD3AUAW
1
z2 =
∞summationdisplay
n=0
(?1)n(n + 1)in
(z?i)n+2,(5B7)
A3 (9 B7),D3B7DGAJAPB0DC w = f(z),A6BYDBA7CYCR Im(z) > 0 B0DCAKAS?B7
|w| < 1,D2COBOABBX f(i) = 0,argfprime(i) = pi2.
BU A5D3B7DGAJAPB0DCAXATA6AODGAD
w = eiθ
parenleftbiggz?λ
z?λ
parenrightbigg
,(Im(λ) > 0),(2B7)
B1ABBXf(i) = 0BFA0A5D3AXB0DCBYAY z = iB0DCAKAS?B7BHAXB7AN w = 0,A5AW
f(z) = eiθ
parenleftbiggz?i
z + i
parenrightbigg
,(3B7)
AYAD fprime(z) = eiθ 2i(z + i)2,
BIB2 fprime(i) = eiθ(?i2).
A5AW
argfprime(i) = argeiθ + arg(?i2) = θ + (?pi2) = pi2,θ = pi,(3B7)
B4DIA5D3B0DCAD
w =?z?iz + i,(1B7)
6
AF(7B7).D8BJf(z) = x2 + g(y) + iv(x,y)DIC0AGBLA1A0D2 f(0) = fprime(0) = 0,D3DF
BLA1g(y),v(x,y)BTf(z),[A7DHA2BEC2 x2 + g(y)DIAZBNBLA1D3 g(y)]
BU AYADu(x,y) = x2 + g(y)DIAZBNBLA1A0A5AW
uxx + uyy = 2 + gprimeprime(y) = 0,
A5AWgprimeprime(y) =?2,AWg(y) =?y2 + C1y + C2,(3B7)
BEC2 C-RB6ALA0B2 vy = ux = 2x,A5AW
v(x,y) =
integraldisplay
2xdy + h(x) = 2xy + h(x),
B3AYADvx =?uy = 2y?C1,B4DIB2
2y + hprime(x) = 2y?C1,
A5AWhprime(x) =?C1,AWh(x) =?C1x + C3,B4DIA0B2
v(x,y) = 2xy?C1x + C3.
A5AW
f(z) = x2?y2 + C1y + C2 + (2xy?C1x + C3)i (3B7)
BEC2f(0) = 0,AWA2 C2 = C3 = 0,BEC2fprime(0) = 0,AWA2 C1 = 0,A5AW
f(z) = x2?y2 + 2xyi,(1B7)
CK (6B7),AVBFf(z)B80 < |z| < 1CUC0AGA0D2 limz→0zf(z) = 1.
(1)BECSA2 z = 0DIf(z)AXATBVBSAYA3
(2)D3Res[f(z),0],[A7DHA2C4CM zf(z)B80 < |z| < 1CUAXCNCBBVA1]
C0 AYAD zf(z) B8 0 < |z| < 1 CUC0AGA0AHD2 limz→0zf(z) = 1,A5AW zf(z) B8BBB7BQ
B5CUAXCNCBBVA1DI
zf(z) = 1 + c1z + c2z2 + c3z3 + ···,(3B7)
B4DI
f(z) = 1z + c1 + c2z + c3z2 + ···,(1B7)
A5AWz = 0DIf(z)AXATBVBSAY (1B7),D2Res[f(z),0] = 1(1B7).
BQBVBZBTBRBY 2005-2006 BYBWBS 1BYBX
〈〈B9?BMA2 〉〉CZCTC5DJDJA9 (A)AFBLARA5
ATA1AAC8A8 (4 × 10 = 40B7)
1,ATB8A1B3AZAXAMCGA4CWDEBCB6AMAQBK
2pi
3 DEB3AZAXB8A1AD 1+i,B9B6B8A1DI
.
BUAUA8AXA0B2 ze2pi3 i = 1 + i,A5AWA2 z = e?2pi3 i(1 + i) =?12 +
√3
2?i
parenleftBigg
1
2 +
√3
2
parenrightBigg
(4B7).
2,BLA1 f(z) = (x2? y2 + ax + by) + i(cxy + 3x + 2y) B8B8CYCRCUAOAOC0AGA0B9
(a,b,c) =,
BU AYADu = x2?y2 + ax + by,v = cxy + 3x + 2y AOAOC7ACA0D2
ux = 2x + a,uy =?2y + b,vx = cy + 3,vy = cx + 2,
COBO C-RABBXA0BI (a,b,c) = (2,2,?3)(4 B7).
3,D8BJC DIBDAMB7BH,|z| = 12,B9
contintegraldisplay
C
ez
z(z? 1)2dz =,
BU BEC2C6AHBRB7BFDGA0B2
contintegraldisplay
C
ez
z(z? 1)2dz =
contintegraldisplay
C
ez
(z? 1)2
z dz = 2pii
bracketleftbigg ez
(z? 1)2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
= 2pii,(4B7)
4,DDC DIBDAMB7BH,|z| = 12,B9BRB7
contintegraldisplay
C
sinz
e? ez dz =,
BUAYADAABRBLA1B8 C CUAOAOC0AGA0BEC2C6AH -BHD9B0CCA0B2
contintegraldisplay
C
sinz
e? ez dz = 0,(4B7)
5,CQBVA1
∞summationdisplay
n=0
(3 + 4i)nznAXA0CEA7C1DIA2,CQBVA1
∞summationdisplay
n=0
parenleftbigg
sin1n
parenrightbigg
zn
AXA0CEA7C1DIA2,
2
BU CQBVA1
∞summationdisplay
n=0
(3 + 4i)nzn AXA0CEA7C1DI
1
5(2 B7),CQBVA1
∞summationdisplay
n=0
parenleftbigg
sin1n
parenrightbigg
zn AXA0CE
A7C1DI 1(2B7).
6,DD C DIAS y = √1?x2 AQ (1,0) AV (0,1) AXATB2B2AMD5AJA0B9
integraldisplay
C
lnzdz =
.
BU AYADAABRBLA1 f(z) = lnz B8AND7 xBIBAA7BIAXD4B5CUC0AGA0BEC2CXB4 - CA
AJCVBMBFDGA0B2
integraldisplay
C
lnzdz = [zlnz?z]
vextendsinglevextendsingle
vextendsinglevextendsingle
z=i
z=1
= ilni?i + 1,(4B7)
7,CNCBBVA1
∞summationdisplay
n=0
1
(z?2)n +
∞summationdisplay
n=0
(z? 2)n
2n AXA0CEB7BQDIA2,
BU A0CEB7BQDIA2 1 < |z? 2| < 2,(4B7)
8,DDf(z) = e
z
z2 + 1,B9Res[f(z),?i] =,
BU AYAD?iDIf(z)AXATBVBSAYA0A5AWA2
Res[f(z),?i] = limz→?i(z + i) e
z
z2 + 1 =
e?ii
2,(4B7)
9,DD z = 0 DI f(z) = z
2
sinz(1? cosz) AX m BVBSAYA0B9 m =,AH
D2Res[f(z),0] =,
BU AYAD z = 0 DI sinz(1? cosz) AXDABVCIAYA0A5AW z = 0 DIf(z) AXATBVBSAYA0
BUm = 1(2B7),AHD2
Res[f(z),0] = limz→0
bracketleftbigg
z · z
2
sinz(1? cosz)
bracketrightbigg
= 2,(2B7)
10,B0DCw = 1 + iz BYB7x2 + y2 < 4B0AKwCYCRDBAX,
BU B8B7BH x2 + y2 = 4 DBARD6DAAYA2?2,2,2i,B8B0DC w = 1 + iz AIA0A6CPAXAL
B7AGDIA2?1 + i2,1 + i2,1?i2,B4DIA0BEC2B7DGAJAPB0DCAXA9B7APA0B0DC w = 1 + iz BY
B7BHx2 + y2 = 4B0AKwCYCRDBAXB7BHA2 u2 + v2 = 12.
3
B3AYAD x2 + y2 < 4 CUAXAY z = 0 B8B0DCAIAXALDIA2 ∞,A5AWB0DC w = 1 + iz BY
B7BHx2 + y2 < 4B0AKwCYCRDBAX u2 + v2 > 12(4B7).
B5A1BWA4AICHBDA8 (4 × 7 = 28B7)
1,BWA4BRB7
contintegraldisplay
C
cosz
z(z2?a2)dz,D0BGA0|a| negationslash= 1,a negationslash= 0,C DIBDAMAS?B7BHA1
C1 AT|a| > 1DEA0AABRBLA1 f(z) =
cosz
z(z2?a2) B8AS?B7CUB2ATBCD1AY z = 0,BE
C2C6AHBRB7BFDGA0AWA2
contintegraldisplay
C
cosz
z(z2?a2)dz = 2pii
bracketleftbigg cosz
z2?a2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
=?2piia2,(3B7)
AT |a| < 1 DEA0AABRBLA1 f(z) = coszz(z2?a2) B8AS?B7CUB2DABCD1AY z = 0,?a,a.
B8C BPDAABBPAIA8BKA0BPAIAKBZAXACD5AJ C1,C2 BNC3,B7AGA8BK 0,?aBNa,BEC2
B8BOACCLB0CCBNC6AHBRB7BFDGA0AWA2
contintegraldisplay
C
cosz
z(z2?a2)dz =
contintegraldisplay
C1
cosz
z(z2?a2)dz +
contintegraldisplay
C2
cosz
z(z2?a2)dz +
contintegraldisplay
C3
cosz
z(z2?a2)dz
= 2pii
bracketleftbigg cosz
z2?a2
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=0
+ 2pii
bracketleftbigg cosz
z(z?a)
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=?a
+ 2pii
bracketleftbigg cosz
z(z + a)
bracketrightbiggvextendsinglevextendsingle
vextendsinglevextendsingle
z=a
= 2pi(cosa? 1)ia2,(4B7)
2,D3BLA1f(z) = sin2zz(z + 1)2 B8C9AMB8CYCRCUAXBGCDD1AYAOAXCJA1A1
BU z = 0DIf(z)AXC7D7D1AYA0 z =?1DIf(z)AX 2BVBSAYA0 z = ∞DIf(z)AX
ABAPD1AY (1B7),A5AW
Res[f(z),0] = 0; (2B7)
Res[f(z),?1] = 11! limz→?1 ddz
bracketleftbigg
(z + 1)2 · sin2zz(z + 1)2
bracketrightbigg
= limz→?1 ddz
bracketleftbiggsin2z
z
bracketrightbigg
=?2cos2 + sin2,(2B7)
BEC2f(z)B8A5B2BDBGCDD1AYAOAXCJA1AXBNBNADCIA0AW
Res[f(z),∞] =?Res[f(z),0]? Res[f(z),?1] = 2cos2?sin2,(2B7)
4
3,BWA4BRB7
contintegraldisplay
C
1
zcoszdz,D0BGC DIBDAMB7BHA2 |z| = 2.
BU AABRBLA1
1
zcosz B8|z| < 2CUB2DABCD1AYA2 z = 0BNz = ±
pi
2,A6CPB1DIATBV
BSAY (2B7),A5AW
Res[f(z),0] = 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=0
= 1; (1B7)
Res
bracketleftBig
f(z),pi2
bracketrightBig
= 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=pi2
=?2pi; (1B7)
Res
bracketleftBig
f(z),?pi2
bracketrightBig
= 1(zcosz)prime
vextendsinglevextendsingle
vextendsinglevextendsingle
z=?pi2
=?2pi,(1B7)
A5AW
contintegraldisplay
C
1
zcoszdz = 2pii
braceleftBig
Res[f(z),0] + Res
bracketleftBig
f(z),pi2
bracketrightBig
+ Res
bracketleftBig
f(z),?pi2
bracketrightBigbracerightBig
= 2pii
parenleftbigg
1? 4pi
parenrightbigg
,(2B7)
4,BWA4BRB7
contintegraldisplay
C
z2e?1z dz,D0BGA0C DIBDAMAS?B7BHA1
C1 AABRBLA1f(z) = z2e?1z B80 < |z| < +∞CUC0AGA0A6AXCNCBBAC3DGDIA2
z2e?1z = z2
parenleftbigg
1? 1z + 12! 1z2? 13! 1z3 + ···
parenrightbigg
= z2?z + 12!? 13! 1z + ··· (5B7)
A5AW
contintegraldisplay
C
z2e?1z dz = 2pii· Res[f(z),0] = 2piic?1 =?13pii,(2B7)
DA (10B7),D3BLA1f(z) = 1z2 B8B7BQB5A2 (1) 0 < |z?i| < 1; (2)1 < |z?i| < +∞
CUAXCNCBBVA1BAC3DGA1
BU BJAXAV z = 0 DIBLA1
1
z2 AXD1AYA0BI f(z) B7AGB8BBCFBCB7BQB5CUDIAOAOC0
AGAXA0C7AWBAC3AKCNCBBVA1A1
5
(1)AT 0 < |z?i| < 1DEA0APDEB2
vextendsinglevextendsingle
vextendsinglevextendsinglez?i
i
vextendsinglevextendsingle
vextendsinglevextendsingle< 1
1
z =
1
z?i + i =
1
i ·
1
1 + z?ii
= 1i
∞summationdisplay
n=0
parenleftbigg
z?ii
parenrightbiggn
=?
∞summationdisplay
n=0
in+1(z?i)n
CFADD3AUAW
1
z2 =
∞summationdisplay
n=1
nin+1(z?i)n?1,(5B7)
(2)AT 1 < |z?i| < +∞DEA0APDEB2
vextendsinglevextendsingle
vextendsinglevextendsingle i
z?i
vextendsinglevextendsingle
vextendsinglevextendsingle< 1,
1
z =
1
z?i + i =
1
z?i ·
1
1 + iz?i
= 1z?i
∞summationdisplay
n=0
(?1)nin
(z?i)n =
∞summationdisplay
n=0
(?1)nin
(z?i)n+1
CFADD3AUAW
1
z2 =
∞summationdisplay
n=0
(?1)n(n + 1)in
(z?i)n+2,(5B7)
A3 (9 B7),D3B7DGAJAPB0DC w = f(z),A6BYDBA7CYCR Im(z) > 0 B0DCAKAS?B7
|w| < 1,D2COBOABBX f(i) = 0,argfprime(i) = pi2.
BU A5D3B7DGAJAPB0DCAXATA6AODGAD
w = eiθ
parenleftbiggz?λ
z?λ
parenrightbigg
,(Im(λ) > 0),(2B7)
B1ABBXf(i) = 0BFA0A5D3AXB0DCBYAY z = iB0DCAKAS?B7BHAXB7AN w = 0,A5AW
f(z) = eiθ
parenleftbiggz?i
z + i
parenrightbigg
,(3B7)
AYAD fprime(z) = eiθ 2i(z + i)2,
BIB2 fprime(i) = eiθ(?i2).
A5AW
argfprime(i) = argeiθ + arg(?i2) = θ + (?pi2) = pi2,θ = pi,(3B7)
B4DIA5D3B0DCAD
w =?z?iz + i,(1B7)
6
AF(7B7).D8BJf(z) = x2 + g(y) + iv(x,y)DIC0AGBLA1A0D2 f(0) = fprime(0) = 0,D3DF
BLA1g(y),v(x,y)BTf(z),[A7DHA2BEC2 x2 + g(y)DIAZBNBLA1D3 g(y)]
BU AYADu(x,y) = x2 + g(y)DIAZBNBLA1A0A5AW
uxx + uyy = 2 + gprimeprime(y) = 0,
A5AWgprimeprime(y) =?2,AWg(y) =?y2 + C1y + C2,(3B7)
BEC2 C-RB6ALA0B2 vy = ux = 2x,A5AW
v(x,y) =
integraldisplay
2xdy + h(x) = 2xy + h(x),
B3AYADvx =?uy = 2y?C1,B4DIB2
2y + hprime(x) = 2y?C1,
A5AWhprime(x) =?C1,AWh(x) =?C1x + C3,B4DIA0B2
v(x,y) = 2xy?C1x + C3.
A5AW
f(z) = x2?y2 + C1y + C2 + (2xy?C1x + C3)i (3B7)
BEC2f(0) = 0,AWA2 C2 = C3 = 0,BEC2fprime(0) = 0,AWA2 C1 = 0,A5AW
f(z) = x2?y2 + 2xyi,(1B7)
CK (6B7),AVBFf(z)B80 < |z| < 1CUC0AGA0D2 limz→0zf(z) = 1.
(1)BECSA2 z = 0DIf(z)AXATBVBSAYA3
(2)D3Res[f(z),0],[A7DHA2C4CM zf(z)B80 < |z| < 1CUAXCNCBBVA1]
C0 AYAD zf(z) B8 0 < |z| < 1 CUC0AGA0AHD2 limz→0zf(z) = 1,A5AW zf(z) B8BBB7BQ
B5CUAXCNCBBVA1DI
zf(z) = 1 + c1z + c2z2 + c3z3 + ···,(3B7)
B4DI
f(z) = 1z + c1 + c2z + c3z2 + ···,(1B7)
A5AWz = 0DIf(z)AXATBVBSAY (1B7),D2Res[f(z),0] = 1(1B7).
