Tips KENZOU PC no problem 2 1 w = f(z) z 1 w w z w = (z z 0 ) b b w = log (z z 0 ) z = z 0 2π 2 z = z 0 w = z 1/2 z = re iθ θ (z = 0) 0 2π 0

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1 Tips KENZOU P no problem 2 w = f(z) z w w z w = (z z ) b b w = log (z z ) z = z 2π 2 z = z w = z /2 z = re iθ θ (z = ) 2π 4π 2 θ θ 2π 4π z r re iθ re i2π = r re i4π = r w r re iθ/2 re iπ = r re i2π = r z θ 2π w w 2 (.) z D

2 w = z /2 = { r e iθ/2 = w ( z D ) r e i(θ/2+π) = r e iθ/2 = w ( z D 2 ) (.) z = w = z /2 w w 2 θ [, 2π ] [, 4π ] z w arg(θ) = θ z arg(z) = θ + 2π z < arg(z) 2π z 2π < arg(z) 4π 2 z D D z z F ig. F ig.2 F ig.3 p z = D D 2 2 z = z = D n D 2 D p z = 2 D D 2 ( D 2 2 D z 2 z w = z /n z = re iθ ( θ < 2π) z b = e b log z w w = z /n = e n log z = e n {log r+i(θ+2mπ)} = r n e i θ n m = r n e i( θ n + 2 n π) m = r n e i( θ n + 4 n π) m = 2 (.2). r n e i( θ n + 2(n ) n π) m = n. w z = n n n n n w = log z z = re iθ ( θ < 2π) w = log r + i(θ + 2nπ) n =, ±, ±2, z = θ θ ± 2π θ ± 4π 2

3 Fig.2 θ θ + 2π Fig.3. x b f(x)dx b (.3) 2 f(x) f(z) = (.4) z + z b f(z)dz (.5).5 z z = z z Fig.4 arg(z) < 2π (.6) Fig.5 R R z b f(z)dz = ( z b f(z) ) (.7) c.7 z b f(z)dz = I + I + I + I (.8) c F ig.4 F ig.5 R 2π l l 2 Re +i Re +i(2π ) l ) arg(z) = + z = xe i = x l 2 arg(z) = 2π z = xe I I I = I = R z b f(z)dz = z b f(z)dz = R x b f(x)dx (.9) ( xe ) R b f(xe )dx e = e b x b f(x)dx (.) R R. ( xe ) b = x b 2 f(z) xe = x R.3 I = e 2πbi I (.) 2 < b < 3

4 R z = Re iθ z b I = z + dz = (Re iθ ) b ir b e ibθ Re iθ + ireiθ dθ = Re iθ dθ (.2) + R lim R R b e ibθ Re iθ + = lim R brb e iθ(b ) = (b < ) z = e iθ lim b e ibθ e iθ + lim b e iθb = (b > ) b ( < b < ).8 z b f(z)dz = I + I = c R x b f(x)dx e 2πbi I e 2πbi ( z b f(z) ) (.3) f(z) = /(z + ) z =.6 z = e πi z b f(z) Res(e πi ) = ( e πi) b = e πib (.4) 3.3 eπb i e 2πb i = 2π i e πb i e πb i = π sin(πb) (.5) 2π < arg(z) arg(z) = 2π + arg(z) = I = I = R R z b f(z)dz = z b f(z)dz = R R (xe ) b f(x)dx e b I (.6) x b f(x)dx I (.7) z = = e πi ( z b e b f(z) ) = e b (e πi ) b = π sin(πb) () (2) x b dx ( < b < ) x i x b dx ( < b < ) (x + ) 2 (3) x b x a dx ( < b < a) + 3 lim (z + z )zb z + z 4

5 (4) x 2 x 4 + dx () z = arg(z) < 2π f(z) = /(z i) Fig.5 z b f(z)dz = I + I + I + I I I I = = (z b f(z) ) (Re iθ ) b (e iθ ) b Re iθ i ireiθ dθ, I = 2π e iθ i ieiθ dθ R I I I arg(z) + z = xe i = x I = R x b f(x)dx ( < b < ) I z arg(z) = 2π z = xe I = R R ( xe ) R b f(x)dx e = e b x b f(x)dx I = I I = e b I ( z b e b f(z) ) z = i z = e iπ/2 Res(e iπ/2 ) = (e iπ/2 ) b = e iπb/2 e b eiπb/2 = e πib e πib e iπb/2 = πi sin(πb) e iπb/2 (2) arg(z) < 2π R I I lim R lim I = I = 2π (Re iθ ) b (Re iθ + ) 2 ireiθ dθ ( e iθ ) b ( e iθ + ) 2 i eiθ dθ (Re iθ ) b (Re iθ + ) 2 ireiθ lim R brb = (b < ) (e iθ ) b (R iθ + ) 2 i eiθ lim R b+ = ( < b) ( < b < ) z = = e πi f(z) = F (z)/(z a) k Res(a) = Res(e πi ) = b(e πi ) b (k )! lim d k z a dz k [(z a)k f(z)] 5

6 e b b(eπi ) b = πb sin(πb) (3)x a = y bx b dx = d(x b ) = d(y b/a ) = (b/a)y b/a dy x b x a + dx = a y (b/a ) + y dy = a y γ dy (γ = b/a) + y. π/ sin πb b b/a < b < < b/a < < b < a ı a sin(πb/a) (4) a = 4 b = 3 π 4 sin(3π/4) = π w = z /2 θ 2 4π w z 2 w w θ [, 2π ] [, 4π ] arg(z) = θ z arg(z) = θ + 2π z F ig.6 z w = z /2 w 2. ) w = z a z a t w = t /2 Fig. z = a 6

7 2) w = z 2 w z =, 5 Fig.7 z z + = r e iθ, z = r 2 e iθ2 z 2 arg(z 2 ) = θ + θ 2 (2.) θ θ 2 < 2π w = r r 2 e i 2 (θ+θ2) (2.2) z Fig.8 θ θ 2 θ, θ 2 = 2π z 2 θ, θ 2 = 2π arg(z 2 ) = 4π w w w = w z z θ θ 2 2π Fig.9 arg(z 2 ) 2π w w 2 w z w w 2 w = r r 2 e i 2 (θ+θ2) (z D ) w = w 2 = r r 2 e i 2 (θ+θ2+2π) (z D 2 ) (2.3) F ig.7 z F ig.8 F ig.9 θ 2 = 2π θ = θ θ 2 z = z = Fig. D D 2 (D 2 D ) F ig. D 2 D w = (z ) /2 (z + ) /2 z = /η ( ) /2 ( ) /2 w = η = η + η ( η)/2 ( + η) /2 η = z = z = 5 z = 7

8 3) w = z /3 z = r e iθ w = r /3 e iθ/3 ( θ < 2π ) z = θ 6π w 3 3 θ θ 2π 4π 6π z r re iθ re i2π = r re i4π = r re i6π = r w r re iθ/3 re i2π/3 re i4π/3 re i2π = r 2 z = z = x ) 6 arg(z) = 2π D arg(z) = 2π + D 2 arg(z) = 4π D arg(z) = 4π + D 3 Fig. arg(z) = 6π D 3 arg(z) = + D 3 z w F ig. D 3 D 2 D D 3 D 2 D y D 3 D 2 D x D 3 D 2 D D 3 D 2 D 4) w = (z ) /2 (z 2) /3 z = /η w = ( ) /2 ( ) /3 η η 2 = η ( 5/6 η)/2 ( 2η) /3 z = 2 η 5/6 η = z = ) η = z = ) 6 z = 2 z = 2 3 z = 6 F ig.2 A B y A B 2 x A B (2 3 ) A B (3 2 ) 6 8

9 Q&A w = z /2 2 w = z /n n n θ Georg Friedrich Bernhard Riemann Hermann Wyle Wyle i i e iθ = cos θ+i sin θ θ = π/2 i = e iπ/2 i log i = log e log i i log i i = e i log i log i = log e i(π/2+2nπ) = i(π/2 + 2nπ),i i = e π/2 2nπ = e π/2 =.2782 i i i i i i z b z b K 2 /2 z b = e b log z (2.4) b b i w = z b N 9

10 z = re iθ θ ( ) w = z b = e b log z = e b log r+i(θ+2nπ) = e i2nπb b(log r+iθ) e e i2nπb b e i2nπb = w b p/q w = z p/q w q b w = e b log z log z w K Ex. z /2 f(z) 7 z2 z (c) z /2 f(z)dz (2.5) z () z 2 z z 2 z /2 2 z z2 z (c) z /2 dz (2.6) z = e i = z 2 = e i 2π = x /2 dx = [ (2/3)x 3/2] = z = e iθ θ 2π Fig.4 7 P (z), Q(z) z f(z) = P (z)/q(z) Q(z)

11 e iθ/2 i e iθ dθ = 2 3 [ e i3θ/2 ] 2π = 4 3 arg(z) = arg(z) = 2π Ex.2 a z Ex.3: dθ a + cos θ = 2π a2 dθ z + cos θ = J = 2 8 Fig.3 (2.7) (a > ) (2.8) dθ z + 2 (z + z ) = dθ (2.9) x 2 dx (2.) + z 2 + log z + dz (2.) z 2 F ig.3 y y (π tan /2) 2 + i (π/4) + i b c a d 2 x x 2 i i (π + tan /2) (7π/4) a b x > z θ a arg(z a 2) = π arg(z a + ) = log z + = log(z + ) log(z 2) z + 2 x 2 log z + z 2 = log z + z 2 + i arg z + z 2 9 a b, b c, c d, d a (2.2). a b ( θ = ) z = x, z + = x +, z 2 = x 2 = 2 x e iπ log z + z 2 = log x + x 2 + i( π) (2.3) 2. b c (θ = 2π) z = + e iθ, z + = e iθ, z 2 = 3 + e iθ 3e iπ log z + z 2 = log e iθ 3 e iθ + i(θ π) (2.4) log ln log

12 3. c d ( θ = 2π) z = x, z + = (x + )e, z 2 = x 2 = 2 x e iπ log z + z 2 = log x + x 2 + i(2π π) (2.5) 4. d a (θ = π π + 2π = 3π) z = 2 + e iθ, z + = 3 + e iθ 3e 2π i, z 2 = e iθ log z + z 2 = log 3 + e iθ e iθ + i(2π θ) (2.6) (2.) J = 2 + { } dx x 2 log x + + x 2 iπ + i { } dx x 2 log x + + x 2 + iπ + i 2 3π π { e iθ e iθ } dθ ( + e iθ ) 2 log + 3 e iθ + i(θ π) { e iθ dθ (2 + e iθ ) 2 log 3 + } eiθ + e iθ + i(2π θ) (2.7) lim log = log 2 lim J = 2. J = z 2 + log z + z 2 dz = dx x 2 = I (2.8) + (z + i)(z i) log z + dz (2.9) z 2 z = ±i z = ±i J = (z + i)(z i) log z + z 2 dz = ( Res(i) + Res( i) + Res( ) ) ( = log i + ) i + log i 2 i 2 + (2.2) Fig.3 + i = 2e iπ/4 i = 2e i7π/4 2 + i = 5e i(π tan (/2)) 2 i = 5e i(π+tan (/2)) (2.2) 2.2 π 4 + tan 2 (2.22) Z 2 h i 2 + x 2 dx = tan x =

13 3 w = /z z = z w w = /z z = w z w z = z f(z) R < z < R f(z) z = Res( ) f(z)dz (3.) 3. z = R f(z) R < z < lim zf(z) z z = /η r(> R) Res( ) = f(z)dz = Res( ) = lim zf(z) (3.2) z z =r η =/r f(/η) ( ) η 2 dη (3.3) z = r η = /r η = /r ( ) f(/η) η = < η < /R F (η) f(/η)(/η 2 ) η = f(/η)( /η 2 )dη = η =/r lim η ηf (η) ResF () = lim η ηf (η) η =/r F (η)dη = ResF () (3.4) f(z) R z ResF () = lim ηf (η) = lim η z z f(z)z2 = lim zf(z) z z =r> f(z) = z n dz = { n= ( ) c n z n (3.5) n n = Res( ) = c (3.6) 3

14 z = f(z) Res( ) 2 f(z) f(z) {z k } A k B f(z) A k + B = k ( f(z)dz = k A k (3.7) 3. f(z)dz = B (3.8) A k + B = ( z = B f(z) z = B = ( ) 39: f(z) = e z (3.9) z = z z = z = f(z) z e z = +! z + 2! z2 + + n! zn + (3.) z z = 4

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy

z f(z) f(z) x, y, u, v, r, θ r > 0 z = x + iy, f = u + iv C γ D f(z) f(z) D f(z) f(z) z, Rm z, z 1.1 z = x + iy = re iθ = r (cos θ + i sin θ) z = x iy z fz fz x, y, u, v, r, θ r > z = x + iy, f = u + iv γ D fz fz D fz fz z, Rm z, z. z = x + iy = re iθ = r cos θ + i sin θ z = x iy = re iθ = r cos θ i sin θ x = z + z = Re z, y = z z = Im z i r = z = z