AUB0DD BKCRBECH
§2.1 BJCQBDCGASB7C2
1,B5ADBDCGASARCGDACOB4
1)AXB7B0B5CL
AWD5AXBYB7w = f(z)B5CLCUAJCX D,z0 BQDDHB0CHB4A0B4 z0 +trianglez
APDB0BFBOA1ASBXC5C0
lim
trianglez→0
f(z0 +trianglez)?f(z0)
trianglez
ATD1A0A2DDAJ f(z)DCz0 BNAR,D4BPC5C0DDAJBQ f(z)DCz0 ASARCG,C9A9
fprime(z0) = dwdz
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglez=z
0
= lim
trianglez→0
f(z0 +trianglez)?f(z0)
trianglez,
ASBXf(z)D1AJCX DA3ARARCUAXA0CPBA f(z)DCDC1BNARA1
BQ 1 BRf(z) = x + 2yiB2BICUAXA4
BJ CMBQ
lim
trianglez→0
f(z +trianglez)?f(z)
trianglez
= lim
trianglez→0
(x +trianglex) + 2(y +triangley)i?x?2yi
trianglez
= lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi
AWz +trianglez CEA3A8C9CU xA0B0DCC1AIC4CU z,ASAYtriangley = 0,C5C0
lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi = limtrianglez→0
trianglex
trianglex = 1.
AWz +trianglez CEA3A8C9CU y A0B0DCC1AIC4CU z,ASAYtrianglex = 0,C5C0
lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi = limtrianglez→0
2triangleyi
triangleyi = 2.
BECIf(z)B0AXB7?ATD1A1
1
2 B0B1BB B5B9B3B7
2) CUAXCWD1CB
CYA2A2D1 z0 CUAXB0BYB7A9B5D1 z0 D1CBA1BEDB?AKD0A1
DG,BRCQAXB7B0B5CLA0?ε > 0,?δ > 0,B0AZAW 0 < |trianglez| < δ AYA0CR
vextendsinglevextendsingle
vextendsinglevextendsinglef(z0 +trianglez)?f(z0)
trianglez?f
prime(z0)
vextendsinglevextendsingle
vextendsinglevextendsingle < ε.
D8
ρ(trianglez) = f(z0 +trianglez)?f(z0)trianglez?fprime(z0),
A2DD lim
trianglez→0
ρ(trianglez) = 0.
CUB2 f(z0 +trianglez)?f(z0) = fprime(z0)trianglez + ρ(trianglez)trianglez.
BECI lim
trianglez→0
f(z0 +trianglez) = f(z0).
BECIf(z)D1z0 CUAXA9B5D1 z0 D1CBA1
BEDB?AKD0A1CZAS f(z) = x + 2yi D1BJA8DHA3ARARD1CBAVARAR?CU
AXA1D8A8A1
BQ 2CDCOBYB7 f(z) = |z|2 B0CUAXCAA1
BJ AXz0 = x0 + iy0 B2AQCHBJB7A1CMBQ
f(z0 +trianglez)?f(z0)
trianglez =
|z0 +trianglez|2?|z0|2
trianglez
= (z0 +trianglez)(z0 +trianglez)?z0z0trianglez
= z0 +trianglez?z0triangleztrianglez,
ASBXz0 = 0,A2DDA0AWtrianglez → 0AYA0AWB1B0C5C0B2D6A1
ASBXz0 negationslash= 0,ASAYA0D8 z0 +trianglez CEDCC1
y?y0 = k(x?x0)
AIC4CU z0,D2CR
trianglez
trianglez =
trianglex?triangleyi
trianglex +triangleyi =
1? triangleytrianglexi
1 + triangleytrianglexi
= 1?ki1 + ki
CWk CRBUA0?AICUCHBPAOB5B0DDA1BECIA0AW trianglez → 0AYA0A7DD
f(z0 +trianglez)?f(z0)
trianglez B0C5C0?ATD1A1
BECIA0 f(z) = |z|2 CND1z = 0ARCUAXA0D1AABFB4B6?CUAXA1
§2.1 B5B9B3B7AYB2B6 3
3) AHAXBDD2A1
(a) (c)prime = 0,AADHcB2BJAIB7A1
(b) (zn)prime = nzn?1,AADHnB2D7D6B7A1
(c) [f(z) ±g(z)]prime = fprime(z) ±gprime(z).
(d) [f(z)g(z)]prime = fprime(z)g(z) + f(z)gprime(z).
(e)
bracketleftbiggf(z)
g(z)
bracketrightbiggprime
= 1g2(z)[fprime(z)g(z)?f(z)gprime(z)],g(z) negationslash= 0.
(f) {f[g(z)]}prime = fprime(w)gprime(z),AADHA0w = g(z).
(g) fprime(z) = 1?prime(w),AADHA0w = f(z)CW z =?(w)B2D2BPC3BQBEBYB7B0
AUDDBYB7A0prime(w) negationslash= 0.
BQ 1AHBXD4BYB7B0AXB7A1
(1) z3 + 2iz; (2) 1z2?1
BJ
(1) f(z) = z3 + 2iz D1BJA8DHAWARARCUAXA0 fprime(z) = 3z2 + 2i;
(2) f(z) = 1z2?1D1BHA1?BQD6AR(C7z negationslash= ±1)CUAXA0fprime(z) =?2z(z2?1)2.
4) BNBHB0BNA4
AXBYB7w = f(z)D1z0 CUAXA0D2CR
trianglew = f(z0 +trianglez)?f(z0) = fprime(z0)trianglez + ρ(trianglez)trianglez,
AADHA0 lim
trianglez→0
ρ(trianglez) = 0,CMAS |ρ(trianglez)trianglez| B2 |trianglez| B0BOCJBTAGC5D3A0BB
fprime(z0)trianglezB2BMAAD3trianglewB0C1CAAFBHA0AJ fprime(z0)trianglezCFBDCGw = f(z)DCAVz0
ASCOB4,C9A9
dw = fprime(z0)trianglez = fprime(z0)dz.
BYB7w = f(z)D1z0 CUAXCWD1 z0 CUBNB2B1CCB0A1
4 B0B1BB B5B9B3B7
2,BJCQBDCGASB7C2
AWD5 ASBXBYB7 f(z)D1z0 C6z0 B0D5CXA3ARARCUAXA0A2DDAJ f(z)DC
z0 BJCQA1
ASBXf(z)D1AJCXDA3DECHB4CMBUA0A2DDAJ f(z)CFDC1ASBJCQBDCG.
ASBXBYB7 f(z)D1z0?CMBUA0A2DDAJ z0 B2f(z)B0C4AVA1
A5 1BYB7D1AJCXA3CMBUB1CCCUD1AJCXA3CUAXA1
A5 2BYB7D1CHB4ARCMBU?B1CCCUD1CHB4ARCUAXA1
BQ 1 CDCOBYB7 f(z) = z
2,g(z) = x + 2yi,h(z) = |z|2,φ(z) = 1
z B0CMBU
CAA1
BJ 1) f(z) = z2 D1BJA8DHA3ARARCUAXA0D2D1BJA8DHA3ARARCMBUA1
2) g(z) = x + 2yi D1BJA8DHA3ARAR?CUAXA0D2D1BJA8DHA3ARAR?CM
BUA1
3) h(z) = |z|2 CND1z = 0ARCUAXA0BTD1BJA8DHA3ARAR?CMBUA1
4) φ(z) = 1z D1BJA8DHA3AQ z = 0BMARARCUAXA0?
dφ(z)
dz =?
1
z2
BTD1BJA8DHA3AQ z = 0BMARARCMBUA1
AWBP 1)D1AJCXDA3CMBUB0D2BPBYB7 f(z),g(x)B0BZA0AHA0C4A0AV (AQ
AMBHA1BQD6B0B4) D1AJCXDA3CMBUA1
2) ASBXBYB7 h = g(z) D1 z A8DHAWB0AJCX D A3CMBUA0BYB7 w = f(h)
D1hA8DHAWB0AJCX GA3CMBUA1ASBXB9 DA3B0DECHBPB4 z,BYB7g(z)B0B9
CODDhB6B5CUG,A2DDBJC1BYB7 w = f[g(z)]D1AJCX DA3CMBUA1
§2.2 B3B7B5B9AYAWBAB8B4 5
§2.2 BDCGBJCQASAJD0CLBH
AWBPD2 AXBYB7f(z) = u(x,y) +iv(x,y)B5CLD1AJCX DA3A0D2f(z)D1
DA3CHB4 z = x + iy CUAXB0ANCFBJCEB2A2 u(x,y)CWv(x,y)D1B4 (x,y)CU
BNA0AD?D1BLB4DBA5CTBV -CWDCBGAM
u
x =
v
y,
u
y =?
v
x.
DGAJB4CY
BRCQCUAXB0B5CLA0CFD8C5C0
lim
trianglez→0
f(z +trianglez)?f(z)
trianglez
ATD1A1
CQCU
f(z +trianglez)?f(z) = u(x +trianglex,y +triangley)?u(x,y)
+i[v(x +trianglex,y +triangley)?v(x,y)]
= triangleu + itrianglev,
CTCMBQu(x,y),v(x,y) D1B4 (x,y)CUBNA0CUDA
triangleu =?u?xtrianglex +?u?ytriangley + ε1trianglex + ε2triangley,
trianglev =?v?xtrianglex +?v?ytriangley + ε3trianglex + ε4triangley,
AADHA0 lim
trianglex →0
triangley →0
εk = 0 (k = 1,2,3,4),CMASCR
f(z +trianglez)?f(z) =
parenleftbigg?u
x + i
v
x
parenrightbigg
trianglex +
parenleftbigg?u
y + i
v
y
parenrightbigg
triangley
+(ε1 + iε3)trianglex + (ε2 + iε4)triangley.
BRCQCTBV -CWDCBGAM
u
y =?
v
x = i
2?v
x,
u
x =
v
y.
BECI
f(z +trianglez)?f(z) =
parenleftbigg?u
x + i
v
x
parenrightbigg
(trianglex + itriangley)
+(ε1 + iε3)trianglex + (ε2 + iε4)triangley.
6 B0B1BB B5B9B3B7
D2CR
f(z +trianglez)?f(z)
trianglez =
u
x + i
v
x
+(ε1 + iε3)trianglextrianglez + (ε2 + iε4)triangleytrianglez.
AWtrianglez → 0AYA0CMBQ
vextendsinglevextendsingle
vextendsinglevextendsingletrianglex
trianglez
vextendsinglevextendsingle
vextendsinglevextendsinglelessorequalslant 1,
vextendsinglevextendsingle
vextendsinglevextendsingletriangley
trianglez
vextendsinglevextendsingle
vextendsinglevextendsinglelessorequalslant 1,
AWB1CSB7B0A8C2D2C3B6AICUD6A1
CMASA0
fprime(z) = lim
trianglez→0
f(z +trianglez)?f(z)
trianglez =
u
x + i
v
x.
D8A8A1
AQBDCG f(z) = u(x,y) + iv(x,y)DCz = x + iy BNARCDA0D7
fprime(z) =?u?x + i?v?x = 1i?u?y +?v?y.
AWBPAZ BYB7f(z) = u(x,y) +iv(x,y)D1AAB5CLCX DA3CMBUB0ANCFBJ
CEB2A2 u(x,y)CWv(x,y)D1DA3CUBNA0AD?DBA5CTBV -CWDCBGAMA1
AWBP (CUBNB0ANBHBJCE) ATBYB7 f(x,y)D1B4P B0A0BPD5CX DA3ATD1
D2BPA7AXB7A0?A7AXB7D1BLB4D1CBA0D2BYB7 f(x,y)D1B4P CUBNA1
(BOB1B7CC (BX) pp,27)
D1A6B4C2DHCFD8DIA2 CHBPCMBUBYB7B0AXBYB7fprime(z)CGB2CMBUB0,CMAS
fprime(z) = ux + ivx = vy?iuy CGB2D1CBB0A1CUB2BSDFCRBXDHB0B5CXA1
AWBPAZ (B) BYB7 f(z) = u(x,y) + iv(x,y) D1AAB5CLCX D A3CMBUB0AN
CFBJCEB2A2 u(x,y)CWv(x,y) D1D A3CRCHCJD1CBA7AXB7A0AD?DBA5CTBV
-CWDCBGAMA1
§2.2 B3B7B5B9AYAWBAB8B4 7
BQ 1A6B5BXD4BYB7D1C0ARCUAXA0D1C0ARCMBUA1
1) f(z) = zRe(z);
2) f(z) = x + yx2 + y2 + i x?yx2 + y2;
3) f(z) = ex(cosy + isiny);
BJ 1)CQf(z) = zRe(z) = x2 + ixy,AZu = x2,v = xy,BECI
u
x = 2x,
u
y = 0,
v
x = y,
v
y = x.
BZAPA0D4BBBPA7AXB7ARARD1CB (CQASDA u,v CUBN),AVB2CNAW x = y = 0
AYA0BGDFAGDBA5CTBV -CWDCBGAMA1
CMASA0BYB7CND1 z = 0CUAXA0D1BJA8DHA3ARAR?CMBUA1
2)CMBQu = x + yx2 + y2,v = x?yx2 + y2,BECI
u
x =
y2?x2?2xy
(x2 + y2)2,
u
y =
x2?y2?2xy
(x2 + y2)2,
v
x =
y2?x2 + 2xy
(x2 + y2)2,
v
y =
y2?x2?2xy
(x2 + y2)2,
CHCJA7AXB7D1 z negationslash= 0 ARD1CB (CQASDA u,v CUBN),AD?DBA5CTBV - CWDCBG
AMA1
CMASA0BYB7D1BJA8DHA3AQ z = 0BMARARCUAXA0D1BJA8DHA3AQ z = 0BM
ARARCMBUA0AD?
fprime(z) =?u?x + i?v?x = (y
2?x2)(1 + i)?2xy(1?i)
(x2 + y2)2 (z negationslash= 0).
3)CMBQu = excosy,v = exsiny,BECI
u
x = e
xcosy,?u
y =?e
xsiny,
v
x = e
xsiny,?v
y = e
xcosy.
CHCJA7AXB7D1BJA8DHA3ARARD1CB (CQASDAu,vCUBN),AD?DBA5CTBV -CWDC
BGAMA1CMASA0BYB7D1BJA8DHA3ARARCMBUA0AD?
fprime(z) =?u?x + i?v?x = ex(cosy + isiny) = f(z).
8 B0B1BB B5B9B3B7
A5A2BMCS -BOBWB2AICF f(z)DCAVz0 BNARASACD0CLBHA0APCK?B3 f(z)
DCz0 BNARASAJB4CLBHA1
BQ 3CDCO f(z) = radicalbig|xy|D1z = 0B0CUAXCAA1
BJCMBQu(x,y) = radicalbig|xy|,v(x,y) = 0,BECI
ux(0,0) = uy(0,0) = vx(0,0) = vy(0,0) = 0,
BBDCAV (0,0)BVA6BMCS -BOBWB2AIA1
APCFA0CBBCBT z CZDICV
braceleftBigg
x = αt
y = βt (α,β AFCMCDCPBS) C7D8BSCDA0
f(z)?f(0)
z?0 =
f(z)
z =
radicalbig|αβ|
α + iβ
B0C5C0?B2BPCHB0A0CUCD f(z)D1B4z = 0B2?CUAXB0A1
BQ 3AX
f(z) =
x3?y3 + i(x3 + y3)
x2 + y2,z negationslash= 0,
0,z = 0,
B3D8f(z)D1CZB4DBA5CTBV -CWDCBGAMA0AV?CUAXA1
DGD8f(z) = u + iv,AZ
u(x,y) =
x3?y3
x2 + y2,(x,y) negationslash= (0,0),
0,(x,y) = (0,0),
v(x,y) =
x3 + y3
x2 + y2,(x,y) negationslash= (0,0),
0,(x,y) = (0,0),
CMBQ
ux(0,0) = limx→0 u(x,0)?u(0,0)x?0 = limx→0 x
3
x3 = 1;
uy(0,0) = limy→0 u(0,y)?u(0,0)y?0 = limy→0?y
3
y3 =?1;
vx(0,0) = limx→0 v(x,0)?v(0,0)x?0 = limx→0 x
3
x3 = 1;
vy(0,0) = limy→0 v(0,y)?v(0,0)y?0 = limy→0 y
3
y3 = 1.
CJD3f(z)DCDBAVBVA6BMCS -BOBWB2AIA1
APCFAQ z CZDICV y = kxC7CXD8BSCDA0
f(z)?f(0)
z?0 =
x3?y3 + i(x3 + y3)
(x + iy)(x2 + y2) →
1?k3 + i(1 + k3)
(1 + ik)(1 + k2)
BECIf(z)D1z = 0?CUAXA1
§2.2 B3B7B5B9AYAWBAB8B4 9
BQ 4 ASBXfprime(z)D1AJCXDARARBQD6A0A2DD f(z)D1D A3BQAIB7A1
DG CMBQ
fprime(z) =?u?x + i?v?x =?v?y?i?u?y ≡ 0,
BT
u
x =
v
x =
v
y =
u
y ≡ 0,
BECIu =AIB7A0v =AIB7A0CMBB f(z)D1DA3BQAIB7A1
BQ 5 ASBX f(z) = u + iv BQCHCMBUBYB7A0? f
prime(z) negationslash= 0,
A2DDAKC1A6
u(x,y) = c1,v(x,y) = c2 A9C3C2D7CGA0AADHA0 c1,c2 BQAIB7A1
DG CQCUf
prime(z) = 1
iuy + vy negationslash= 0,BTuy CWvy A9?ANBQD6A1
ASBXD1AKC1CGB4AR uy CW vy B6?BQD6A0BRCQCNBYB7AHAXBDD2DAA0
AKC1 u(x,y) = c1 B0C7D9B2 k1 =?ux/uy; AKC1 v(x,y) = c2 B0C7D9B2
k2 =?vx/vy,CYCPCTBV -CWDCBGAMAZ
k1 ·k2 = (?ux/uy)· (?vx/vy) = (?vy/uy)· (uy/vy) =?1.
ASBXD1AKC1CGB4AR uy CW vy CRCHBPBQD6A0D2D7CHBPA9?BQD6A0ASAY
CJDAD2A6DHB0AKC1D1CGB4ARB0ADC1CHBJB2B9A8B0A0D7CHBJB2ACDCB0A0BG
DFARC3C2D7CGA1
10 B0B1BB B5B9B3B7
§2.3 AKATBDCG
A6CKA5AZAABYB7DHB0CHC6AICPB0AOB1BYB7BLBVAYBJAABYB7B0AFC8A1
A2CKAUBPBGDHA2 1,B5CLA3 2,CADGA3 3,CMBUCAA1
1,A0CGBDCG
1)AWD5:B9CUBJAAB7 z = x + iy,AJ
ez = ex(cosy + isiny)
BQz B0A0CGBDCG.
2)BJCQCY
a) f(z)D1BJA8DHA3ARARCUAXA0? (ez)prime = ez.
3)CYA2
a)CABDB5CXA2
ez1 ·ez2 = ez1+z2.
DGC0A2AX z1 = x1 + iy1,z2 = x2 + iy2,BRCQB5CLA0CR
ez1 ·ez2 = ex1(cosy1 + isiny1)·ex2(cosy2 + isiny2)
= ex1+x2[cos(y1 + y2) + isin(y1 + y2)]
= ez1+z2.
b)DJA9CA
ez+2kpii = ez ·e2kpii = ez.
§2.3 AXAZB3B7 11
2,AXCGBDCG
1)AWD5:A5DBA5BGAM
ew = z (z negationslash= 0)
B0BYB7w = f(z)AJBQAXCGBDCG.
C6BJAMB2AI,D8 w = u + iv,z = reiθ,A2DD
eu+iv = reiθ =? eueiv = reiθ,
BECI u = lnr,v = θ.
CMAS w = ln|z| + iArgz.
C9A9
Lnz = ln|z| + iArgz.
D6D8 LnzCFAYDJBDCGA0BXBRB8DJAG 2piiASDECGABA1
BWB5AADH ArgzALA1DD argzB0A2CHC3A0AJA9 LnzB0A4DJA0C9A9lnz,
D2CR
lnz = ln|z| + iargz.
BQ 1 AHLn2,Ln(?1)CIC6BGDFC2COB0A1DDA1
BJ CMBQ
Ln2 = ln2 + iArg2 = ln2 + 2kpii,
BGB0A1DDB2 ln2.
CMBQ
Ln(?1) = ln1 + iArg(?1) = (2k + 1)pii,
BGB0A1DDB2 ln(?1) = pii.
12 B0B1BB B5B9B3B7
2)CYA2:
a) AWz = x > 0AYA0Lnz B0A1DD lnz = lnx.
b)
Ln(z1z2) = Lnz1 + Lnz2,
Ln
parenleftbiggz
1
z2
parenrightbigg
= Lnz1?Lnz2.
DGA2
Ln(z1z2) = ln|z1z2|+ iArgz1z2
= ln|z1||z2|+ i(Argz1 + Argz2)
= (ln|z1| + iArgz1) + (ln|z2| + iArgz2)
= Lnz1 + Lnz2.
Ln
parenleftbiggz
1
z2
parenrightbigg
= ln
vextendsinglevextendsingle
vextendsinglevextendsinglez1
z2
vextendsinglevextendsingle
vextendsinglevextendsingle + iArgz1
z2
= ln|z1||z
2|
+ i(Argz1?Argz2)
= (ln|z1|?ln|z2|) + i(Argz1?Argz2)
= (ln|z1| + iArgz1)?(ln|z2|+ iArgz2)
= Lnz1?Lnz2.
A2A2BXDHB0B1B1?D0AKD0A2
Lnzn = nLnz,
Ln n√z = 1nLnz.
§2.3 AXAZB3B7 13
3)BJCQCY:
CTBLBUA4DJ lnz ASBJCQCYA1
CMBQlnz = ln|z|+iargz,BBln|z|AQCZB4BMD1AABGB4B6B2D1CBB0A0argz
D1AQCZB4CWBKAZA0CIBMB6D1CBA1
CUB2A0w = lnz DCALDBAVDAB6CEA3ASC3BZC1BJCQA0C5(CQBEBYB7B0AH
AXBDD2)
dlnz
dz =
1
dew
dw
= 1ew = 1z.
CMBQ Lnz = lnz + 2kpii,BECI Lnz ASB9B8B4DHDCALDBAVDAB6CEA3ASC3
BZC1BJCQA0?C5D7CWCMASARCGDJA1
3,AHBYab DABYBDCG
1)AWD5:AXaBQ?BQD6B0CHBPBJB7A0b BQAQCKCHBPBJB7A0B5CLALDGBQ
ab = ebLna.
CQCU LnaB2BADDB0A0CMBB ab CGB2BADDB0A1
2)CYA2:
a) AB bAOASAMALA0 ab AIARAAAQACAUA1C7A2
ab = ebLna = eb[ln|a|+i(arga+2kpi)]
= eb(ln|a|+iarga)+2kbpii
= eb(ln|a|+iarga) ·e2kbpiibracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright
= eb(ln|a|+iarga) = eblna,
b) AB b = pq(p AG q AOAHAVACASAMA0 q > 0) ALA0 ab AIAR q AFACAUA1
C7A2
ab = epqln|a|+ipq(arga+2kpi)
= epqln|a|
bracketleftbigg
cospq(arga + 2kpi) + isinpq(arga + 2kpi)
bracketrightbigg
,
AADHA2k = 0,1,···,(q?1).
c) A8A9ATANA0AQA7?AJ ab AIARAPAKADAFAUA1
14 B0B1BB B5B9B3B7
BHACB2A0AW bBQD7D6B7 nAYA0BRCQB5CL
an = enLna = eLna+Lna+···+Lna
= eLna ·eLna ·····eLna
= a·a·····a.
AWb = 1n AYA0CR
a1n = e1nLna = e1nln|a|
parenleftbigg
cosarga + 2kpin + isinarga + 2kpin
parenrightbigg
= n
radicalbig
|a|
parenleftbigg
cosarga + 2kpin + isinarga + 2kpin
parenrightbigg
= n√a
AADHA0 k = 0,1,2,···,(n?1).
A4AWBEB6A0AQ b CPDFDECG nBGB4CG
1
n CDCFDA a AS nANBYBG aAS n
ANBAASD4D5CNCAD2A1A1
3)BJCQCY:
a) zn DCB5C3BZC1CFAODJBJCQBDCGA0C5
(zn)prime = nzn?1.
b)CMBQ
z 1n = n√z = e1nLnz
BECIz 1n B2CHBPBADDBYB7A0CRCR nBPBHD9A1
CQCULnzB0DEBPBHD9D1AQAMCZB4BZBKAZA0B0BJA8DHA3B2CMBUA0CJD3
z 1n ASBXB8B4DHDCALC9DBAVBFB6CEA3ASB5C3BZC1D1CFBJCQASA0C5
(z 1n)prime = 1nz 1n?1.
c)zb(AQAM b = nCW
1
n D2DIAFCV) B2CHBPBADDBYB7A0CRBTAGBADDA1 z
b
ASBXB8B4DHDCALC9DBAVBFB6CEA3ASB5C3BZC1D1CFBJCQASA0C5
(zb)prime = bzb?1.
§2.3 AXAZB3B7 15
4,CCBIBDCGBFCIC8BDCG
1)AWD5:
cosz = e
iz + e?iz
2,sinz =
eiz?e?iz
2i,
2)CYA2:
a)DJA9CAA2 cos(z + 2pi) = cosz,sin(z + 2pi) = sinz.
b)ABA5CAA2 cos(?z) = cosz,sin(?z) =?sinz.
c)CRBUCVBYBZD7BYB0C8BCBSB1
cos(z1 + z2) = cosz1cosz2?sinz1sinz2,
sin(z1 + z2) = sinz1cosz2?cosz1sinz2,
sin2z + cos2z = 1.
A5A2 CADG|sinz|lessorequalslant 1BZ|cosz|lessorequalslant 1D1BJB7BFBOA3?D0AKD0A1
3)BJCQCY:
AUCHBYB7 cosz BZ sinz B6B2BJA8DHA3B0CMBUBYB7A0?CR
(cosz)prime =?sinz,(sinz)prime = cosz.
16 B0B1BB B5B9B3B7
5,B1CCBIBDCGDAB1CIC8BDCG
1)AWD5,BEAUCHBYB7B5CLBQAUCHBYB7B0BEBYB7A1AX
z = cosw,
A2DDAJ wBQz B0B1D9CUBDCG,C9A9
w = Arccosz,
CQz = cosw = 12(eiw + e?iw)AZ
e2iw?2zeiw + 1 = 0,
BGB0BRBQ eiw = z +√z2?1,
AADHA0
√z2?1
COCXCMBQB8DDBYB7A1CMASA0D2B7ALB9B7A0AZ
Arccosz =?iLn(z +
radicalbig
z2?1).
§2.3 AXAZB3B7 17
2,(pp,66)BXD4BYB7C0ARCUAXA4C0ARCMBUA4
3) f(z) = 2x3 + 3y3i;
BJCMBQu = 2x3,v = 3y3,AD?
u
x = 6x
2,?u
y = 0,
v
x = 0,
v
y = 9y
2.
A7AXB7D1CB (CQASDAu,vCUBN),AVCTBV-CWDCBGAMux = vy,uy =?vx CNAW
6x2 = 9y2 AYAKD0A1
BTf(z)DFD1 6x
2 = 9y2
AWCUAXA0D1BJA8DHAWARAR?CMBUA1
6,(pp,66)A6B8BXD4DJBIB0D5CBA1
5)ASBX u(x,y)BZv(x,y)CUAX (DEA7AXB7ATD1),A2DD f(z) = u + iv CG
CUAXA3
BJ?D7AOA1BYB7 u = 2xBZv = y CSCUAXA0AV f(z) = u + iv = 2x + iy
D1BJA8DHAWARAR?CUAXA1
6) AX f(z) = u + iv D1AJCX D A3B2CMBUB0A1ASBX u B2AZAIB7A0A2DD
f(z)D1D6BP DA3B2AIB7A3ASBX vB2AZAIB7A0A2DD f(z)D1D6BP DA3B2AI
B7A1
BJf(z)D1DA3CMBUA0D2 u,v D1DA3CUBN?DBA5CTBV -CWDCBGAMA2
ux = vy,uy =?vx.
BECIAWuB2AZAIB7AYA0CR ux = uy = 0,CUB2
vy = ux = 0,vx =?uy = 0,
BTf(z)D1DA3B2AIB7A1
18 B0B1BB B5B9B3B7
10,(pp,67)D8DIA2ASBXBYB7 f(z) = u+ivD1AJCXDA3CMBUA0ADDBA5
BXD4BJCEDBCHA0A2DD f(z)B2AIB7A1
2) f(z)D1DA3CMBUA3
4) argf(z)D1DA3B2CHBPAIB7A3
BJA2f(z) D1DA3CMBUA0D2
ux = vy,uy =?vx,(1)
2)ASBXf(z) = u?iv D1DA3CMBUA0D2CRCTBV -CWDCBGAM
ux =?vy,uy = vx.
CYCP (1)B1A0D2ux = uy = vx = vy = 0,BECIf(z)B2AIB7A1
4)ASBXargf(z)D1DA3B2CHBPAIB7A0C9 argf(z) = C,D2
v
u = tanC(BQAIB7A0C9BQ C
prime)
CUB2v = Cprimeu,D2CR
vx = Cprimeux,vy = Cprimeuy.
D0BRCQ (1)CUAZux = uy = vx = vy = 0,BECIf(z)B2AIB7A1
11,(pp,67)BXD4BUBWB2BID7AOA4
1) ez = ez; 3) sinz = sinz
BJ 1)CMBQ ez = ex+iy = ex(cosy + isiny).
BECI ez = ex(cosy?isiny).
CTCMBQ
ez = ex?iy
= ex[cos(?y) + isin(?y)]
= ex(cosy?isiny).
BTez = ez.
3)CMBQsinz = 12i(eiz?e?iz),BECI (CYCP (1)B0CLDA)
sinz = 12i(eiz?e?iz) = 1?2i(eiz?e?iz)
= 1?2i(e?iz?eiz) = sinz.
§2.3 AXAZB3B7 19
12,(pp,67)D3APBXD4BGAMB0ANAFCMA2
1) sinz = 0;
BJA2sinz = 0B1CCCU
1
2i(e
iz?e?iz) = 0,
C7e2iz = 1,CUB2
2iz = Ln1 = ln1 + i(0 + 2kpi) = 2kpii,
BECIBGAMB0ANAFCMBQ z = kpi(k = 0,±1,±2,···).
13,(pp,67)D8DIA2
1) cos(z1 + z2) = cosz1cosz2?sinz1sinz2;
sin(z1 + z2) = sinz1cosz2 + cosz1sinz2;
2) cos2z + sin2z = 1;
3) sin2z = 2sinzcosz;
5) sin
parenleftBigz
2?z
parenrightBig
= cosz; cos(z + pi) = cosz;
DG(1)
cosz1cosz2?sinz1sinz2
= e
iz1 + e?iz1
2 ·
eiz2 + e?iz2
2?
eiz1?e?iz1
2i ·
eiz2?e?iz2
2i
= e
i(z1+z2) + e?i(z1+z2)
2 = cos(z1 + z2).
BKCXA2 sin(z1 + z2) = sinz1cosz2 + cosz1sinz2.
(2)
sin2z + cos2z
=
parenleftbiggeiz?e?iz
2i
parenrightbigg2
+
parenleftbiggeiz + e?iz
2
parenrightbigg2
=?14(e2iz?2 + e?2iz) + 14(e2iz + 2 + e?2iz) = 1.
(3) D1 (1) DHB1B1 sin(z1 + z2) = sinz1cosz2 + cosz1sinz2 AL z1 = z2 = z
C7AZA1
(5)CYCP (1)B0CLDAA0
sin
parenleftBigpi
2?z
parenrightBig
= sinpi2cosz + cospi2sinz = cosz;
cos(z + pi) = coszcospi?sinzsinpi =?cosz.
20 B0B1BB B5B9B3B7
17,(pp,68)BADIBXD4B1B1B2BID7AOA1
1) Lnz2 = 2Lnz; 2) Ln√z = 12Lnz,
BJ D8z = reiθ,r = |z|,θ = argz.
1)?D7AOA1CMBQ
Lnz2 = lnr2 + i(2θ + 2kpi) = 2lnr + i(2θ + 2kpi)
(k = 0,±1,±2,···,)
BB
2Lnz = 2{lnr + i(θ + 2k1pi)} = 2lnr + i(2θ + 4k1pi)
(k1 = 0,±1,±2,···,)
BECI Lnz2 B0DDA7 2Lnz B0DDBAA0B1B1?D7AOA1
2)?D7AOA1CMBQ
1
2Lnz =
1
2{lnr + i(θ + 2kpi)} =
1
2lnr + i
parenleftbiggθ
2 + kpi
parenrightbigg
(k = 0,±1,±2,···,)
D7BMA0CMBQ
√z = √reθ+2lpi
2 i(l = 0,1),
C7
√z
CRD2BPBRA2
√reθ
2i
BZ
√z = √reθ+2pi
2 i.
CUB2 Ln√z CRD2A7DDA0BHACB2
B3CHA7
Ln(√reθ2i) = ln√r + i
parenleftbiggθ
2 + 2k1pi
parenrightbigg
= ln√r + i
parenleftbiggθ
2 + 2k1pi
parenrightbigg
(k1 = 0,±1,±2,···)
B3BCA7
Ln(√reθ+2pi2 i) = ln√r + i
parenleftbiggθ + 2pi
2 + 2k2pi
parenrightbigg
= 12lnr + i
bracketleftbiggθ
2 + (2k2 + 1)pi
bracketrightbigg
(k2 = 0,±1,±2,···)
BDAPAW kB2A5B7AYA0
1
2LnzCWB3CHA7DDC2BKA0AW kB2ABB7AYA0
1
2Lnz
CWB3BCA7DDC2BKA0AV
1
2Lnz CWLn
√z
B2D2BP?BKB0BTAGDDBYB7A1
§2.3 AXAZB3B7 21
18,(pp,68)AH(1 + i)i B0DDA1
BJA2
(1 + i)i = eiLn(1+i) = ei(ln|1+i|+i[arg(1+i)+2kpi])
= ei{ln
√2+i(pi
4+2kpi)}
= e?(pi4+2kpi)+ln
√2i
= e?(pi4+2kpi)(cosln√2 + isinln√2)
(k = 0,±1,±2,···)
9,D8DICTBV -CWDCBGAMB0C5AAABC8B1B2A2
u
r =
1
r
v
θ,
v
r =?
1
r
u
θ.
DGCQDCCHAAABCWC5AAABB0BUBWA2 x = rcosθ,y = rsinθ,CIC6
u = u(x,y) = u(r,θ),v = v(x,y) = v(r,θ),BRCQBACYBYB7B0BJC1BYB7AHAX
BDD2A0
u
r =
u
x
x
r +
u
y
y
r = cosθ
u
x + sinθ
u
y
u
θ =
u
x
x
θ +
u
y
y
θ =?rsinθ
u
x + rcosθ
u
y
v
r =
v
x
x
r +
v
y
y
r = cosθ
v
x + sinθ
v
y
v
θ =
v
x
x
θ +
v
y
y
θ =?rsinθ
v
x + rcosθ
v
y,
CYCP
u
x =
v
y,
u
y =?
v
x,A7CICIAWBQB1A0C7AZC5AAABC8B1B0CTBV -CWDC
BGAMA2
u
r =
1
r
v
θ,
v
r =?
1
r
u
θ.
§2.1 BJCQBDCGASB7C2
1,B5ADBDCGASARCGDACOB4
1)AXB7B0B5CL
AWD5AXBYB7w = f(z)B5CLCUAJCX D,z0 BQDDHB0CHB4A0B4 z0 +trianglez
APDB0BFBOA1ASBXC5C0
lim
trianglez→0
f(z0 +trianglez)?f(z0)
trianglez
ATD1A0A2DDAJ f(z)DCz0 BNAR,D4BPC5C0DDAJBQ f(z)DCz0 ASARCG,C9A9
fprime(z0) = dwdz
vextendsinglevextendsingle
vextendsinglevextendsingle
vextendsinglez=z
0
= lim
trianglez→0
f(z0 +trianglez)?f(z0)
trianglez,
ASBXf(z)D1AJCX DA3ARARCUAXA0CPBA f(z)DCDC1BNARA1
BQ 1 BRf(z) = x + 2yiB2BICUAXA4
BJ CMBQ
lim
trianglez→0
f(z +trianglez)?f(z)
trianglez
= lim
trianglez→0
(x +trianglex) + 2(y +triangley)i?x?2yi
trianglez
= lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi
AWz +trianglez CEA3A8C9CU xA0B0DCC1AIC4CU z,ASAYtriangley = 0,C5C0
lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi = limtrianglez→0
trianglex
trianglex = 1.
AWz +trianglez CEA3A8C9CU y A0B0DCC1AIC4CU z,ASAYtrianglex = 0,C5C0
lim
trianglez→0
trianglex + 2triangleyi
trianglex +triangleyi = limtrianglez→0
2triangleyi
triangleyi = 2.
BECIf(z)B0AXB7?ATD1A1
1
2 B0B1BB B5B9B3B7
2) CUAXCWD1CB
CYA2A2D1 z0 CUAXB0BYB7A9B5D1 z0 D1CBA1BEDB?AKD0A1
DG,BRCQAXB7B0B5CLA0?ε > 0,?δ > 0,B0AZAW 0 < |trianglez| < δ AYA0CR
vextendsinglevextendsingle
vextendsinglevextendsinglef(z0 +trianglez)?f(z0)
trianglez?f
prime(z0)
vextendsinglevextendsingle
vextendsinglevextendsingle < ε.
D8
ρ(trianglez) = f(z0 +trianglez)?f(z0)trianglez?fprime(z0),
A2DD lim
trianglez→0
ρ(trianglez) = 0.
CUB2 f(z0 +trianglez)?f(z0) = fprime(z0)trianglez + ρ(trianglez)trianglez.
BECI lim
trianglez→0
f(z0 +trianglez) = f(z0).
BECIf(z)D1z0 CUAXA9B5D1 z0 D1CBA1
BEDB?AKD0A1CZAS f(z) = x + 2yi D1BJA8DHA3ARARD1CBAVARAR?CU
AXA1D8A8A1
BQ 2CDCOBYB7 f(z) = |z|2 B0CUAXCAA1
BJ AXz0 = x0 + iy0 B2AQCHBJB7A1CMBQ
f(z0 +trianglez)?f(z0)
trianglez =
|z0 +trianglez|2?|z0|2
trianglez
= (z0 +trianglez)(z0 +trianglez)?z0z0trianglez
= z0 +trianglez?z0triangleztrianglez,
ASBXz0 = 0,A2DDA0AWtrianglez → 0AYA0AWB1B0C5C0B2D6A1
ASBXz0 negationslash= 0,ASAYA0D8 z0 +trianglez CEDCC1
y?y0 = k(x?x0)
AIC4CU z0,D2CR
trianglez
trianglez =
trianglex?triangleyi
trianglex +triangleyi =
1? triangleytrianglexi
1 + triangleytrianglexi
= 1?ki1 + ki
CWk CRBUA0?AICUCHBPAOB5B0DDA1BECIA0AW trianglez → 0AYA0A7DD
f(z0 +trianglez)?f(z0)
trianglez B0C5C0?ATD1A1
BECIA0 f(z) = |z|2 CND1z = 0ARCUAXA0D1AABFB4B6?CUAXA1
§2.1 B5B9B3B7AYB2B6 3
3) AHAXBDD2A1
(a) (c)prime = 0,AADHcB2BJAIB7A1
(b) (zn)prime = nzn?1,AADHnB2D7D6B7A1
(c) [f(z) ±g(z)]prime = fprime(z) ±gprime(z).
(d) [f(z)g(z)]prime = fprime(z)g(z) + f(z)gprime(z).
(e)
bracketleftbiggf(z)
g(z)
bracketrightbiggprime
= 1g2(z)[fprime(z)g(z)?f(z)gprime(z)],g(z) negationslash= 0.
(f) {f[g(z)]}prime = fprime(w)gprime(z),AADHA0w = g(z).
(g) fprime(z) = 1?prime(w),AADHA0w = f(z)CW z =?(w)B2D2BPC3BQBEBYB7B0
AUDDBYB7A0prime(w) negationslash= 0.
BQ 1AHBXD4BYB7B0AXB7A1
(1) z3 + 2iz; (2) 1z2?1
BJ
(1) f(z) = z3 + 2iz D1BJA8DHAWARARCUAXA0 fprime(z) = 3z2 + 2i;
(2) f(z) = 1z2?1D1BHA1?BQD6AR(C7z negationslash= ±1)CUAXA0fprime(z) =?2z(z2?1)2.
4) BNBHB0BNA4
AXBYB7w = f(z)D1z0 CUAXA0D2CR
trianglew = f(z0 +trianglez)?f(z0) = fprime(z0)trianglez + ρ(trianglez)trianglez,
AADHA0 lim
trianglez→0
ρ(trianglez) = 0,CMAS |ρ(trianglez)trianglez| B2 |trianglez| B0BOCJBTAGC5D3A0BB
fprime(z0)trianglezB2BMAAD3trianglewB0C1CAAFBHA0AJ fprime(z0)trianglezCFBDCGw = f(z)DCAVz0
ASCOB4,C9A9
dw = fprime(z0)trianglez = fprime(z0)dz.
BYB7w = f(z)D1z0 CUAXCWD1 z0 CUBNB2B1CCB0A1
4 B0B1BB B5B9B3B7
2,BJCQBDCGASB7C2
AWD5 ASBXBYB7 f(z)D1z0 C6z0 B0D5CXA3ARARCUAXA0A2DDAJ f(z)DC
z0 BJCQA1
ASBXf(z)D1AJCXDA3DECHB4CMBUA0A2DDAJ f(z)CFDC1ASBJCQBDCG.
ASBXBYB7 f(z)D1z0?CMBUA0A2DDAJ z0 B2f(z)B0C4AVA1
A5 1BYB7D1AJCXA3CMBUB1CCCUD1AJCXA3CUAXA1
A5 2BYB7D1CHB4ARCMBU?B1CCCUD1CHB4ARCUAXA1
BQ 1 CDCOBYB7 f(z) = z
2,g(z) = x + 2yi,h(z) = |z|2,φ(z) = 1
z B0CMBU
CAA1
BJ 1) f(z) = z2 D1BJA8DHA3ARARCUAXA0D2D1BJA8DHA3ARARCMBUA1
2) g(z) = x + 2yi D1BJA8DHA3ARAR?CUAXA0D2D1BJA8DHA3ARAR?CM
BUA1
3) h(z) = |z|2 CND1z = 0ARCUAXA0BTD1BJA8DHA3ARAR?CMBUA1
4) φ(z) = 1z D1BJA8DHA3AQ z = 0BMARARCUAXA0?
dφ(z)
dz =?
1
z2
BTD1BJA8DHA3AQ z = 0BMARARCMBUA1
AWBP 1)D1AJCXDA3CMBUB0D2BPBYB7 f(z),g(x)B0BZA0AHA0C4A0AV (AQ
AMBHA1BQD6B0B4) D1AJCXDA3CMBUA1
2) ASBXBYB7 h = g(z) D1 z A8DHAWB0AJCX D A3CMBUA0BYB7 w = f(h)
D1hA8DHAWB0AJCX GA3CMBUA1ASBXB9 DA3B0DECHBPB4 z,BYB7g(z)B0B9
CODDhB6B5CUG,A2DDBJC1BYB7 w = f[g(z)]D1AJCX DA3CMBUA1
§2.2 B3B7B5B9AYAWBAB8B4 5
§2.2 BDCGBJCQASAJD0CLBH
AWBPD2 AXBYB7f(z) = u(x,y) +iv(x,y)B5CLD1AJCX DA3A0D2f(z)D1
DA3CHB4 z = x + iy CUAXB0ANCFBJCEB2A2 u(x,y)CWv(x,y)D1B4 (x,y)CU
BNA0AD?D1BLB4DBA5CTBV -CWDCBGAM
u
x =
v
y,
u
y =?
v
x.
DGAJB4CY
BRCQCUAXB0B5CLA0CFD8C5C0
lim
trianglez→0
f(z +trianglez)?f(z)
trianglez
ATD1A1
CQCU
f(z +trianglez)?f(z) = u(x +trianglex,y +triangley)?u(x,y)
+i[v(x +trianglex,y +triangley)?v(x,y)]
= triangleu + itrianglev,
CTCMBQu(x,y),v(x,y) D1B4 (x,y)CUBNA0CUDA
triangleu =?u?xtrianglex +?u?ytriangley + ε1trianglex + ε2triangley,
trianglev =?v?xtrianglex +?v?ytriangley + ε3trianglex + ε4triangley,
AADHA0 lim
trianglex →0
triangley →0
εk = 0 (k = 1,2,3,4),CMASCR
f(z +trianglez)?f(z) =
parenleftbigg?u
x + i
v
x
parenrightbigg
trianglex +
parenleftbigg?u
y + i
v
y
parenrightbigg
triangley
+(ε1 + iε3)trianglex + (ε2 + iε4)triangley.
BRCQCTBV -CWDCBGAM
u
y =?
v
x = i
2?v
x,
u
x =
v
y.
BECI
f(z +trianglez)?f(z) =
parenleftbigg?u
x + i
v
x
parenrightbigg
(trianglex + itriangley)
+(ε1 + iε3)trianglex + (ε2 + iε4)triangley.
6 B0B1BB B5B9B3B7
D2CR
f(z +trianglez)?f(z)
trianglez =
u
x + i
v
x
+(ε1 + iε3)trianglextrianglez + (ε2 + iε4)triangleytrianglez.
AWtrianglez → 0AYA0CMBQ
vextendsinglevextendsingle
vextendsinglevextendsingletrianglex
trianglez
vextendsinglevextendsingle
vextendsinglevextendsinglelessorequalslant 1,
vextendsinglevextendsingle
vextendsinglevextendsingletriangley
trianglez
vextendsinglevextendsingle
vextendsinglevextendsinglelessorequalslant 1,
AWB1CSB7B0A8C2D2C3B6AICUD6A1
CMASA0
fprime(z) = lim
trianglez→0
f(z +trianglez)?f(z)
trianglez =
u
x + i
v
x.
D8A8A1
AQBDCG f(z) = u(x,y) + iv(x,y)DCz = x + iy BNARCDA0D7
fprime(z) =?u?x + i?v?x = 1i?u?y +?v?y.
AWBPAZ BYB7f(z) = u(x,y) +iv(x,y)D1AAB5CLCX DA3CMBUB0ANCFBJ
CEB2A2 u(x,y)CWv(x,y)D1DA3CUBNA0AD?DBA5CTBV -CWDCBGAMA1
AWBP (CUBNB0ANBHBJCE) ATBYB7 f(x,y)D1B4P B0A0BPD5CX DA3ATD1
D2BPA7AXB7A0?A7AXB7D1BLB4D1CBA0D2BYB7 f(x,y)D1B4P CUBNA1
(BOB1B7CC (BX) pp,27)
D1A6B4C2DHCFD8DIA2 CHBPCMBUBYB7B0AXBYB7fprime(z)CGB2CMBUB0,CMAS
fprime(z) = ux + ivx = vy?iuy CGB2D1CBB0A1CUB2BSDFCRBXDHB0B5CXA1
AWBPAZ (B) BYB7 f(z) = u(x,y) + iv(x,y) D1AAB5CLCX D A3CMBUB0AN
CFBJCEB2A2 u(x,y)CWv(x,y) D1D A3CRCHCJD1CBA7AXB7A0AD?DBA5CTBV
-CWDCBGAMA1
§2.2 B3B7B5B9AYAWBAB8B4 7
BQ 1A6B5BXD4BYB7D1C0ARCUAXA0D1C0ARCMBUA1
1) f(z) = zRe(z);
2) f(z) = x + yx2 + y2 + i x?yx2 + y2;
3) f(z) = ex(cosy + isiny);
BJ 1)CQf(z) = zRe(z) = x2 + ixy,AZu = x2,v = xy,BECI
u
x = 2x,
u
y = 0,
v
x = y,
v
y = x.
BZAPA0D4BBBPA7AXB7ARARD1CB (CQASDA u,v CUBN),AVB2CNAW x = y = 0
AYA0BGDFAGDBA5CTBV -CWDCBGAMA1
CMASA0BYB7CND1 z = 0CUAXA0D1BJA8DHA3ARAR?CMBUA1
2)CMBQu = x + yx2 + y2,v = x?yx2 + y2,BECI
u
x =
y2?x2?2xy
(x2 + y2)2,
u
y =
x2?y2?2xy
(x2 + y2)2,
v
x =
y2?x2 + 2xy
(x2 + y2)2,
v
y =
y2?x2?2xy
(x2 + y2)2,
CHCJA7AXB7D1 z negationslash= 0 ARD1CB (CQASDA u,v CUBN),AD?DBA5CTBV - CWDCBG
AMA1
CMASA0BYB7D1BJA8DHA3AQ z = 0BMARARCUAXA0D1BJA8DHA3AQ z = 0BM
ARARCMBUA0AD?
fprime(z) =?u?x + i?v?x = (y
2?x2)(1 + i)?2xy(1?i)
(x2 + y2)2 (z negationslash= 0).
3)CMBQu = excosy,v = exsiny,BECI
u
x = e
xcosy,?u
y =?e
xsiny,
v
x = e
xsiny,?v
y = e
xcosy.
CHCJA7AXB7D1BJA8DHA3ARARD1CB (CQASDAu,vCUBN),AD?DBA5CTBV -CWDC
BGAMA1CMASA0BYB7D1BJA8DHA3ARARCMBUA0AD?
fprime(z) =?u?x + i?v?x = ex(cosy + isiny) = f(z).
8 B0B1BB B5B9B3B7
A5A2BMCS -BOBWB2AICF f(z)DCAVz0 BNARASACD0CLBHA0APCK?B3 f(z)
DCz0 BNARASAJB4CLBHA1
BQ 3CDCO f(z) = radicalbig|xy|D1z = 0B0CUAXCAA1
BJCMBQu(x,y) = radicalbig|xy|,v(x,y) = 0,BECI
ux(0,0) = uy(0,0) = vx(0,0) = vy(0,0) = 0,
BBDCAV (0,0)BVA6BMCS -BOBWB2AIA1
APCFA0CBBCBT z CZDICV
braceleftBigg
x = αt
y = βt (α,β AFCMCDCPBS) C7D8BSCDA0
f(z)?f(0)
z?0 =
f(z)
z =
radicalbig|αβ|
α + iβ
B0C5C0?B2BPCHB0A0CUCD f(z)D1B4z = 0B2?CUAXB0A1
BQ 3AX
f(z) =
x3?y3 + i(x3 + y3)
x2 + y2,z negationslash= 0,
0,z = 0,
B3D8f(z)D1CZB4DBA5CTBV -CWDCBGAMA0AV?CUAXA1
DGD8f(z) = u + iv,AZ
u(x,y) =
x3?y3
x2 + y2,(x,y) negationslash= (0,0),
0,(x,y) = (0,0),
v(x,y) =
x3 + y3
x2 + y2,(x,y) negationslash= (0,0),
0,(x,y) = (0,0),
CMBQ
ux(0,0) = limx→0 u(x,0)?u(0,0)x?0 = limx→0 x
3
x3 = 1;
uy(0,0) = limy→0 u(0,y)?u(0,0)y?0 = limy→0?y
3
y3 =?1;
vx(0,0) = limx→0 v(x,0)?v(0,0)x?0 = limx→0 x
3
x3 = 1;
vy(0,0) = limy→0 v(0,y)?v(0,0)y?0 = limy→0 y
3
y3 = 1.
CJD3f(z)DCDBAVBVA6BMCS -BOBWB2AIA1
APCFAQ z CZDICV y = kxC7CXD8BSCDA0
f(z)?f(0)
z?0 =
x3?y3 + i(x3 + y3)
(x + iy)(x2 + y2) →
1?k3 + i(1 + k3)
(1 + ik)(1 + k2)
BECIf(z)D1z = 0?CUAXA1
§2.2 B3B7B5B9AYAWBAB8B4 9
BQ 4 ASBXfprime(z)D1AJCXDARARBQD6A0A2DD f(z)D1D A3BQAIB7A1
DG CMBQ
fprime(z) =?u?x + i?v?x =?v?y?i?u?y ≡ 0,
BT
u
x =
v
x =
v
y =
u
y ≡ 0,
BECIu =AIB7A0v =AIB7A0CMBB f(z)D1DA3BQAIB7A1
BQ 5 ASBX f(z) = u + iv BQCHCMBUBYB7A0? f
prime(z) negationslash= 0,
A2DDAKC1A6
u(x,y) = c1,v(x,y) = c2 A9C3C2D7CGA0AADHA0 c1,c2 BQAIB7A1
DG CQCUf
prime(z) = 1
iuy + vy negationslash= 0,BTuy CWvy A9?ANBQD6A1
ASBXD1AKC1CGB4AR uy CW vy B6?BQD6A0BRCQCNBYB7AHAXBDD2DAA0
AKC1 u(x,y) = c1 B0C7D9B2 k1 =?ux/uy; AKC1 v(x,y) = c2 B0C7D9B2
k2 =?vx/vy,CYCPCTBV -CWDCBGAMAZ
k1 ·k2 = (?ux/uy)· (?vx/vy) = (?vy/uy)· (uy/vy) =?1.
ASBXD1AKC1CGB4AR uy CW vy CRCHBPBQD6A0D2D7CHBPA9?BQD6A0ASAY
CJDAD2A6DHB0AKC1D1CGB4ARB0ADC1CHBJB2B9A8B0A0D7CHBJB2ACDCB0A0BG
DFARC3C2D7CGA1
10 B0B1BB B5B9B3B7
§2.3 AKATBDCG
A6CKA5AZAABYB7DHB0CHC6AICPB0AOB1BYB7BLBVAYBJAABYB7B0AFC8A1
A2CKAUBPBGDHA2 1,B5CLA3 2,CADGA3 3,CMBUCAA1
1,A0CGBDCG
1)AWD5:B9CUBJAAB7 z = x + iy,AJ
ez = ex(cosy + isiny)
BQz B0A0CGBDCG.
2)BJCQCY
a) f(z)D1BJA8DHA3ARARCUAXA0? (ez)prime = ez.
3)CYA2
a)CABDB5CXA2
ez1 ·ez2 = ez1+z2.
DGC0A2AX z1 = x1 + iy1,z2 = x2 + iy2,BRCQB5CLA0CR
ez1 ·ez2 = ex1(cosy1 + isiny1)·ex2(cosy2 + isiny2)
= ex1+x2[cos(y1 + y2) + isin(y1 + y2)]
= ez1+z2.
b)DJA9CA
ez+2kpii = ez ·e2kpii = ez.
§2.3 AXAZB3B7 11
2,AXCGBDCG
1)AWD5:A5DBA5BGAM
ew = z (z negationslash= 0)
B0BYB7w = f(z)AJBQAXCGBDCG.
C6BJAMB2AI,D8 w = u + iv,z = reiθ,A2DD
eu+iv = reiθ =? eueiv = reiθ,
BECI u = lnr,v = θ.
CMAS w = ln|z| + iArgz.
C9A9
Lnz = ln|z| + iArgz.
D6D8 LnzCFAYDJBDCGA0BXBRB8DJAG 2piiASDECGABA1
BWB5AADH ArgzALA1DD argzB0A2CHC3A0AJA9 LnzB0A4DJA0C9A9lnz,
D2CR
lnz = ln|z| + iargz.
BQ 1 AHLn2,Ln(?1)CIC6BGDFC2COB0A1DDA1
BJ CMBQ
Ln2 = ln2 + iArg2 = ln2 + 2kpii,
BGB0A1DDB2 ln2.
CMBQ
Ln(?1) = ln1 + iArg(?1) = (2k + 1)pii,
BGB0A1DDB2 ln(?1) = pii.
12 B0B1BB B5B9B3B7
2)CYA2:
a) AWz = x > 0AYA0Lnz B0A1DD lnz = lnx.
b)
Ln(z1z2) = Lnz1 + Lnz2,
Ln
parenleftbiggz
1
z2
parenrightbigg
= Lnz1?Lnz2.
DGA2
Ln(z1z2) = ln|z1z2|+ iArgz1z2
= ln|z1||z2|+ i(Argz1 + Argz2)
= (ln|z1| + iArgz1) + (ln|z2| + iArgz2)
= Lnz1 + Lnz2.
Ln
parenleftbiggz
1
z2
parenrightbigg
= ln
vextendsinglevextendsingle
vextendsinglevextendsinglez1
z2
vextendsinglevextendsingle
vextendsinglevextendsingle + iArgz1
z2
= ln|z1||z
2|
+ i(Argz1?Argz2)
= (ln|z1|?ln|z2|) + i(Argz1?Argz2)
= (ln|z1| + iArgz1)?(ln|z2|+ iArgz2)
= Lnz1?Lnz2.
A2A2BXDHB0B1B1?D0AKD0A2
Lnzn = nLnz,
Ln n√z = 1nLnz.
§2.3 AXAZB3B7 13
3)BJCQCY:
CTBLBUA4DJ lnz ASBJCQCYA1
CMBQlnz = ln|z|+iargz,BBln|z|AQCZB4BMD1AABGB4B6B2D1CBB0A0argz
D1AQCZB4CWBKAZA0CIBMB6D1CBA1
CUB2A0w = lnz DCALDBAVDAB6CEA3ASC3BZC1BJCQA0C5(CQBEBYB7B0AH
AXBDD2)
dlnz
dz =
1
dew
dw
= 1ew = 1z.
CMBQ Lnz = lnz + 2kpii,BECI Lnz ASB9B8B4DHDCALDBAVDAB6CEA3ASC3
BZC1BJCQA0?C5D7CWCMASARCGDJA1
3,AHBYab DABYBDCG
1)AWD5:AXaBQ?BQD6B0CHBPBJB7A0b BQAQCKCHBPBJB7A0B5CLALDGBQ
ab = ebLna.
CQCU LnaB2BADDB0A0CMBB ab CGB2BADDB0A1
2)CYA2:
a) AB bAOASAMALA0 ab AIARAAAQACAUA1C7A2
ab = ebLna = eb[ln|a|+i(arga+2kpi)]
= eb(ln|a|+iarga)+2kbpii
= eb(ln|a|+iarga) ·e2kbpiibracehtipupleft bracehtipdownrightbracehtipdownleft bracehtipupright
= eb(ln|a|+iarga) = eblna,
b) AB b = pq(p AG q AOAHAVACASAMA0 q > 0) ALA0 ab AIAR q AFACAUA1
C7A2
ab = epqln|a|+ipq(arga+2kpi)
= epqln|a|
bracketleftbigg
cospq(arga + 2kpi) + isinpq(arga + 2kpi)
bracketrightbigg
,
AADHA2k = 0,1,···,(q?1).
c) A8A9ATANA0AQA7?AJ ab AIARAPAKADAFAUA1
14 B0B1BB B5B9B3B7
BHACB2A0AW bBQD7D6B7 nAYA0BRCQB5CL
an = enLna = eLna+Lna+···+Lna
= eLna ·eLna ·····eLna
= a·a·····a.
AWb = 1n AYA0CR
a1n = e1nLna = e1nln|a|
parenleftbigg
cosarga + 2kpin + isinarga + 2kpin
parenrightbigg
= n
radicalbig
|a|
parenleftbigg
cosarga + 2kpin + isinarga + 2kpin
parenrightbigg
= n√a
AADHA0 k = 0,1,2,···,(n?1).
A4AWBEB6A0AQ b CPDFDECG nBGB4CG
1
n CDCFDA a AS nANBYBG aAS n
ANBAASD4D5CNCAD2A1A1
3)BJCQCY:
a) zn DCB5C3BZC1CFAODJBJCQBDCGA0C5
(zn)prime = nzn?1.
b)CMBQ
z 1n = n√z = e1nLnz
BECIz 1n B2CHBPBADDBYB7A0CRCR nBPBHD9A1
CQCULnzB0DEBPBHD9D1AQAMCZB4BZBKAZA0B0BJA8DHA3B2CMBUA0CJD3
z 1n ASBXB8B4DHDCALC9DBAVBFB6CEA3ASB5C3BZC1D1CFBJCQASA0C5
(z 1n)prime = 1nz 1n?1.
c)zb(AQAM b = nCW
1
n D2DIAFCV) B2CHBPBADDBYB7A0CRBTAGBADDA1 z
b
ASBXB8B4DHDCALC9DBAVBFB6CEA3ASB5C3BZC1D1CFBJCQASA0C5
(zb)prime = bzb?1.
§2.3 AXAZB3B7 15
4,CCBIBDCGBFCIC8BDCG
1)AWD5:
cosz = e
iz + e?iz
2,sinz =
eiz?e?iz
2i,
2)CYA2:
a)DJA9CAA2 cos(z + 2pi) = cosz,sin(z + 2pi) = sinz.
b)ABA5CAA2 cos(?z) = cosz,sin(?z) =?sinz.
c)CRBUCVBYBZD7BYB0C8BCBSB1
cos(z1 + z2) = cosz1cosz2?sinz1sinz2,
sin(z1 + z2) = sinz1cosz2?cosz1sinz2,
sin2z + cos2z = 1.
A5A2 CADG|sinz|lessorequalslant 1BZ|cosz|lessorequalslant 1D1BJB7BFBOA3?D0AKD0A1
3)BJCQCY:
AUCHBYB7 cosz BZ sinz B6B2BJA8DHA3B0CMBUBYB7A0?CR
(cosz)prime =?sinz,(sinz)prime = cosz.
16 B0B1BB B5B9B3B7
5,B1CCBIBDCGDAB1CIC8BDCG
1)AWD5,BEAUCHBYB7B5CLBQAUCHBYB7B0BEBYB7A1AX
z = cosw,
A2DDAJ wBQz B0B1D9CUBDCG,C9A9
w = Arccosz,
CQz = cosw = 12(eiw + e?iw)AZ
e2iw?2zeiw + 1 = 0,
BGB0BRBQ eiw = z +√z2?1,
AADHA0
√z2?1
COCXCMBQB8DDBYB7A1CMASA0D2B7ALB9B7A0AZ
Arccosz =?iLn(z +
radicalbig
z2?1).
§2.3 AXAZB3B7 17
2,(pp,66)BXD4BYB7C0ARCUAXA4C0ARCMBUA4
3) f(z) = 2x3 + 3y3i;
BJCMBQu = 2x3,v = 3y3,AD?
u
x = 6x
2,?u
y = 0,
v
x = 0,
v
y = 9y
2.
A7AXB7D1CB (CQASDAu,vCUBN),AVCTBV-CWDCBGAMux = vy,uy =?vx CNAW
6x2 = 9y2 AYAKD0A1
BTf(z)DFD1 6x
2 = 9y2
AWCUAXA0D1BJA8DHAWARAR?CMBUA1
6,(pp,66)A6B8BXD4DJBIB0D5CBA1
5)ASBX u(x,y)BZv(x,y)CUAX (DEA7AXB7ATD1),A2DD f(z) = u + iv CG
CUAXA3
BJ?D7AOA1BYB7 u = 2xBZv = y CSCUAXA0AV f(z) = u + iv = 2x + iy
D1BJA8DHAWARAR?CUAXA1
6) AX f(z) = u + iv D1AJCX D A3B2CMBUB0A1ASBX u B2AZAIB7A0A2DD
f(z)D1D6BP DA3B2AIB7A3ASBX vB2AZAIB7A0A2DD f(z)D1D6BP DA3B2AI
B7A1
BJf(z)D1DA3CMBUA0D2 u,v D1DA3CUBN?DBA5CTBV -CWDCBGAMA2
ux = vy,uy =?vx.
BECIAWuB2AZAIB7AYA0CR ux = uy = 0,CUB2
vy = ux = 0,vx =?uy = 0,
BTf(z)D1DA3B2AIB7A1
18 B0B1BB B5B9B3B7
10,(pp,67)D8DIA2ASBXBYB7 f(z) = u+ivD1AJCXDA3CMBUA0ADDBA5
BXD4BJCEDBCHA0A2DD f(z)B2AIB7A1
2) f(z)D1DA3CMBUA3
4) argf(z)D1DA3B2CHBPAIB7A3
BJA2f(z) D1DA3CMBUA0D2
ux = vy,uy =?vx,(1)
2)ASBXf(z) = u?iv D1DA3CMBUA0D2CRCTBV -CWDCBGAM
ux =?vy,uy = vx.
CYCP (1)B1A0D2ux = uy = vx = vy = 0,BECIf(z)B2AIB7A1
4)ASBXargf(z)D1DA3B2CHBPAIB7A0C9 argf(z) = C,D2
v
u = tanC(BQAIB7A0C9BQ C
prime)
CUB2v = Cprimeu,D2CR
vx = Cprimeux,vy = Cprimeuy.
D0BRCQ (1)CUAZux = uy = vx = vy = 0,BECIf(z)B2AIB7A1
11,(pp,67)BXD4BUBWB2BID7AOA4
1) ez = ez; 3) sinz = sinz
BJ 1)CMBQ ez = ex+iy = ex(cosy + isiny).
BECI ez = ex(cosy?isiny).
CTCMBQ
ez = ex?iy
= ex[cos(?y) + isin(?y)]
= ex(cosy?isiny).
BTez = ez.
3)CMBQsinz = 12i(eiz?e?iz),BECI (CYCP (1)B0CLDA)
sinz = 12i(eiz?e?iz) = 1?2i(eiz?e?iz)
= 1?2i(e?iz?eiz) = sinz.
§2.3 AXAZB3B7 19
12,(pp,67)D3APBXD4BGAMB0ANAFCMA2
1) sinz = 0;
BJA2sinz = 0B1CCCU
1
2i(e
iz?e?iz) = 0,
C7e2iz = 1,CUB2
2iz = Ln1 = ln1 + i(0 + 2kpi) = 2kpii,
BECIBGAMB0ANAFCMBQ z = kpi(k = 0,±1,±2,···).
13,(pp,67)D8DIA2
1) cos(z1 + z2) = cosz1cosz2?sinz1sinz2;
sin(z1 + z2) = sinz1cosz2 + cosz1sinz2;
2) cos2z + sin2z = 1;
3) sin2z = 2sinzcosz;
5) sin
parenleftBigz
2?z
parenrightBig
= cosz; cos(z + pi) = cosz;
DG(1)
cosz1cosz2?sinz1sinz2
= e
iz1 + e?iz1
2 ·
eiz2 + e?iz2
2?
eiz1?e?iz1
2i ·
eiz2?e?iz2
2i
= e
i(z1+z2) + e?i(z1+z2)
2 = cos(z1 + z2).
BKCXA2 sin(z1 + z2) = sinz1cosz2 + cosz1sinz2.
(2)
sin2z + cos2z
=
parenleftbiggeiz?e?iz
2i
parenrightbigg2
+
parenleftbiggeiz + e?iz
2
parenrightbigg2
=?14(e2iz?2 + e?2iz) + 14(e2iz + 2 + e?2iz) = 1.
(3) D1 (1) DHB1B1 sin(z1 + z2) = sinz1cosz2 + cosz1sinz2 AL z1 = z2 = z
C7AZA1
(5)CYCP (1)B0CLDAA0
sin
parenleftBigpi
2?z
parenrightBig
= sinpi2cosz + cospi2sinz = cosz;
cos(z + pi) = coszcospi?sinzsinpi =?cosz.
20 B0B1BB B5B9B3B7
17,(pp,68)BADIBXD4B1B1B2BID7AOA1
1) Lnz2 = 2Lnz; 2) Ln√z = 12Lnz,
BJ D8z = reiθ,r = |z|,θ = argz.
1)?D7AOA1CMBQ
Lnz2 = lnr2 + i(2θ + 2kpi) = 2lnr + i(2θ + 2kpi)
(k = 0,±1,±2,···,)
BB
2Lnz = 2{lnr + i(θ + 2k1pi)} = 2lnr + i(2θ + 4k1pi)
(k1 = 0,±1,±2,···,)
BECI Lnz2 B0DDA7 2Lnz B0DDBAA0B1B1?D7AOA1
2)?D7AOA1CMBQ
1
2Lnz =
1
2{lnr + i(θ + 2kpi)} =
1
2lnr + i
parenleftbiggθ
2 + kpi
parenrightbigg
(k = 0,±1,±2,···,)
D7BMA0CMBQ
√z = √reθ+2lpi
2 i(l = 0,1),
C7
√z
CRD2BPBRA2
√reθ
2i
BZ
√z = √reθ+2pi
2 i.
CUB2 Ln√z CRD2A7DDA0BHACB2
B3CHA7
Ln(√reθ2i) = ln√r + i
parenleftbiggθ
2 + 2k1pi
parenrightbigg
= ln√r + i
parenleftbiggθ
2 + 2k1pi
parenrightbigg
(k1 = 0,±1,±2,···)
B3BCA7
Ln(√reθ+2pi2 i) = ln√r + i
parenleftbiggθ + 2pi
2 + 2k2pi
parenrightbigg
= 12lnr + i
bracketleftbiggθ
2 + (2k2 + 1)pi
bracketrightbigg
(k2 = 0,±1,±2,···)
BDAPAW kB2A5B7AYA0
1
2LnzCWB3CHA7DDC2BKA0AW kB2ABB7AYA0
1
2Lnz
CWB3BCA7DDC2BKA0AV
1
2Lnz CWLn
√z
B2D2BP?BKB0BTAGDDBYB7A1
§2.3 AXAZB3B7 21
18,(pp,68)AH(1 + i)i B0DDA1
BJA2
(1 + i)i = eiLn(1+i) = ei(ln|1+i|+i[arg(1+i)+2kpi])
= ei{ln
√2+i(pi
4+2kpi)}
= e?(pi4+2kpi)+ln
√2i
= e?(pi4+2kpi)(cosln√2 + isinln√2)
(k = 0,±1,±2,···)
9,D8DICTBV -CWDCBGAMB0C5AAABC8B1B2A2
u
r =
1
r
v
θ,
v
r =?
1
r
u
θ.
DGCQDCCHAAABCWC5AAABB0BUBWA2 x = rcosθ,y = rsinθ,CIC6
u = u(x,y) = u(r,θ),v = v(x,y) = v(r,θ),BRCQBACYBYB7B0BJC1BYB7AHAX
BDD2A0
u
r =
u
x
x
r +
u
y
y
r = cosθ
u
x + sinθ
u
y
u
θ =
u
x
x
θ +
u
y
y
θ =?rsinθ
u
x + rcosθ
u
y
v
r =
v
x
x
r +
v
y
y
r = cosθ
v
x + sinθ
v
y
v
θ =
v
x
x
θ +
v
y
y
θ =?rsinθ
v
x + rcosθ
v
y,
CYCP
u
x =
v
y,
u
y =?
v
x,A7CICIAWBQB1A0C7AZC5AAABC8B1B0CTBV -CWDC
BGAMA2
u
r =
1
r
v
θ,
v
r =?
1
r
u
θ.
